Physics Book - User contributions [en]

Field of a Charged Ball

2020-05-11T17:53:49Z

Sydney: /* History */

==The Main Idea==
In this section, we will discuss the electric field of a solid sphere. Such a sphere has charge distributed throughout the volume (rather than only on the surface), and can be modeled by several layers of concentric, charged spherical shells. Calculating the electric field both outside and inside the sphere will be addressed.

[[File:Wiki_pic.JPG|400px|right|thumb|Figure 1: A sphere with uniformly distributed charge: note that this can be thought of as infinitely thin concentric charged shells]]
===A Mathematical Model===

First we must determine the relationship between r, the radius of the observation point from the center of the sphere, and R, the radius of the sphere itself.

Here, it is necessary to determine whether the observation point is outside or inside the sphere.

If r>R, then we are outside the sphere. As a result, the sphere can be simply treated as if the sphere were a point charge located at the center of the sphere. Hence, the electric field at any point outside of the radius of the sphere can be calculated using the formula for the electric field of a point charge.

<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>
when r>R, and R is the radius of the sphere.

However, when r<R, the observation location is inside of the sphere and the sphere overall must be thought of as infinitely many concentric charged shells inside of each other. All of the shells with a radius larger than that of the observation location do not contribute to the electric field at the observation location. This phenomena is because the electric field produced by these larger shells cancels out at any given point inside of itself. As a result only the shells smaller than the radius of the observation location need to be accounted for.
To find <math>\vec E_{net} </math>, add the contributions to the electric field from the inner shells.
After adding the contributions of each inner shell, you should have an electric field equal to:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{\Delta Q}{r^2}</math>

We find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

We find that:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

The charge inside the sphere is proportional to r. When r=R, we again treat the sphere as a point charge located at the center of the sphere.

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2} </math>

'''Looking Ahead, Gauss's Law'''

Later in this wikibook, you will learn about Gauss's Law. This will make calculating the electric field easier.

The formula for Gauss's Law is:

<math>\oint \vec E \bullet \hat n dA = \frac{1}{\epsilon_0}\Sigma Q_{\text {inside the surface}} </math>

Using this, one can calculate the the electric field when r<R for a solid sphere charged throughout.

<math>\vec E * 4\pi*r^2= \frac{1}{\epsilon_0}Q\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} </math>

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

===A Computational Model===

This demonstration using VPython shows the field of a charged ball with a fixed radius. Try changing the radius to see what happens!

https://trinket.io/glowscript/48f6efd07a

# GlowScript 1.1 VPython
## constants
oofpez = 9e9
qproton = 1.6e-19
scalefactor = 2e-20
Radius =3.5
r=3

## objects
particle = sphere(pos=vector(0,0,0), opacity =0.5, radius=Radius, color=color.blue)
## initial values
obslocation = vector(r,0,0)
##calculate position vector
rvector = obslocation - particle.pos
arrow1 = arrow(pos=particle.pos, axis=rvector, color=color.green)
rmag=mag(rvector)

##colored arrow

if r<= Radius:
E1= oofpez*(qproton)/(Radius)**3 *rmag* vector(1,0,0)
E2= oofpez*(qproton)/(Radius)**3 *rmag* vector(-1,0,0)
E3= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,1,0)
E4= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,-1,0)
E5= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,1)
E6= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,-1)
ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)
if r>Radius:
E1= oofpez*(qproton)/(rmag)**2 *vector(1,0,0)
E2= oofpez*(qproton)/(rmag)**2 *vector(-1,0,0)
E3= oofpez*(qproton)/(rmag)**2 * vector(0,1,0)
E4= oofpez*(qproton)/(rmag)**2 *vector(0,-1,0)
E5= oofpez*(qproton)/(rmag)**2 *vector(0,0,1)
E6= oofpez*(qproton)/(rmag)**2 * vector(0,0,-1)
ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)

==Examples==

===Simple===
[[File:Approx_point.png|200px|right|thumb|Figure 2: This is an example of applying the equation for finding the electric field of a point charge]]
Describe the pattern of the electric field of charged ball from far away.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

If we observe a charged ball from far away, we can say that r>R. Because we are outside of the sphere, the ball can essentially be viewed as a point charge. Thus the electric field can be defined by:
<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>

</div></div>

===Middling===
[[File:Concentric_sphere.png|200px|right|thumb|Figure 3: This is very similar to problem 1 and involves calculating the field of a point charge.]]
A sphere is charged throughout it's volume with a charge of Q= 6e-5 C. The radius of the this sphere is R=10. Find the electric field created by a sphere of radius r=4.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Step 1: Cut up the sphere into shells.

step 2: We know that r<R.

Next find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r = 9e9 * \frac{6*10^{-5} C}{(10m)^3}*4m = 2160 N/C </math>

</div></div>

===Difficult===
A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge −e. (The actual charge density in the ground state is nonuniform.)

[[File:circle_outline.png|200px|thumb|left|Figure 4: This circle represents a hydrogen atom. Use the radius and total charge to find alpha ]]

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

For the uniform-density model, calculate the polarizability α of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud

(s « R in the diagram). (Use the following as necessary: R and ε0.)

Useful equations:

<math> \vec p = \alpha *E</math>

<math> \vec p = Q*s </math>

Assume an applied electric field of strength E. This electric field polarized the hydrogen atom. Now there is a spherical charge of radius s. Use the volume ratio, and then use the useful equations to find <math> \alpha </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{e}{s^2}\frac{\frac{4}{3} \pi s^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s </math>

<math>\alpha = \frac{e*s}{\frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s} = 4\pi \epsilon_0 R^3 </math>

</div></div>

==Connectedness==
It may be particularly useful to discuss real-life applications of a charged solid sphere. Two examples stem from the structure of an atom. The nucleus of an atom is packed very tightly so that we can consider the charge to be uniformly distributed. The electron cloud also can be viewed as a packed spherical region of charged. Of course the radii of these structures are very small; radii are about 10e-15 m and 10e-10 m for nuclei and electron clouds respectively!

==History==
(should be completed by a student)

== See also ==

===Further reading===

*[[Point Charge]]

*[[Electric Dipole]]

*[http://www.physicsbook.gatech.edu/Gauss%27s_Flux_Theorem Gauss's Flux Theorem]

===External Links===

==References==

*http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html

*https://www.physicsforums.com/threads/calculate-the-polarizability-a-lpha-of-atomic-hydrogen-in-terms-of-r.339994/

*Matter and Interactions, 4th Edition: 1-2

Charged Capacitor

2020-05-11T17:52:14Z

Sydney: /* History */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
<br>This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

===A Mathematical Model===

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
<br>The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
<br>

'''Derivation'''

[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
<br>Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
<br><br>Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
<br>If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
<br>

'''Gauss's Law'''

[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
<br>Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
<br>Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
<br>Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

===A Computational Model===

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
<br><br>
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capacitance is 3 F.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
<br><math>A = \frac{Cd}{\varepsilon_{0}}</math>
<br><math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
<br><br>

</div></div>

[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]

Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V</math>
<br><math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
<br>

</div></div>

===Middling===

[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
<br>To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
<br>To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.

</div></div>

[[File:qwertt.png|200px|thumb|Figure 6: Middling problem #2]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
<br>Since <math>C = \frac{Q}{V}</math>,
<br>the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
<br>and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
<br>The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>

</div></div>

===Difficult===

A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

[[File:zxcv.png|300px|thumb|right|Figure 7: Difficult problem]]
<br><br>
<br>First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
<br> Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
<br><math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
<br><math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
<br><br>Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
<br><math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

</div></div>

==Connectedness==
(should be completed by a student)

==History==
(should be completed by a student)

==See also==

===Further reading===

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
<br><math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
<br><math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
<br><math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
<br><br>
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
<br><br>
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2020-05-11T17:51:52Z

Sydney: /* Connectedness */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
<br>This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

===A Mathematical Model===

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
<br>The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
<br>

'''Derivation'''

[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
<br>Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
<br><br>Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
<br>If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
<br>

'''Gauss's Law'''

[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
<br>Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
<br>Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
<br>Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

===A Computational Model===

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
<br><br>
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capacitance is 3 F.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
<br><math>A = \frac{Cd}{\varepsilon_{0}}</math>
<br><math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
<br><br>

</div></div>

[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]

Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V</math>
<br><math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
<br>

</div></div>

===Middling===

[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
<br>To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
<br>To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.

</div></div>

[[File:qwertt.png|200px|thumb|Figure 6: Middling problem #2]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
<br>Since <math>C = \frac{Q}{V}</math>,
<br>the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
<br>and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
<br>The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>

</div></div>

===Difficult===

A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

[[File:zxcv.png|300px|thumb|right|Figure 7: Difficult problem]]
<br><br>
<br>First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
<br> Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
<br><math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
<br><math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
<br><br>Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
<br><math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

</div></div>

==Connectedness==
(should be completed by a student)

==History==
ADD MORE INFO ABOUT HISTORY HERE

==See also==

===Further reading===

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
<br><math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
<br><math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
<br><math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
<br><br>
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
<br><br>
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Conductor and Charged Insulator

2020-05-11T17:49:38Z

Sydney: /* History */

Claimed by Amanda Barber Spring 2018

Okay – so we know that '''charged particles''' exist out there (shout out to those positrons and electrons for being “lit” when they annihilate) and that objects can either be negatively charged, positively charged, or neutral depending on the ratio of charged particles that object has. For example, if I happen to observe some random thing in nature that has 5 protons and 4 electrons – that thing has a net charge of +1. But what we haven’t really explored '''''how something''''' (like a '''conductor''' or an '''insulator''') '''''becomes positively, negatively, or neutrally charged''''' or '''''what happens when something becomes charged'''''. Guess what? That’s what this wiki is about! And by the time you’ve finished reading this, you should have a better understanding of how one can become “charged up” (in the context of E&M physics, not in the context of 6 Gawd Drizzy Drake preparing to release the hottest diss track of the decade in response to “twitter-finger” beef initiated by former rapper Meek Mill in the Summer of 2015).

==The Main Idea==

Okay, so remember in that short description above when I bolded and italicized the phrase “how something can become positively, negatively, or neutrally charged”? The processes by which this can occur are called '''charging''' and '''discharging'''. The definitions are simple: charging – how an object becomes positively or negatively charged; discharging – how a positively or negatively charged object becomes neutral. What happens to an object as a result of charging or discharging depends on the nature of that object and whether or not the object is a conductor or an insulator. All materials are made of atoms that contain positive and negative charges (protons and electrons). While all materials contain these two basic units, the charge distribution patterns change depending on the microscopic behavior of the atom's movement in an electric field. These differences have created two distinct classes of materials: conductors and insulators. This article will explore the differences between a charged conductor and a charged insulator. But before we go further into the specifics regarding conductors and insulators, let us further discuss the means by which charging can occur: '''conduction''' and '''induction'''. Both processes involve objects becoming charged by exchanging, gaining, or losing mobile charged particles. In most examples, objects will become charged by gaining or losing electrons, but that is not always the case – other mobile charged particles can contribute to an objects net charge, such as mobile potassium or sodium ions.

'''Charging by Conduction'''

Okay, so y’all remember that example in lecture where the professor kept pulling tape apart? And the lab that followed up afterwards? That’s what charging by induction is all about! (No worries if you don’t remember or if you haven’t gotten to that part of the course yet – I’m not gonna leave you hanging and just assume you know and/or remember what that was all about. I gotcha back homie! Checkout this link: https://www.youtube.com/watch?v=O7siDnbEuko

[[File:Charging tape.png|thumb|Charges on strips of tape as they are pulled apart from http://p3server.pa.msu.edu/coursewiki/doku.php?id=184_notes:charging_discharging]]

As illustrated in the pulling tape exercise, charging by conduction is all about '''making contact in order to transfer charged particles'''. When the two strips of tape are quickly ripped apart, mobile charged particles (electrons) are transferred from one strip of tape to the other – leaving one strip positively charged and the other strip negatively charged. What mobile charged particles are transferred (such as electrons) and where those mobile charged particles go depends on chemical makeup of the materials involved in the process of charging by conduction. For example, when one is charging by conduction with plastic and wool – the plastic gains electrons to become negatively charged. But when one is charging by conduction using plastic and silk, the plastic loses electrons to become positively charged.

'''Charging by Induction'''

Charging by induction is all about '''using a charged object to separate mobile charged particles between two other neutral objects in contact with one another'''. First, the two neutral objects are in physical contact with each other outside the presence of the charged object. Next, the two objects are subject to the presence of a charged object, causing mobile charged particles to separate among the two connected objects, resulting in excess charge building up on the opposing surfaces of the two connected objects. Then, the two connected objects are separated while still subject to the presence of the charged object. Finally, the separated objects removed from the presence of the charged object and the excess charges distribute across the entire surfaces of each object, resulting in what were initially two neutral objects now being charged.

[[File:Charging induction.png|thumb|Charging two neutral conductors by induction from http://p3server.pa.msu.edu/coursewiki/doku.php?id=184_notes:charging_discharging]]

Here's a video that might also help explain charging by induction: https://www.youtube.com/watch?v=cMM6hZiWnig

[[File:Inductiontransfer.gif]]

'''Discharging'''

Discharging is the opposite of charging: the process by which charged objects lose their excess charge and become neutral. There are a number of ways in which charged objects can discharge. For example, charged tape can be discharged by water in the surrounding environment (over time) or by using your finger and touching the tape. When we touch charged tape, our bodies act as electrical grounds. An electrical ground is huge pool of charged particles that remains neutral when small amounts of charged particles are added or taken away

[[File:05172.png]]

'''Charging Insulators'''

[[File:Insulator and conductor.png|thumb|Charging an insulator vs charging a conductor from http://p3server.pa.msu.edu/coursewiki/doku.php?id=184_notes:charging_discharging]]

Insulators can only be charged by conduction. When an insulator is charged by conduction, the charged particles remain at the initial point of contact throughout time until the insulator is discharged.

When a charged object enters the immediate region of an insulator, the result is polarization of the atoms and molecules within the insulator as seen in the example to the right (only while in the presence of that charged object). Since insulators impede the movement of charged particles, two insulators cannot be charged by induction in the presence of a charged object.

Here’s a neat little simulator to see how an insulator (a balloon) can become charged by conduction and see what happens afterwards:
https://phet.colorado.edu/en/simulation/balloons-and-static-electricity

'''Charging Conductors'''

When conductors are charged by conduction, the excess charges distribute evenly across the entire surface of the conductor over time since conductors enable charged particles to move freely. While the inside electric field remains zero. This all happens because of Coulomb's Law. It insists that charges lay as far away from each other as atom-ly (you get it? like human-ly) possible. If you charge a conductor, the conductor becomes charged with whatever type of charged was used. i.e. i charge a metal ball with a negative charge, thus the ball becomes negatively charged.

Here's a fun thing. You ever heard of the Van de Graaff? That metal ball thing that shocks you? It is an electrostatic generator that uses a belt of sorts to accumulate charge. It creates very big electric potentials so as you go to touch it, it discharges onto you, shocking you.

Remember how I kept using the word “object” when describing the neutral things needed in explaining charging by induction? Yea, so those objects are always going to be conductors. So charging by induction is always charging conductors by induction.

[[File:Inschargedist.gif|center|375]]

===A Mathematical Model===

===A Computational Model===

==Examples==
[[File:Screen Shot 2018-04-08 at 4.16.06 PM.png|300 px]]
[[File:Screen Shot 2018-04-08 at 4.27.27 PM.png|300 px]]
[[File:Screen Shot 2018-04-08 at 4.34.27 PM.png|300 px]]

===Example Problem===

A neutral metal sphere enters a capacitor, as shown. Show the charge distribution on the sphere. If the sphere was instead made of plastic, show the the charge distribution.

[[File:Screen Shot 2018-04-08 at 4.37.07 PM.png|300 px]]

===Solution===
For a metal sphere:

[[File:Screen Shot 2018-04-08 at 4.38.17 PM.png|300 px]]

Since metals are conductors, there are mobile electrons within the sphere that are free to move about the sphere when they feel the electrostatic force from the capacitor.

For a plastic sphere:

[[File:Screen Shot 2018-04-08 at 4.39.59 PM.png|300 px]]

Plastic is an insulator, so the individual molecules will become polarized due to the capacitor and become dipoles.

==Connectedness==

[[File:Lightning bedning.png|thumb|Mako bending lightning to save Republic City in Avatar Legend of Korra]]

Okay, so I’m going to get super fan boy nerdy for a second and connect what we’ve learned about charging and discharging to my favorite cartoon of all time: Avatar The Last Airbender. For those of you not familiar with this amazing series – these resources might be help create some context (I highly recommend watching the original series and the spinoff with Avatar Korra): http://www.nick.com/avatar-the-last-airbender/

Within the world of Avatar and bending, each bending art has one or more specialized bending techniques that are considered to be very rare, highly coveted skills. Lightening bending (or lightning generation) is a special technique within the art of firebending. Lightning bending is an extremely difficult technique to master, as it requires the bender to be at total peace and void of all emotion (which is why it was always impossible for Zuko to learn from Uncle Iroh because that boy had so many emotional issues – which is completely understandable considering his circumstances).

Lightning bending connects to charging and discharging in a few ways. Lightning (and lightning generation) in and of itself alone is a representation of electrical discharge. Lightning is created when two dense pools of opposite charges reach out and connect to one another, creating a channel for electrical transfer of charge that results in each pool becoming neutral once that transfer is complete.

[[File:Strike1.gif]] [[File:Strike2.gif]] [[File:Strike3.gif]]

Photos from free U.S. government resource: http://www.srh.noaa.gov/jetstream/lightning/lightning.html

Before we continue, please watch the following video to see lightning bending in action: https://www.youtube.com/watch?v=6htOzNpBJv8

So let’s break down what we see in the video and how it relates to the physics of charging and discharging. Before we even see Uncle Iroh generate visible lightning, he “charges himself” by, “separating the energies of yin and yang” (according to Avatar mythology). In other words, Uncle Iroh separates mobile negatively charged particles (yin) from mobile positively charged particles (yang) within his body and immediate surroundings to set the stage for electrical charge transfer. When the amount of charge Uncle Iroh has built up in each pool is great enough to overcome the air and his body’s insulation of electric flow, this is the moment we begin to actually see lightning. The generated pools of charge connect and create a channel where charges begin to flow between pools in order for each pool to discharge and become neutral.

So all in all, in order for a firebender to lighting bend, the bender must be adept at charging and discharging to perform the technique. Once the lighting is generated, the bender then guides the discharging electric flow of energy in a desired direction (more than likely at an opponent in order to zap them).

Note: all representations of anything related to Avatar The Last Airbender are property of Nickelodeon and they rights are reserved.

While insulators are not useful for transferring charge, they do serve a critical role in electrostatic experiments and demonstrations. Conductive objects are often mounted upon insulating objects. This arrangement of a conductor on top of an insulator prevents charge from being transferred from the conductive object to its surroundings.

==History==
(should be completed by a student)

==See also==

===Further Reading===
*[http://www.physicsbook.gatech.edu/Charge_Transfer Charge Transfer]
*[http://www.physicsbook.gatech.edu/Insulators Insulators]
*[http://www.physicsbook.gatech.edu/Conductivity Conductivity]
*[http://www.physicsbook.gatech.edu/Polarization_of_a_conductor Polarization of a Conductor]

===External links===
*[http://www.schoolphysics.co.uk/age16-19/Electricity%20and%20magnetism/Electrostatics/text/Electric_charge_distribution/index.html Explanation of why charge concentrates at a point on a conductor]

==References==
*[http://www.physicsclassroom.com/class/estatics/Lesson-1/Conductors-and-Insulators http://www.physicsclassroom.com/class/estatics/Lesson-1/Conductors-and-Insulators]
*http://www.schoolphysics.co.uk/age16-19/Electricity%20and%20magnetism/Electrostatics/text/Electric_charge_distribution/index.html
*http://p3server.pa.msu.edu/coursewiki/doku.php?id=184_notes:charging_discharging
*http://avatar.wikia.com/wiki/Specialized_bending_techniques
*http://www.srh.noaa.gov/jetstream/lightning/lightning.html

Electrostatic Discharge

2020-05-11T17:48:47Z

Sydney: /* Examples */

Edited by Jonathan Ledet (Spring 2017)

Electrostatic Discharge (ESD) is the swift transfer of charges between objects at different potentials. ESD commonly occurs after the accumulation of static electricity on a material and can have devastating effects on solid-state electronics.

==The Main Idea==

All materials (both insulators and conductors) are defined on the Triboelectric series, which essentially rates that material's affinity for electrons. When an object lower on the series is touched by an object higher on the series, the lower object will acquire a negative charge. When the objects are separated, their charges are equal and opposite- just waiting for the opportunity to discharge. When the objects finally do discharge, a high voltage spark can form, turning the bridging air to plasma or frying circuits. This process is referred to as tribocharging.

You may have seen a demonstration in class involving a glass rod and a piece of cloth, or rubbed a balloon on your hair and observed its wacky behavior. Both of these are examples of the Triboelectric effect. Additionally, although friction greatly increases the magnitude of the effect, it is not required for the effect to occur. Only an initial contact and subsequent separation are required; friction only amplifies the effect due to the increased contact and separation of molecules on the microscopic level.

Electrostatic Discharge can be caused by an electrical breakdown, a short circuit, and--most commonly--Tribocharging. ESD is measured using an electrostatic voltmeter.

Sparks and lightning are visible Electrostatic Discharge events, but only represent part of the threat of ESD. For example, such a high voltage can be very damaging to the delicate pathways and components on circuit boards.

ESD events, like a spark from a human hand, allow current to travel to the ground through electronic devices, burning holes in integrated circuits and dealing heat damage to the circuit board. This can happen when working barehanded (without an electrostatic wrist strap) with circuit boards and other sensitive electronic equipment, when negatively charged synthetic materials are on or near sensitive electronic equipment, or due to the fast movement of air near electronic equipment.

There are some preventative methods taken to discourage Electrostatic Discharge. During the process in which electronic components are assembled most manufacturers implement Electrostatic Discharge Protected Areas (EPA). These areas are specially designed to prevent the build up of charge on the components, workers, and all other conductive materials. To protect against Electrostatic Discharge during transit, antistatic bags act as Faraday Cages to protect sensitive devices.

===The Physics Principals and Visual Aids===

Triboelectric Series
*Celluloid
*Sulfer
*Rubber
*Copper, Brass
*Amber
*Wood
*Cotton
*Human Skin
*Silk
*Cat Fur
*Wool
*Glass
*Rabbit Fur
*Asbestos

The objects near the top (of the list above) will tend to gain negative charges, while those below them will gain positive charges. The law of Conservation of Charge is followed.

[[File:5255360854_b0db025afb.jpg]]

First, a tube made out of plastic is charged when rubbed with synthetic fur (top pictures). The tube, which is now charged, is brought close to neutral paper bits on the table (bottom left). You can see that the tube and paper now attract each other and that this attraction is strong enough to lift the pieces of paper off the table.

==Examples==
(should be completed by a student)

===Simple===

===Middling===

===Difficult===

==Connectedness==
The concept of Electrostatic Discharge (ESD) can be applied to many practical scenarios; however, one of the most notable instances of ESD is in the electronics manufacturing industry. Due to the fact that charges can accumulate relatively easily in electronics manufacturing, companies often have to take ESD precautions by making employees wear special clothing and by designing their workbenches and flooring out of special material which prevents electrostatic build up.

Electrostatic Discharge is connected to chemical engineering in that there is a great deal of equipment in plants that has the potential to acquire electrostatic charge. As a result, it is vital to safety to be knowledgeable and cautious when dealing with such equipment in order to reduce work related injuries.

Electrostatic Discharge also has connections to Aerospace Engineering by how potentially dangerous it is to not guard against ESD in environments in which many different electronic devices are integrated into air and spacecraft. Aerospace research agencies, such as NASA, JAXA, or RKA, and corporation, such as Lockheed Martin, have implemented ESD protected areas and grounded workbenches, mandated the use of protective equipment, introduced protocol in which charge generating materials cannot be worn in protected areas, and have audits and inspections to make sure important aerospace electronics stay protected from ESD.

Additionally, ESD poses a threat in the operation of air and spacecraft due to its volatile nature and possible interaction with fuel. When fueling aircraft, it is mandated procedure to ground the air frame to the fuel truck in order to prevent a spark from jumping between the metal surfaces.

One interesting way that ESD can be used is to create sparks. When the dielectric between two oppositely charged sources is damaged or if the electric field due to the build up charge exceeds the dielectric, then ESD can occur by means of a spark through the air or other dielectric medium.

==History==
The phenomena of electrostatic discharge (ESD) has been known for a very long time, dating all the way back to the ancient Greeks. While ESD was mostly considered a non-factor throughout most of history, due to the growth of solid state electronics in the 1950's companies and researchers have had to account for and study in greater detail the phenomena of ESD. As the prevalence of electronics increased people began to notice the ESD could have very negative effects on certain components, causing them to short-circuit or malfunction. The 1960s and 70s were characterized by companies discovering methods and techniques to test for ESD, some of which included the Human Body Discharge Model and the Horizontal Coupling Plate. In the 1980's the release of the IBM personal computer saw an increased need for materials with resistance to ESD and thus the focus shifted towards maximizing the ability of electronic components to avoid ESD and subsequent malfunctioning. Since the 1980's and well into modern times, companies have been working to hone and refine this process in order to maximize the efficacy of their circuit components by increasing their ability to resist ESD when in use.

== See also ==
*[[Charge Transfer]]

*[[Charge Motion in Metals]]

*[[Polarization]]

*[https://en.wikipedia.org/wiki/Michael_Faraday Micheal Faraday]

===Further reading===

*Electro Static Discharge: Understand, Simulate, and Fix ESD Problems, 3rd Edition

===External links===

*http://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Friction

==References==

*Aaq.auburn.edu,. 'Summary | Academy Of Aerospace Quality'. N.p., 2015. Web. 5 Dec. 2015. (http://aaq.auburn.edu/node/277)

*Hoolihan, Daniel D. "A Brief History of Electrostatic Discharge Testing of Electronic Products Read More: Http://incompliancemag.com/article/a-brief-history-of-electrostatic-discharge-testing-of-electronic-products/#ixzz4R5MRIIIQ Follow Us: @incompliancemag on Twitter | Incompliancemag on Facebook." INCompliance. ECM Consulting, 01 Mar. 2014. Web. 25 Nov. 2016.

*Michaels, Ken, AYMAN ZAHER, and Stan Herron. 'Electrostatic Discharge: Causes, Effects, And Solutions'. Ecmweb.com. N.p., 2013. Web. 6 Dec. 2015. (http://ecmweb.com/content/electrostatic-discharge-causes-effects-and-solutions)

*Physicsclassroom.com,. 'Charging By Friction'. N.p., 2015. Web. 6 Dec. 2015. (http://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Friction)

[[Category: Fields]]

Charge Transfer

2020-05-11T17:46:29Z

Sydney: /* A Mathematical Model */

claimed by '''Fehmeen Tariq Spring 2020'''

If a charged conductor comes in contact, or is in close enough proximity, with another conductor, it is possible to transfer this charge to the second conductor. This process is called '''charge transfer'''. However, the ''Law of Conservation of Charge'' states that charges cannot be created or destroyed. Charge cannot be created, the presence of a negative charge is merely the effect of an object gaining electrons from another material. Since charge cannot be created or destroyed, and is just the transfer of electrons between materials, the magnitude of the charge transfer between two objects will be equivalent. For however much the charge of one object increases during charge transfer, the other must decrease the same amount. There are multiple ways that charge can be transferred such as through direct contact (ie friction), induction, and conduction.

==The Main Idea==
'''Insulators vs Conductors'''

[[File:inschargedist.gif|thumb|500px|Charges transferred to an insulator remains at the location of transfer.]]

In an '''insulator''', electrons are bounded tightly to atoms, which prevents charged particles from moving through the material. If charge is transferred to an insulator at a given location, the charge will remain at the location that the transfer occurred. Examples of insulators include rubber and air.

Within '''conductors''', on the other hand, electrons are able to flow freely from particle to particle. When charge is transferred to a conductor, the charge is distributed evenly across the surface of the object via ''electron movement''. The electrons will be distributed until the repelling force between the excess electrons is minimized. This is the main difference between insulators and conductors: insulators do not have mobile charged particles whereas conductors have mobile charged particles that allow for charge transfer through the free movement of electrons. Examples of conductors include metals and salt water.

'''Charge by Friction'''

Certain objects and materials have a greater attraction to electrons than others. For example, rubber is highly attracted to electrons, whereas fur or hair has a lower attraction. Therefore, if you take a balloon and rub it on your head, electrons previously in the atoms of the hair will be pulled to the atoms of the rubber balloon. This creates an electron imbalance which makes the balloon negatively charged and the hair positively charged. This creates the effect where the hair will stand up and be pulled towards the balloon since the opposing charges make the two objects attracted. Likewise, two charged balloons will repel each other. Another example of this is rubbing a glass rod with silk. The glass rod will become positively charged and the silk will become negatively charged; this means that electrons were transferred from the glass rod to the silk, since protons are not removed from the nuclei. Rubbing two objects together is not necessary for charge transfer, but because rubbing creates more points of contact between two objects, it facilitates charge transfer.

'''Transfer Charges by Conduction'''

As shown in the previous section, electrons move from one object to another through points of contact; this is especially true among metals. Charging by conduction requires the contact of a charged metal, positive or negative, to a neutral metal. If a negatively charged object touches the neutral metal, the excess electrons will flow through the neutral object. Since electrons repel each other and the negatively charged metal has a buildup of electrons, a certain number of excess electrons will flow out and spread across the neutral object when given an outlet in the form of the neutral metal. This process leaves both metals negatively charged. The same process occurs with positively charged objects touching neutral metals.

[[File:Conductiontransfer.gif|450 px]]

'''Transfer Charges by Induction'''

Unlike the transfer of charges by conduction, objects can transfer charges by induction without making contact. When an object is charged, it has an electric field. This electric field will repel or attract electrons in another object. This electron movement is called transfer of charges by induction. A neutral object can be charged by another charged object through a process called '''polarization'''. This is when electrons in the object are repelled or attracted to one side of the object by the charged second object. For example, if a negatively charged sphere is placed near a neutral sphere, the electrons in the neutral sphere will be repelled by the charged sphere. The neutral sphere is now polarized, with one side of it being negatively charged and the other side being positively charged. The negatively charged side of the sphere can be removed through grounding or with a conductor. Once removed, the originally neutral sphere will now be positively charged. Another example of induction is the balloon and black pepper experiment. A balloon can be given a negative charge by rubbing it on hair. When the balloon is placed near grounded black pepper, the black pepper particles will be polarized so that they become positively charged on top and will be attracted to the negatively charged balloon.

[[File:Indtransfer.gif|500 px]]

===A Mathematical Model===
(should be completed by a student)

===A Computational Model===

This [https://phet.colorado.edu/en/simulation/balloons website] is a great model for charge transfer. It shows charge transfer between a sweater and a balloon to demonstrate static electricity. Try using it without the charges showing and guess where they will go.

[[File:Charge_transfer_model_pic.PNG|400 px]]

==Examples==

===Simple===

A student said, "When you touch a charged piece of metal, the metal is no longer charged; all the charge on the metal is neutralized." As a practical matter, this is nearly correct, but it isn't exactly right. What's wrong with saying that all the charge on the metal is neutralized? (Q15 from page 579 of ''Matter and Interactions'')

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Touching a charged piece of metal actually causes charge transfer. Electrons move from whichever item is more positively charged, or more lacking in electrons. What the student said is not entirely correct because the metal loses it's charge by the movement of electrons (negative charges).

</div></div>

===Middling===

You take two invisible tapes of some unknown brand, which we will call U and L, stick them together, and discharge the pair before pulling them apart and hanging them from the edge of your desk. When you bring an uncharged plastic pen within 10 cm of either the U tape or the L tape you see a slight attraction. Next you rub the pen through your hair, which is known to charge the pen negatively. Now you find that if you bring the charged pen within 8 cm of the L tape you see a slight repulsion, and if you bring the pen within 12 cm of the U tape you see a slight attraction. Briefly explain all of your observations. (Q21 from page 580 of ''Matter and Interactions'')

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

If a plastic pen is rubbed through your hair, it gets negatively charged. There is a slight repulsion with L-tape and negatively charged plastic pen. So the charge on the L-tape should be positive. There is a slight attraction with U-tape and negatively charged plastic pen. So the charge on the U-tape should be positive.

</div></div>

===Difficult===

You run your finger along the slick side of a positively charged tape, and then observe that the tape is no longer attracted to your hand. Which of the following are not plausible explanations for this observation? Check all that apply. (1) Sodium ions (Na+) from the salt water on your skin move onto the tape, leaving the tape with a zero (or very small) net charge. (2) Electrons from the mobile electron sea in your hand move onto the tape, leaving the tape with a zero (or very small) net charge. (3) Chloride ions (Cl-) from the salt water on your skin move onto the tape, leaving the tape with a zero (or very small) net charge. (4) Protons are pulled out of the nuclei of atoms in the tape and move onto your finger. (P61 from page 585 of ''Matter and Interactions'')

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

When Sodium ions (Na+) from the salt water on your skin move onto the tape, they provide additional positive charge to the already positive charged tape. So 1) is not a plausible explanation.

Human hand is considered neutral so no electrons travel from the hand to the positively charged tape so the net charge remains the same as before. So 2) is not a plausible explanation.

Chlorine particles are negative and they neutralize the positive tape. So tape then has no net charge and can't attract the hand. So 3) is a plausible explanation.

There is no chance of protons getting pulled out of the atoms nucleus as they are bound very strongly by the strong nuclear force. So 4) is not a plausible explanation.

</div></div>

==Connectedness==
(should be completed by a student)

==History==
(should be completed by a student)

==See also==

===Further Reading===
[[Charge Motion in Metals]]

[[Polarization]]

===External Links===
More on [https://byjus.com/physics/charge-transfer/ charge transfer].

More on [https://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Conduction charging by conduction].

This [https://www.brightstorm.com/science/physics/electricity/charge-transfer-electroscope/ video] talks about methods of charge transfer and using an electroscope to measure charge.

==References==

Internet resources on this topic:

http://www.physicsclassroom.com/class/estatics/Lesson-1/Conductors-and-Insulators

http://www.physicsclassroom.com/class/estatics/Lesson-1/Charge-Interactions

Chabay, Ruth W., Bruce Sherwood. Matter and Interactions, Volume II: Electric and Magnetic Interactions, 4th Edition. Wiley, 19/2015. VitalBook file.

[[Category:Which Category did you place this in?]]

Charge Transfer

2020-05-11T17:46:06Z

Sydney: /* History */

claimed by '''Fehmeen Tariq Spring 2020'''

If a charged conductor comes in contact, or is in close enough proximity, with another conductor, it is possible to transfer this charge to the second conductor. This process is called '''charge transfer'''. However, the ''Law of Conservation of Charge'' states that charges cannot be created or destroyed. Charge cannot be created, the presence of a negative charge is merely the effect of an object gaining electrons from another material. Since charge cannot be created or destroyed, and is just the transfer of electrons between materials, the magnitude of the charge transfer between two objects will be equivalent. For however much the charge of one object increases during charge transfer, the other must decrease the same amount. There are multiple ways that charge can be transferred such as through direct contact (ie friction), induction, and conduction.

==The Main Idea==
'''Insulators vs Conductors'''

[[File:inschargedist.gif|thumb|500px|Charges transferred to an insulator remains at the location of transfer.]]

In an '''insulator''', electrons are bounded tightly to atoms, which prevents charged particles from moving through the material. If charge is transferred to an insulator at a given location, the charge will remain at the location that the transfer occurred. Examples of insulators include rubber and air.

Within '''conductors''', on the other hand, electrons are able to flow freely from particle to particle. When charge is transferred to a conductor, the charge is distributed evenly across the surface of the object via ''electron movement''. The electrons will be distributed until the repelling force between the excess electrons is minimized. This is the main difference between insulators and conductors: insulators do not have mobile charged particles whereas conductors have mobile charged particles that allow for charge transfer through the free movement of electrons. Examples of conductors include metals and salt water.

'''Charge by Friction'''

Certain objects and materials have a greater attraction to electrons than others. For example, rubber is highly attracted to electrons, whereas fur or hair has a lower attraction. Therefore, if you take a balloon and rub it on your head, electrons previously in the atoms of the hair will be pulled to the atoms of the rubber balloon. This creates an electron imbalance which makes the balloon negatively charged and the hair positively charged. This creates the effect where the hair will stand up and be pulled towards the balloon since the opposing charges make the two objects attracted. Likewise, two charged balloons will repel each other. Another example of this is rubbing a glass rod with silk. The glass rod will become positively charged and the silk will become negatively charged; this means that electrons were transferred from the glass rod to the silk, since protons are not removed from the nuclei. Rubbing two objects together is not necessary for charge transfer, but because rubbing creates more points of contact between two objects, it facilitates charge transfer.

'''Transfer Charges by Conduction'''

As shown in the previous section, electrons move from one object to another through points of contact; this is especially true among metals. Charging by conduction requires the contact of a charged metal, positive or negative, to a neutral metal. If a negatively charged object touches the neutral metal, the excess electrons will flow through the neutral object. Since electrons repel each other and the negatively charged metal has a buildup of electrons, a certain number of excess electrons will flow out and spread across the neutral object when given an outlet in the form of the neutral metal. This process leaves both metals negatively charged. The same process occurs with positively charged objects touching neutral metals.

[[File:Conductiontransfer.gif|450 px]]

'''Transfer Charges by Induction'''

Unlike the transfer of charges by conduction, objects can transfer charges by induction without making contact. When an object is charged, it has an electric field. This electric field will repel or attract electrons in another object. This electron movement is called transfer of charges by induction. A neutral object can be charged by another charged object through a process called '''polarization'''. This is when electrons in the object are repelled or attracted to one side of the object by the charged second object. For example, if a negatively charged sphere is placed near a neutral sphere, the electrons in the neutral sphere will be repelled by the charged sphere. The neutral sphere is now polarized, with one side of it being negatively charged and the other side being positively charged. The negatively charged side of the sphere can be removed through grounding or with a conductor. Once removed, the originally neutral sphere will now be positively charged. Another example of induction is the balloon and black pepper experiment. A balloon can be given a negative charge by rubbing it on hair. When the balloon is placed near grounded black pepper, the black pepper particles will be polarized so that they become positively charged on top and will be attracted to the negatively charged balloon.

[[File:Indtransfer.gif|500 px]]

===A Mathematical Model===

===A Computational Model===

This [https://phet.colorado.edu/en/simulation/balloons website] is a great model for charge transfer. It shows charge transfer between a sweater and a balloon to demonstrate static electricity. Try using it without the charges showing and guess where they will go.

[[File:Charge_transfer_model_pic.PNG|400 px]]

==Examples==

===Simple===

A student said, "When you touch a charged piece of metal, the metal is no longer charged; all the charge on the metal is neutralized." As a practical matter, this is nearly correct, but it isn't exactly right. What's wrong with saying that all the charge on the metal is neutralized? (Q15 from page 579 of ''Matter and Interactions'')

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Touching a charged piece of metal actually causes charge transfer. Electrons move from whichever item is more positively charged, or more lacking in electrons. What the student said is not entirely correct because the metal loses it's charge by the movement of electrons (negative charges).

</div></div>

===Middling===

You take two invisible tapes of some unknown brand, which we will call U and L, stick them together, and discharge the pair before pulling them apart and hanging them from the edge of your desk. When you bring an uncharged plastic pen within 10 cm of either the U tape or the L tape you see a slight attraction. Next you rub the pen through your hair, which is known to charge the pen negatively. Now you find that if you bring the charged pen within 8 cm of the L tape you see a slight repulsion, and if you bring the pen within 12 cm of the U tape you see a slight attraction. Briefly explain all of your observations. (Q21 from page 580 of ''Matter and Interactions'')

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

If a plastic pen is rubbed through your hair, it gets negatively charged. There is a slight repulsion with L-tape and negatively charged plastic pen. So the charge on the L-tape should be positive. There is a slight attraction with U-tape and negatively charged plastic pen. So the charge on the U-tape should be positive.

</div></div>

===Difficult===

You run your finger along the slick side of a positively charged tape, and then observe that the tape is no longer attracted to your hand. Which of the following are not plausible explanations for this observation? Check all that apply. (1) Sodium ions (Na+) from the salt water on your skin move onto the tape, leaving the tape with a zero (or very small) net charge. (2) Electrons from the mobile electron sea in your hand move onto the tape, leaving the tape with a zero (or very small) net charge. (3) Chloride ions (Cl-) from the salt water on your skin move onto the tape, leaving the tape with a zero (or very small) net charge. (4) Protons are pulled out of the nuclei of atoms in the tape and move onto your finger. (P61 from page 585 of ''Matter and Interactions'')

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

When Sodium ions (Na+) from the salt water on your skin move onto the tape, they provide additional positive charge to the already positive charged tape. So 1) is not a plausible explanation.

Human hand is considered neutral so no electrons travel from the hand to the positively charged tape so the net charge remains the same as before. So 2) is not a plausible explanation.

Chlorine particles are negative and they neutralize the positive tape. So tape then has no net charge and can't attract the hand. So 3) is a plausible explanation.

There is no chance of protons getting pulled out of the atoms nucleus as they are bound very strongly by the strong nuclear force. So 4) is not a plausible explanation.

</div></div>

==Connectedness==
(should be completed by a student)

==History==
(should be completed by a student)

==See also==

===Further Reading===
[[Charge Motion in Metals]]

[[Polarization]]

===External Links===
More on [https://byjus.com/physics/charge-transfer/ charge transfer].

More on [https://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Conduction charging by conduction].

This [https://www.brightstorm.com/science/physics/electricity/charge-transfer-electroscope/ video] talks about methods of charge transfer and using an electroscope to measure charge.

==References==

Internet resources on this topic:

http://www.physicsclassroom.com/class/estatics/Lesson-1/Conductors-and-Insulators

http://www.physicsclassroom.com/class/estatics/Lesson-1/Charge-Interactions

Chabay, Ruth W., Bruce Sherwood. Matter and Interactions, Volume II: Electric and Magnetic Interactions, 4th Edition. Wiley, 19/2015. VitalBook file.

[[Category:Which Category did you place this in?]]

Charge Transfer

2020-05-11T17:45:51Z

Sydney: /* Connectedness */

claimed by '''Fehmeen Tariq Spring 2020'''

If a charged conductor comes in contact, or is in close enough proximity, with another conductor, it is possible to transfer this charge to the second conductor. This process is called '''charge transfer'''. However, the ''Law of Conservation of Charge'' states that charges cannot be created or destroyed. Charge cannot be created, the presence of a negative charge is merely the effect of an object gaining electrons from another material. Since charge cannot be created or destroyed, and is just the transfer of electrons between materials, the magnitude of the charge transfer between two objects will be equivalent. For however much the charge of one object increases during charge transfer, the other must decrease the same amount. There are multiple ways that charge can be transferred such as through direct contact (ie friction), induction, and conduction.

==The Main Idea==
'''Insulators vs Conductors'''

[[File:inschargedist.gif|thumb|500px|Charges transferred to an insulator remains at the location of transfer.]]

In an '''insulator''', electrons are bounded tightly to atoms, which prevents charged particles from moving through the material. If charge is transferred to an insulator at a given location, the charge will remain at the location that the transfer occurred. Examples of insulators include rubber and air.

Within '''conductors''', on the other hand, electrons are able to flow freely from particle to particle. When charge is transferred to a conductor, the charge is distributed evenly across the surface of the object via ''electron movement''. The electrons will be distributed until the repelling force between the excess electrons is minimized. This is the main difference between insulators and conductors: insulators do not have mobile charged particles whereas conductors have mobile charged particles that allow for charge transfer through the free movement of electrons. Examples of conductors include metals and salt water.

'''Charge by Friction'''

Certain objects and materials have a greater attraction to electrons than others. For example, rubber is highly attracted to electrons, whereas fur or hair has a lower attraction. Therefore, if you take a balloon and rub it on your head, electrons previously in the atoms of the hair will be pulled to the atoms of the rubber balloon. This creates an electron imbalance which makes the balloon negatively charged and the hair positively charged. This creates the effect where the hair will stand up and be pulled towards the balloon since the opposing charges make the two objects attracted. Likewise, two charged balloons will repel each other. Another example of this is rubbing a glass rod with silk. The glass rod will become positively charged and the silk will become negatively charged; this means that electrons were transferred from the glass rod to the silk, since protons are not removed from the nuclei. Rubbing two objects together is not necessary for charge transfer, but because rubbing creates more points of contact between two objects, it facilitates charge transfer.

'''Transfer Charges by Conduction'''

As shown in the previous section, electrons move from one object to another through points of contact; this is especially true among metals. Charging by conduction requires the contact of a charged metal, positive or negative, to a neutral metal. If a negatively charged object touches the neutral metal, the excess electrons will flow through the neutral object. Since electrons repel each other and the negatively charged metal has a buildup of electrons, a certain number of excess electrons will flow out and spread across the neutral object when given an outlet in the form of the neutral metal. This process leaves both metals negatively charged. The same process occurs with positively charged objects touching neutral metals.

[[File:Conductiontransfer.gif|450 px]]

'''Transfer Charges by Induction'''

Unlike the transfer of charges by conduction, objects can transfer charges by induction without making contact. When an object is charged, it has an electric field. This electric field will repel or attract electrons in another object. This electron movement is called transfer of charges by induction. A neutral object can be charged by another charged object through a process called '''polarization'''. This is when electrons in the object are repelled or attracted to one side of the object by the charged second object. For example, if a negatively charged sphere is placed near a neutral sphere, the electrons in the neutral sphere will be repelled by the charged sphere. The neutral sphere is now polarized, with one side of it being negatively charged and the other side being positively charged. The negatively charged side of the sphere can be removed through grounding or with a conductor. Once removed, the originally neutral sphere will now be positively charged. Another example of induction is the balloon and black pepper experiment. A balloon can be given a negative charge by rubbing it on hair. When the balloon is placed near grounded black pepper, the black pepper particles will be polarized so that they become positively charged on top and will be attracted to the negatively charged balloon.

[[File:Indtransfer.gif|500 px]]

===A Mathematical Model===

===A Computational Model===

This [https://phet.colorado.edu/en/simulation/balloons website] is a great model for charge transfer. It shows charge transfer between a sweater and a balloon to demonstrate static electricity. Try using it without the charges showing and guess where they will go.

[[File:Charge_transfer_model_pic.PNG|400 px]]

==Examples==

===Simple===

A student said, "When you touch a charged piece of metal, the metal is no longer charged; all the charge on the metal is neutralized." As a practical matter, this is nearly correct, but it isn't exactly right. What's wrong with saying that all the charge on the metal is neutralized? (Q15 from page 579 of ''Matter and Interactions'')

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Touching a charged piece of metal actually causes charge transfer. Electrons move from whichever item is more positively charged, or more lacking in electrons. What the student said is not entirely correct because the metal loses it's charge by the movement of electrons (negative charges).

</div></div>

===Middling===

You take two invisible tapes of some unknown brand, which we will call U and L, stick them together, and discharge the pair before pulling them apart and hanging them from the edge of your desk. When you bring an uncharged plastic pen within 10 cm of either the U tape or the L tape you see a slight attraction. Next you rub the pen through your hair, which is known to charge the pen negatively. Now you find that if you bring the charged pen within 8 cm of the L tape you see a slight repulsion, and if you bring the pen within 12 cm of the U tape you see a slight attraction. Briefly explain all of your observations. (Q21 from page 580 of ''Matter and Interactions'')

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

If a plastic pen is rubbed through your hair, it gets negatively charged. There is a slight repulsion with L-tape and negatively charged plastic pen. So the charge on the L-tape should be positive. There is a slight attraction with U-tape and negatively charged plastic pen. So the charge on the U-tape should be positive.

</div></div>

===Difficult===

You run your finger along the slick side of a positively charged tape, and then observe that the tape is no longer attracted to your hand. Which of the following are not plausible explanations for this observation? Check all that apply. (1) Sodium ions (Na+) from the salt water on your skin move onto the tape, leaving the tape with a zero (or very small) net charge. (2) Electrons from the mobile electron sea in your hand move onto the tape, leaving the tape with a zero (or very small) net charge. (3) Chloride ions (Cl-) from the salt water on your skin move onto the tape, leaving the tape with a zero (or very small) net charge. (4) Protons are pulled out of the nuclei of atoms in the tape and move onto your finger. (P61 from page 585 of ''Matter and Interactions'')

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

When Sodium ions (Na+) from the salt water on your skin move onto the tape, they provide additional positive charge to the already positive charged tape. So 1) is not a plausible explanation.

Human hand is considered neutral so no electrons travel from the hand to the positively charged tape so the net charge remains the same as before. So 2) is not a plausible explanation.

Chlorine particles are negative and they neutralize the positive tape. So tape then has no net charge and can't attract the hand. So 3) is a plausible explanation.

There is no chance of protons getting pulled out of the atoms nucleus as they are bound very strongly by the strong nuclear force. So 4) is not a plausible explanation.

</div></div>

==Connectedness==
(should be completed by a student)

==History==

==See also==

===Further Reading===
[[Charge Motion in Metals]]

[[Polarization]]

===External Links===
More on [https://byjus.com/physics/charge-transfer/ charge transfer].

More on [https://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Conduction charging by conduction].

This [https://www.brightstorm.com/science/physics/electricity/charge-transfer-electroscope/ video] talks about methods of charge transfer and using an electroscope to measure charge.

==References==

Internet resources on this topic:

http://www.physicsclassroom.com/class/estatics/Lesson-1/Conductors-and-Insulators

http://www.physicsclassroom.com/class/estatics/Lesson-1/Charge-Interactions

Chabay, Ruth W., Bruce Sherwood. Matter and Interactions, Volume II: Electric and Magnetic Interactions, 4th Edition. Wiley, 19/2015. VitalBook file.

[[Category:Which Category did you place this in?]]

Conductors

2020-05-11T17:44:55Z

Sydney: /* Difficult */

Claimed by Sydney Triplett Spring 2020.

Conductor are materials that allow electric current to travel with little resistance throughout. This is related to the structure of the atoms of the conductor. In this section, we will look at what a conductor is, why it is this way, and the applications.

==The Main Idea==

Conductors are defined as a material that allows for charged particles to move easily throughout. Charges placed on the surface of a conductor will not simply stay in one spot or only spread over the surface, but will immediately spread evenly throughout the conductor- given there are no interfering forces. If the conductor is in an electric field, for example, it will cause the (negative) charges to move in the opposite direction of the field.

Electric current flows by the net movement of electric charge. This can be by electrons, ions, or other charged particles.

Conductors allow for easy movement of charged particles because of the structure of the atoms. The outermost electrons in atoms that make up conductors are only loosely bound and allow for more interaction with other particles. These electrons that are free to move about the conductor are called the "sea of electrons". The sea of electrons is therefore able to move in response to other charges that are put on the conductor or in response to an electric field that the conductor is within. A conductor in an electric field will have an almost instantaneous rearranging of electrons so that there is a net zero electric field within the conductor. The external electric field induces an equal and opposite electric field within the conductor, so the two fields cancel out for a net zero electric field.

[[File:Cross section and length image.gif|right|180 px]] There are some factors that can change the conductance of a conductor. Shape and size, for example, affect the conductance of an object. A thicker(larger cross sectional area A as shown in the diagram) piece will be a better conductor than a thinner piece of the same material and other dimensions. This is the same concept that a thicker piece of wire allows for greater current flow. The larger cross sectional area allows for more flow of charge carriers. A shorter conductor will also conduct better since it has less resistance than a longer piece. Conductance itself can also change conductivity. In actively conducting electric current, the conductor heats up. This is secretly the third factor affecting conductance: temperature. Changes in temperature can cause the same object to have a different conductance under otherwise identical conditions. The most well known example of this is glass. Glass is more of an insulator at typical to cold temperatures, but becomes a good conductor at higher temperatures. Generally, metals are better conductors at cooler temperatures. This is because an increase in temperature is an increase in energy, specifically for electrons.

[[File:Conductor chart.gif|600 px]]

===A Mathematical Model===

Ohm's Law of J = σE can be used to model the relationship of conductivity to electric current density where J is electric current density, σ is conductivity of the material, and E is electric field. This is a generalized form of the well known V = IR.

σ is larger for better conductors like metals and saltwater. For "perfect" conductors, σ approaches infinity. In this case, E would be zero since the current density J cannot be infinity.

Materials are generally divided into three categories based on σ:
*Lossless Materials: σ = 0
*Lossy Materials: σ > 0
*Conductors: σ >> 0

Below is a breakdown of how conductivity is calculated. This could be considered a formula for conductivity, but it would be more accurate to think of it as a definition.

[[File:Electric conductivity equation.jpg|400 px]]

Conductivity can also be explained as the inverse of resistivity. σ = 1/ρ where ρ is resistivity.

===A Computational Model===

[[File:Conductor computational model.PNG|right|400 px]]

[http://www.physics-chemistry-interactive-flash-animation.com/electricity_electromagnetism_interactive/electric_conductors_insulators.htm This simple interactive] is a great way to see which real objects are made of conducting or insulating material. Try to guess which objects will allow for flow of electricity before you test with the interactive.

https://phet.colorado.edu/en/simulation/semiconductor

==Examples==

===Simple===
Material A has a resitivity of <math> 5.90 \cdot 10^{-8} Ω \cdot m </math> and Material B has a conductivity of <math> 1.00 \cdot 10^7 S/m </math>. They are the same size and temperature. Which is a better conductor?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can change Material A's resitivity to conductivity with the formula σ = 1/ρ. σ = <math>\frac{1}{(5.90 \cdot 10^{-8} Ω \cdot m)} = 1.69 \cdot 10^7 S/m</math>. Since Material A has a greater conductivity, it is a better conductor.

These numbers are the real conductivities of Zinc(Material A) and Iron(Material B).

</div></div>

===Middling===
A negatively charged iron block is placed in a region where there is an electric field downward (in the -y direction). What will be the charge distribution of the iron block in this field? (Problem 47 from Matter and Interactions, page 583)

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

[[File:Conductor question diagram.jpg|300 px]]

Remember that in the direction of an electric field is traditionally the direction of positive charge movement. Since the iron block is a conductor, it has electrons that are free to move and will travel opposite the direction of the electric field. This will leave an excess of positive charge at the bottom of the block and an excess of negative charge at the top of the block, as shown in the diagram.

</div></div>

===Difficult===
This should be completed by a student

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

solution

</div></div>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Modern research on electricity and conductors starts in the 1700s. Many different scientists contributed to the research that led to the understanding and use of conductors. Stephan Gray was one of the first of these, first studying the idea of electricity and then conductors. In Gray's time, the general consensus was that "electric virtue" was a quality that some materials could attain and others could not. Some materials, like glass, could acquire electric virtue by friction, while others, like metal, could be given electric virtue by contact with a charged object. Gray tested this theory with many different types of material, even with a child(who did work as a conductor).

Dufay began research on the same topics of charge transfer and conductance just after Gray. He lengthened the list of objects that could be given electric virtue by friction. Dufay also named the growing categories. "Electrical bodies" were what we call insulators and "non-electric bodies" were what we know as conductors. While this seems backwards from the way we think about insulators and conductors, it comes from the idea that electric virtue was intrinsic to insulators because charge could be induced on these simply by friction, while conductors can only come to have a charge by contact with a charged insulator.

Only when Benjamin Franklin came around some later did the ideology and vocabulary make a big switch. Franklin suggested that electricity is not created by electrical bodies through friction, but that it is a fluid shared by all bodies and can pass between them. Franklin also caused the shift in language from non-electric bodies to conductors and electric bodies to non-conductors.

There is some evidence that ancient Egyptians also used electricity.

== See also ==

===Further reading===

To learn more about conductors:
*[[Polarization of a conductor]]
*[[Charged Conductor and Charged Insulator]]
*[[Field of a Charged Ball]]

===External links===

*https://www.youtube.com/watch?v=BY8ZPobU8B0
*https://www.khanacademy.org/science/ap-physics-1/ap-electric-charge-electric-force-and-voltage/conservation-of-charge-ap/v/conductors-and-insulators
*https://www.youtube.com/watch?v=PafSqL1riS4

==References==

*https://en.wikipedia.org/wiki/Electrical_conductor

*https://www.thoughtco.com/examples-of-electrical-conductors-and-insulators-608315

*https://www.rpi.edu/dept/phys/ScIT/InformationProcessing/semicond/sc_glossary/scglossary.htm

*http://maxwells-equations.com/materials/conductivity.php

*https://en.wikipedia.org/wiki/Ohm%27s_law

*https://www.youtube.com/watch?v=BY8ZPobU8B0

*https://www.physicsclassroom.com/class/estatics/Lesson-1/Conductors-and-Insulators

*http://histoires-de-sciences.over-blog.fr/2018/04/history-of-electricity.the-discovery-of-conductors-and-insulators-by-gray-dufay-and-franklin.html

*Benjamin, Park. A History of Electricity (The Intellectual Rise of Electricity) from Antiquity to the Days of Benjamin Franklin. J. Wiley & Sons, 1898. Google Books, https://books.google.com/books?hl=en&lr=&id=K2dDAAAAIAAJ&oi=fnd&pg=PR1&dq=ancient egypt electricity&ots=edMffcocC0&sig=zFX9kUz2FKoPcPTCf8YVId2AjhQ#v=onepage&q&f=false.

*https://www.thoughtco.com/table-of-electrical-resistivity-conductivity-608499

[[Category:Which Category did you place this in?]]

Conductors

2020-05-06T16:46:24Z

Sydney: /* See also */

Claimed by Sydney Triplett Spring 2020.

Conductor are materials that allow electric current to travel with little resistance throughout. This is related to the structure of the atoms of the conductor. In this section, we will look at what a conductor is, why it is this way, and the applications.

==The Main Idea==

Conductors are defined as a material that allows for charged particles to move easily throughout. Charges placed on the surface of a conductor will not simply stay in one spot or only spread over the surface, but will immediately spread evenly throughout the conductor- given there are no interfering forces. If the conductor is in an electric field, for example, it will cause the (negative) charges to move in the opposite direction of the field.

Electric current flows by the net movement of electric charge. This can be by electrons, ions, or other charged particles.

Conductors allow for easy movement of charged particles because of the structure of the atoms. The outermost electrons in atoms that make up conductors are only loosely bound and allow for more interaction with other particles. These electrons that are free to move about the conductor are called the "sea of electrons". The sea of electrons is therefore able to move in response to other charges that are put on the conductor or in response to an electric field that the conductor is within. A conductor in an electric field will have an almost instantaneous rearranging of electrons so that there is a net zero electric field within the conductor. The external electric field induces an equal and opposite electric field within the conductor, so the two fields cancel out for a net zero electric field.

[[File:Cross section and length image.gif|right|180 px]] There are some factors that can change the conductance of a conductor. Shape and size, for example, affect the conductance of an object. A thicker(larger cross sectional area A as shown in the diagram) piece will be a better conductor than a thinner piece of the same material and other dimensions. This is the same concept that a thicker piece of wire allows for greater current flow. The larger cross sectional area allows for more flow of charge carriers. A shorter conductor will also conduct better since it has less resistance than a longer piece. Conductance itself can also change conductivity. In actively conducting electric current, the conductor heats up. This is secretly the third factor affecting conductance: temperature. Changes in temperature can cause the same object to have a different conductance under otherwise identical conditions. The most well known example of this is glass. Glass is more of an insulator at typical to cold temperatures, but becomes a good conductor at higher temperatures. Generally, metals are better conductors at cooler temperatures. This is because an increase in temperature is an increase in energy, specifically for electrons.

[[File:Conductor chart.gif|600 px]]

===A Mathematical Model===

Ohm's Law of J = σE can be used to model the relationship of conductivity to electric current density where J is electric current density, σ is conductivity of the material, and E is electric field. This is a generalized form of the well known V = IR.

σ is larger for better conductors like metals and saltwater. For "perfect" conductors, σ approaches infinity. In this case, E would be zero since the current density J cannot be infinity.

Materials are generally divided into three categories based on σ:
*Lossless Materials: σ = 0
*Lossy Materials: σ > 0
*Conductors: σ >> 0

Below is a breakdown of how conductivity is calculated. This could be considered a formula for conductivity, but it would be more accurate to think of it as a definition.

[[File:Electric conductivity equation.jpg|400 px]]

Conductivity can also be explained as the inverse of resistivity. σ = 1/ρ where ρ is resistivity.

===A Computational Model===

[[File:Conductor computational model.PNG|right|400 px]]

[http://www.physics-chemistry-interactive-flash-animation.com/electricity_electromagnetism_interactive/electric_conductors_insulators.htm This simple interactive] is a great way to see which real objects are made of conducting or insulating material. Try to guess which objects will allow for flow of electricity before you test with the interactive.

https://phet.colorado.edu/en/simulation/semiconductor

==Examples==

===Simple===
Material A has a resitivity of <math> 5.90 \cdot 10^{-8} Ω \cdot m </math> and Material B has a conductivity of <math> 1.00 \cdot 10^7 S/m </math>. They are the same size and temperature. Which is a better conductor?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can change Material A's resitivity to conductivity with the formula σ = 1/ρ. σ = <math>\frac{1}{(5.90 \cdot 10^{-8} Ω \cdot m)} = 1.69 \cdot 10^7 S/m</math>. Since Material A has a greater conductivity, it is a better conductor.

These numbers are the real conductivities of Zinc(Material A) and Iron(Material B).

</div></div>

===Middling===
A negatively charged iron block is placed in a region where there is an electric field downward (in the -y direction). What will be the charge distribution of the iron block in this field? (Problem 47 from Matter and Interactions, page 583)

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

[[File:Conductor question diagram.jpg|300 px]]

Remember that in the direction of an electric field is traditionally the direction of positive charge movement. Since the iron block is a conductor, it has electrons that are free to move and will travel opposite the direction of the electric field. This will leave an excess of positive charge at the bottom of the block and an excess of negative charge at the top of the block, as shown in the diagram.

</div></div>

===Difficult===
question

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

solution

</div></div>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Modern research on electricity and conductors starts in the 1700s. Many different scientists contributed to the research that led to the understanding and use of conductors. Stephan Gray was one of the first of these, first studying the idea of electricity and then conductors. In Gray's time, the general consensus was that "electric virtue" was a quality that some materials could attain and others could not. Some materials, like glass, could acquire electric virtue by friction, while others, like metal, could be given electric virtue by contact with a charged object. Gray tested this theory with many different types of material, even with a child(who did work as a conductor).

Dufay began research on the same topics of charge transfer and conductance just after Gray. He lengthened the list of objects that could be given electric virtue by friction. Dufay also named the growing categories. "Electrical bodies" were what we call insulators and "non-electric bodies" were what we know as conductors. While this seems backwards from the way we think about insulators and conductors, it comes from the idea that electric virtue was intrinsic to insulators because charge could be induced on these simply by friction, while conductors can only come to have a charge by contact with a charged insulator.

Only when Benjamin Franklin came around some later did the ideology and vocabulary make a big switch. Franklin suggested that electricity is not created by electrical bodies through friction, but that it is a fluid shared by all bodies and can pass between them. Franklin also caused the shift in language from non-electric bodies to conductors and electric bodies to non-conductors.

There is some evidence that ancient Egyptians also used electricity.

== See also ==

===Further reading===

To learn more about conductors:
*[[Polarization of a conductor]]
*[[Charged Conductor and Charged Insulator]]
*[[Field of a Charged Ball]]

===External links===

*https://www.youtube.com/watch?v=BY8ZPobU8B0
*https://www.khanacademy.org/science/ap-physics-1/ap-electric-charge-electric-force-and-voltage/conservation-of-charge-ap/v/conductors-and-insulators
*https://www.youtube.com/watch?v=PafSqL1riS4

==References==

*https://en.wikipedia.org/wiki/Electrical_conductor

*https://www.thoughtco.com/examples-of-electrical-conductors-and-insulators-608315

*https://www.rpi.edu/dept/phys/ScIT/InformationProcessing/semicond/sc_glossary/scglossary.htm

*http://maxwells-equations.com/materials/conductivity.php

*https://en.wikipedia.org/wiki/Ohm%27s_law

*https://www.youtube.com/watch?v=BY8ZPobU8B0

*https://www.physicsclassroom.com/class/estatics/Lesson-1/Conductors-and-Insulators

*http://histoires-de-sciences.over-blog.fr/2018/04/history-of-electricity.the-discovery-of-conductors-and-insulators-by-gray-dufay-and-franklin.html

*Benjamin, Park. A History of Electricity (The Intellectual Rise of Electricity) from Antiquity to the Days of Benjamin Franklin. J. Wiley & Sons, 1898. Google Books, https://books.google.com/books?hl=en&lr=&id=K2dDAAAAIAAJ&oi=fnd&pg=PR1&dq=ancient egypt electricity&ots=edMffcocC0&sig=zFX9kUz2FKoPcPTCf8YVId2AjhQ#v=onepage&q&f=false.

*https://www.thoughtco.com/table-of-electrical-resistivity-conductivity-608499

[[Category:Which Category did you place this in?]]

Conductors

2020-05-06T16:35:06Z

Sydney: /* The Main Idea */

Claimed by Sydney Triplett Spring 2020.

Conductor are materials that allow electric current to travel with little resistance throughout. This is related to the structure of the atoms of the conductor. In this section, we will look at what a conductor is, why it is this way, and the applications.

==The Main Idea==

Conductors are defined as a material that allows for charged particles to move easily throughout. Charges placed on the surface of a conductor will not simply stay in one spot or only spread over the surface, but will immediately spread evenly throughout the conductor- given there are no interfering forces. If the conductor is in an electric field, for example, it will cause the (negative) charges to move in the opposite direction of the field.

Electric current flows by the net movement of electric charge. This can be by electrons, ions, or other charged particles.

Conductors allow for easy movement of charged particles because of the structure of the atoms. The outermost electrons in atoms that make up conductors are only loosely bound and allow for more interaction with other particles. These electrons that are free to move about the conductor are called the "sea of electrons". The sea of electrons is therefore able to move in response to other charges that are put on the conductor or in response to an electric field that the conductor is within. A conductor in an electric field will have an almost instantaneous rearranging of electrons so that there is a net zero electric field within the conductor. The external electric field induces an equal and opposite electric field within the conductor, so the two fields cancel out for a net zero electric field.

[[File:Cross section and length image.gif|right|180 px]] There are some factors that can change the conductance of a conductor. Shape and size, for example, affect the conductance of an object. A thicker(larger cross sectional area A as shown in the diagram) piece will be a better conductor than a thinner piece of the same material and other dimensions. This is the same concept that a thicker piece of wire allows for greater current flow. The larger cross sectional area allows for more flow of charge carriers. A shorter conductor will also conduct better since it has less resistance than a longer piece. Conductance itself can also change conductivity. In actively conducting electric current, the conductor heats up. This is secretly the third factor affecting conductance: temperature. Changes in temperature can cause the same object to have a different conductance under otherwise identical conditions. The most well known example of this is glass. Glass is more of an insulator at typical to cold temperatures, but becomes a good conductor at higher temperatures. Generally, metals are better conductors at cooler temperatures. This is because an increase in temperature is an increase in energy, specifically for electrons.

[[File:Conductor chart.gif|600 px]]

===A Mathematical Model===

Ohm's Law of J = σE can be used to model the relationship of conductivity to electric current density where J is electric current density, σ is conductivity of the material, and E is electric field. This is a generalized form of the well known V = IR.

σ is larger for better conductors like metals and saltwater. For "perfect" conductors, σ approaches infinity. In this case, E would be zero since the current density J cannot be infinity.

Materials are generally divided into three categories based on σ:
*Lossless Materials: σ = 0
*Lossy Materials: σ > 0
*Conductors: σ >> 0

Below is a breakdown of how conductivity is calculated. This could be considered a formula for conductivity, but it would be more accurate to think of it as a definition.

[[File:Electric conductivity equation.jpg|400 px]]

Conductivity can also be explained as the inverse of resistivity. σ = 1/ρ where ρ is resistivity.

===A Computational Model===

[[File:Conductor computational model.PNG|right|400 px]]

[http://www.physics-chemistry-interactive-flash-animation.com/electricity_electromagnetism_interactive/electric_conductors_insulators.htm This simple interactive] is a great way to see which real objects are made of conducting or insulating material. Try to guess which objects will allow for flow of electricity before you test with the interactive.

https://phet.colorado.edu/en/simulation/semiconductor

==Examples==

===Simple===
Material A has a resitivity of <math> 5.90 \cdot 10^{-8} Ω \cdot m </math> and Material B has a conductivity of <math> 1.00 \cdot 10^7 S/m </math>. They are the same size and temperature. Which is a better conductor?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can change Material A's resitivity to conductivity with the formula σ = 1/ρ. σ = <math>\frac{1}{(5.90 \cdot 10^{-8} Ω \cdot m)} = 1.69 \cdot 10^7 S/m</math>. Since Material A has a greater conductivity, it is a better conductor.

These numbers are the real conductivities of Zinc(Material A) and Iron(Material B).

</div></div>

===Middling===
A negatively charged iron block is placed in a region where there is an electric field downward (in the -y direction). What will be the charge distribution of the iron block in this field? (Problem 47 from Matter and Interactions, page 583)

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

[[File:Conductor question diagram.jpg|300 px]]

Remember that in the direction of an electric field is traditionally the direction of positive charge movement. Since the iron block is a conductor, it has electrons that are free to move and will travel opposite the direction of the electric field. This will leave an excess of positive charge at the bottom of the block and an excess of negative charge at the top of the block, as shown in the diagram.

</div></div>

===Difficult===
question

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

solution

</div></div>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Modern research on electricity and conductors starts in the 1700s. Many different scientists contributed to the research that led to the understanding and use of conductors. Stephan Gray was one of the first of these, first studying the idea of electricity and then conductors. In Gray's time, the general consensus was that "electric virtue" was a quality that some materials could attain and others could not. Some materials, like glass, could acquire electric virtue by friction, while others, like metal, could be given electric virtue by contact with a charged object. Gray tested this theory with many different types of material, even with a child(who did work as a conductor).

Dufay began research on the same topics of charge transfer and conductance just after Gray. He lengthened the list of objects that could be given electric virtue by friction. Dufay also named the growing categories. "Electrical bodies" were what we call insulators and "non-electric bodies" were what we know as conductors. While this seems backwards from the way we think about insulators and conductors, it comes from the idea that electric virtue was intrinsic to insulators because charge could be induced on these simply by friction, while conductors can only come to have a charge by contact with a charged insulator.

Only when Benjamin Franklin came around some later did the ideology and vocabulary make a big switch. Franklin suggested that electricity is not created by electrical bodies through friction, but that it is a fluid shared by all bodies and can pass between them. Franklin also caused the shift in language from non-electric bodies to conductors and electric bodies to non-conductors.

There is some evidence that ancient Egyptians also used electricity.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

https://www.youtube.com/watch?v=BY8ZPobU8B0

==References==

*https://en.wikipedia.org/wiki/Electrical_conductor

*https://www.thoughtco.com/examples-of-electrical-conductors-and-insulators-608315

*https://www.rpi.edu/dept/phys/ScIT/InformationProcessing/semicond/sc_glossary/scglossary.htm

*http://maxwells-equations.com/materials/conductivity.php

*https://en.wikipedia.org/wiki/Ohm%27s_law

*https://www.youtube.com/watch?v=BY8ZPobU8B0

*https://www.physicsclassroom.com/class/estatics/Lesson-1/Conductors-and-Insulators

*http://histoires-de-sciences.over-blog.fr/2018/04/history-of-electricity.the-discovery-of-conductors-and-insulators-by-gray-dufay-and-franklin.html

*Benjamin, Park. A History of Electricity (The Intellectual Rise of Electricity) from Antiquity to the Days of Benjamin Franklin. J. Wiley & Sons, 1898. Google Books, https://books.google.com/books?hl=en&lr=&id=K2dDAAAAIAAJ&oi=fnd&pg=PR1&dq=ancient egypt electricity&ots=edMffcocC0&sig=zFX9kUz2FKoPcPTCf8YVId2AjhQ#v=onepage&q&f=false.

*https://www.thoughtco.com/table-of-electrical-resistivity-conductivity-608499

[[Category:Which Category did you place this in?]]

Field of a Charged Ball

2020-05-05T21:46:00Z

Sydney: /* Further reading */

==The Main Idea==
In this section, we will discuss the electric field of a solid sphere. Such a sphere has charge distributed throughout the volume (rather than only on the surface), and can be modeled by several layers of concentric, charged spherical shells. Calculating the electric field both outside and inside the sphere will be addressed.

[[File:Wiki_pic.JPG|400px|right|thumb|Figure 1: A sphere with uniformly distributed charge: note that this can be thought of as infinitely thin concentric charged shells]]
===A Mathematical Model===

First we must determine the relationship between r, the radius of the observation point from the center of the sphere, and R, the radius of the sphere itself.

Here, it is necessary to determine whether the observation point is outside or inside the sphere.

If r>R, then we are outside the sphere. As a result, the sphere can be simply treated as if the sphere were a point charge located at the center of the sphere. Hence, the electric field at any point outside of the radius of the sphere can be calculated using the formula for the electric field of a point charge.

<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>
when r>R, and R is the radius of the sphere.

However, when r<R, the observation location is inside of the sphere and the sphere overall must be thought of as infinitely many concentric charged shells inside of each other. All of the shells with a radius larger than that of the observation location do not contribute to the electric field at the observation location. This phenomena is because the electric field produced by these larger shells cancels out at any given point inside of itself. As a result only the shells smaller than the radius of the observation location need to be accounted for.
To find <math>\vec E_{net} </math>, add the contributions to the electric field from the inner shells.
After adding the contributions of each inner shell, you should have an electric field equal to:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{\Delta Q}{r^2}</math>

We find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

We find that:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

The charge inside the sphere is proportional to r. When r=R, we again treat the sphere as a point charge located at the center of the sphere.

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2} </math>

'''Looking Ahead, Gauss's Law'''

Later in this wikibook, you will learn about Gauss's Law. This will make calculating the electric field easier.

The formula for Gauss's Law is:

<math>\oint \vec E \bullet \hat n dA = \frac{1}{\epsilon_0}\Sigma Q_{\text {inside the surface}} </math>

Using this, one can calculate the the electric field when r<R for a solid sphere charged throughout.

<math>\vec E * 4\pi*r^2= \frac{1}{\epsilon_0}Q\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} </math>

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

===A Computational Model===

This demonstration using VPython shows the field of a charged ball with a fixed radius. Try changing the radius to see what happens!

https://trinket.io/glowscript/48f6efd07a

# GlowScript 1.1 VPython
## constants
oofpez = 9e9
qproton = 1.6e-19
scalefactor = 2e-20
Radius =3.5
r=3

## objects
particle = sphere(pos=vector(0,0,0), opacity =0.5, radius=Radius, color=color.blue)
## initial values
obslocation = vector(r,0,0)
##calculate position vector
rvector = obslocation - particle.pos
arrow1 = arrow(pos=particle.pos, axis=rvector, color=color.green)
rmag=mag(rvector)

##colored arrow

if r<= Radius:
E1= oofpez*(qproton)/(Radius)**3 *rmag* vector(1,0,0)
E2= oofpez*(qproton)/(Radius)**3 *rmag* vector(-1,0,0)
E3= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,1,0)
E4= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,-1,0)
E5= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,1)
E6= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,-1)
ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)
if r>Radius:
E1= oofpez*(qproton)/(rmag)**2 *vector(1,0,0)
E2= oofpez*(qproton)/(rmag)**2 *vector(-1,0,0)
E3= oofpez*(qproton)/(rmag)**2 * vector(0,1,0)
E4= oofpez*(qproton)/(rmag)**2 *vector(0,-1,0)
E5= oofpez*(qproton)/(rmag)**2 *vector(0,0,1)
E6= oofpez*(qproton)/(rmag)**2 * vector(0,0,-1)
ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)

==Examples==

===Simple===
[[File:Approx_point.png|200px|right|thumb|Figure 2: This is an example of applying the equation for finding the electric field of a point charge]]
Describe the pattern of the electric field of charged ball from far away.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

If we observe a charged ball from far away, we can say that r>R. Because we are outside of the sphere, the ball can essentially be viewed as a point charge. Thus the electric field can be defined by:
<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>

</div></div>

===Middling===
[[File:Concentric_sphere.png|200px|right|thumb|Figure 3: This is very similar to problem 1 and involves calculating the field of a point charge.]]
A sphere is charged throughout it's volume with a charge of Q= 6e-5 C. The radius of the this sphere is R=10. Find the electric field created by a sphere of radius r=4.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Step 1: Cut up the sphere into shells.

step 2: We know that r<R.

Next find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r = 9e9 * \frac{6*10^{-5} C}{(10m)^3}*4m = 2160 N/C </math>

</div></div>

===Difficult===
A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge −e. (The actual charge density in the ground state is nonuniform.)

[[File:circle_outline.png|200px|thumb|left|Figure 4: This circle represents a hydrogen atom. Use the radius and total charge to find alpha ]]

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

For the uniform-density model, calculate the polarizability α of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud

(s « R in the diagram). (Use the following as necessary: R and ε0.)

Useful equations:

<math> \vec p = \alpha *E</math>

<math> \vec p = Q*s </math>

Assume an applied electric field of strength E. This electric field polarized the hydrogen atom. Now there is a spherical charge of radius s. Use the volume ratio, and then use the useful equations to find <math> \alpha </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{e}{s^2}\frac{\frac{4}{3} \pi s^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s </math>

<math>\alpha = \frac{e*s}{\frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s} = 4\pi \epsilon_0 R^3 </math>

</div></div>

==Connectedness==
It may be particularly useful to discuss real-life applications of a charged solid sphere. Two examples stem from the structure of an atom. The nucleus of an atom is packed very tightly so that we can consider the charge to be uniformly distributed. The electron cloud also can be viewed as a packed spherical region of charged. Of course the radii of these structures are very small; radii are about 10e-15 m and 10e-10 m for nuclei and electron clouds respectively!

==History==
ADD MORE ABOUT HISTORY HERE

== See also ==

===Further reading===

*[[Point Charge]]

*[[Electric Dipole]]

*[http://www.physicsbook.gatech.edu/Gauss%27s_Flux_Theorem Gauss's Flux Theorem]

===External Links===

==References==

*http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html

*https://www.physicsforums.com/threads/calculate-the-polarizability-a-lpha-of-atomic-hydrogen-in-terms-of-r.339994/

*Matter and Interactions, 4th Edition: 1-2

Field of a Charged Ball

2020-05-05T21:42:52Z

Sydney: /* Examples */

==The Main Idea==
In this section, we will discuss the electric field of a solid sphere. Such a sphere has charge distributed throughout the volume (rather than only on the surface), and can be modeled by several layers of concentric, charged spherical shells. Calculating the electric field both outside and inside the sphere will be addressed.

[[File:Wiki_pic.JPG|400px|right|thumb|Figure 1: A sphere with uniformly distributed charge: note that this can be thought of as infinitely thin concentric charged shells]]
===A Mathematical Model===

First we must determine the relationship between r, the radius of the observation point from the center of the sphere, and R, the radius of the sphere itself.

Here, it is necessary to determine whether the observation point is outside or inside the sphere.

If r>R, then we are outside the sphere. As a result, the sphere can be simply treated as if the sphere were a point charge located at the center of the sphere. Hence, the electric field at any point outside of the radius of the sphere can be calculated using the formula for the electric field of a point charge.

<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>
when r>R, and R is the radius of the sphere.

However, when r<R, the observation location is inside of the sphere and the sphere overall must be thought of as infinitely many concentric charged shells inside of each other. All of the shells with a radius larger than that of the observation location do not contribute to the electric field at the observation location. This phenomena is because the electric field produced by these larger shells cancels out at any given point inside of itself. As a result only the shells smaller than the radius of the observation location need to be accounted for.
To find <math>\vec E_{net} </math>, add the contributions to the electric field from the inner shells.
After adding the contributions of each inner shell, you should have an electric field equal to:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{\Delta Q}{r^2}</math>

We find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

We find that:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

The charge inside the sphere is proportional to r. When r=R, we again treat the sphere as a point charge located at the center of the sphere.

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2} </math>

'''Looking Ahead, Gauss's Law'''

Later in this wikibook, you will learn about Gauss's Law. This will make calculating the electric field easier.

The formula for Gauss's Law is:

<math>\oint \vec E \bullet \hat n dA = \frac{1}{\epsilon_0}\Sigma Q_{\text {inside the surface}} </math>

Using this, one can calculate the the electric field when r<R for a solid sphere charged throughout.

<math>\vec E * 4\pi*r^2= \frac{1}{\epsilon_0}Q\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} </math>

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

===A Computational Model===

This demonstration using VPython shows the field of a charged ball with a fixed radius. Try changing the radius to see what happens!

https://trinket.io/glowscript/48f6efd07a

# GlowScript 1.1 VPython
## constants
oofpez = 9e9
qproton = 1.6e-19
scalefactor = 2e-20
Radius =3.5
r=3

## objects
particle = sphere(pos=vector(0,0,0), opacity =0.5, radius=Radius, color=color.blue)
## initial values
obslocation = vector(r,0,0)
##calculate position vector
rvector = obslocation - particle.pos
arrow1 = arrow(pos=particle.pos, axis=rvector, color=color.green)
rmag=mag(rvector)

##colored arrow

if r<= Radius:
E1= oofpez*(qproton)/(Radius)**3 *rmag* vector(1,0,0)
E2= oofpez*(qproton)/(Radius)**3 *rmag* vector(-1,0,0)
E3= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,1,0)
E4= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,-1,0)
E5= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,1)
E6= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,-1)
ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)
if r>Radius:
E1= oofpez*(qproton)/(rmag)**2 *vector(1,0,0)
E2= oofpez*(qproton)/(rmag)**2 *vector(-1,0,0)
E3= oofpez*(qproton)/(rmag)**2 * vector(0,1,0)
E4= oofpez*(qproton)/(rmag)**2 *vector(0,-1,0)
E5= oofpez*(qproton)/(rmag)**2 *vector(0,0,1)
E6= oofpez*(qproton)/(rmag)**2 * vector(0,0,-1)
ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)

==Examples==

===Simple===
[[File:Approx_point.png|200px|right|thumb|Figure 2: This is an example of applying the equation for finding the electric field of a point charge]]
Describe the pattern of the electric field of charged ball from far away.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

If we observe a charged ball from far away, we can say that r>R. Because we are outside of the sphere, the ball can essentially be viewed as a point charge. Thus the electric field can be defined by:
<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>

</div></div>

===Middling===
[[File:Concentric_sphere.png|200px|right|thumb|Figure 3: This is very similar to problem 1 and involves calculating the field of a point charge.]]
A sphere is charged throughout it's volume with a charge of Q= 6e-5 C. The radius of the this sphere is R=10. Find the electric field created by a sphere of radius r=4.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Step 1: Cut up the sphere into shells.

step 2: We know that r<R.

Next find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r = 9e9 * \frac{6*10^{-5} C}{(10m)^3}*4m = 2160 N/C </math>

</div></div>

===Difficult===
A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge −e. (The actual charge density in the ground state is nonuniform.)

[[File:circle_outline.png|200px|thumb|left|Figure 4: This circle represents a hydrogen atom. Use the radius and total charge to find alpha ]]

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

For the uniform-density model, calculate the polarizability α of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud

(s « R in the diagram). (Use the following as necessary: R and ε0.)

Useful equations:

<math> \vec p = \alpha *E</math>

<math> \vec p = Q*s </math>

Assume an applied electric field of strength E. This electric field polarized the hydrogen atom. Now there is a spherical charge of radius s. Use the volume ratio, and then use the useful equations to find <math> \alpha </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{e}{s^2}\frac{\frac{4}{3} \pi s^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s </math>

<math>\alpha = \frac{e*s}{\frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s} = 4\pi \epsilon_0 R^3 </math>

</div></div>

==Connectedness==
It may be particularly useful to discuss real-life applications of a charged solid sphere. Two examples stem from the structure of an atom. The nucleus of an atom is packed very tightly so that we can consider the charge to be uniformly distributed. The electron cloud also can be viewed as a packed spherical region of charged. Of course the radii of these structures are very small; radii are about 10e-15 m and 10e-10 m for nuclei and electron clouds respectively!

==History==
ADD MORE ABOUT HISTORY HERE

== See also ==

===Further reading===

*http://www.physicsbook.gatech.edu/Point_Charge

*http://www.physicsbook.gatech.edu/Electric_Dipole

*http://www.physicsbook.gatech.edu/Gauss%27s_Flux_Theorem

===External Links===

==References==

*http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html

*https://www.physicsforums.com/threads/calculate-the-polarizability-a-lpha-of-atomic-hydrogen-in-terms-of-r.339994/

*Matter and Interactions, 4th Edition: 1-2

Field of a Charged Ball

2020-05-05T21:40:04Z

Sydney:

==The Main Idea==
In this section, we will discuss the electric field of a solid sphere. Such a sphere has charge distributed throughout the volume (rather than only on the surface), and can be modeled by several layers of concentric, charged spherical shells. Calculating the electric field both outside and inside the sphere will be addressed.

[[File:Wiki_pic.JPG|400px|right|thumb|Figure 1: A sphere with uniformly distributed charge: note that this can be thought of as infinitely thin concentric charged shells]]
===A Mathematical Model===

First we must determine the relationship between r, the radius of the observation point from the center of the sphere, and R, the radius of the sphere itself.

Here, it is necessary to determine whether the observation point is outside or inside the sphere.

If r>R, then we are outside the sphere. As a result, the sphere can be simply treated as if the sphere were a point charge located at the center of the sphere. Hence, the electric field at any point outside of the radius of the sphere can be calculated using the formula for the electric field of a point charge.

<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>
when r>R, and R is the radius of the sphere.

However, when r<R, the observation location is inside of the sphere and the sphere overall must be thought of as infinitely many concentric charged shells inside of each other. All of the shells with a radius larger than that of the observation location do not contribute to the electric field at the observation location. This phenomena is because the electric field produced by these larger shells cancels out at any given point inside of itself. As a result only the shells smaller than the radius of the observation location need to be accounted for.
To find <math>\vec E_{net} </math>, add the contributions to the electric field from the inner shells.
After adding the contributions of each inner shell, you should have an electric field equal to:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{\Delta Q}{r^2}</math>

We find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

We find that:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

The charge inside the sphere is proportional to r. When r=R, we again treat the sphere as a point charge located at the center of the sphere.

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2} </math>

'''Looking Ahead, Gauss's Law'''

Later in this wikibook, you will learn about Gauss's Law. This will make calculating the electric field easier.

The formula for Gauss's Law is:

<math>\oint \vec E \bullet \hat n dA = \frac{1}{\epsilon_0}\Sigma Q_{\text {inside the surface}} </math>

Using this, one can calculate the the electric field when r<R for a solid sphere charged throughout.

<math>\vec E * 4\pi*r^2= \frac{1}{\epsilon_0}Q\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} </math>

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

===A Computational Model===

This demonstration using VPython shows the field of a charged ball with a fixed radius. Try changing the radius to see what happens!

https://trinket.io/glowscript/48f6efd07a

# GlowScript 1.1 VPython
## constants
oofpez = 9e9
qproton = 1.6e-19
scalefactor = 2e-20
Radius =3.5
r=3

## objects
particle = sphere(pos=vector(0,0,0), opacity =0.5, radius=Radius, color=color.blue)
## initial values
obslocation = vector(r,0,0)
##calculate position vector
rvector = obslocation - particle.pos
arrow1 = arrow(pos=particle.pos, axis=rvector, color=color.green)
rmag=mag(rvector)

##colored arrow

if r<= Radius:
E1= oofpez*(qproton)/(Radius)**3 *rmag* vector(1,0,0)
E2= oofpez*(qproton)/(Radius)**3 *rmag* vector(-1,0,0)
E3= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,1,0)
E4= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,-1,0)
E5= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,1)
E6= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,-1)
ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)
if r>Radius:
E1= oofpez*(qproton)/(rmag)**2 *vector(1,0,0)
E2= oofpez*(qproton)/(rmag)**2 *vector(-1,0,0)
E3= oofpez*(qproton)/(rmag)**2 * vector(0,1,0)
E4= oofpez*(qproton)/(rmag)**2 *vector(0,-1,0)
E5= oofpez*(qproton)/(rmag)**2 *vector(0,0,1)
E6= oofpez*(qproton)/(rmag)**2 * vector(0,0,-1)
ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)

==Examples==

===Simple===
[[File:Approx_point.png|200px|right|thumb|Figure 2: This is an example of applying the equation for finding the electric field of a point charge]]
1. Describe the pattern of the electric field of charged ball from far away.

ANS: If we observe a charged ball from far away, we can say that r>R. Because we are outside of the sphere, the ball can essentially be viewed as a point charge. Thus the electric field can be defined by:
<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>

===Middling===
[[File:Concentric_sphere.png|200px|right|thumb|Figure 3: This is very similar to problem 1 and involves calculating the field of a point charge.]]
A sphere is charged throughout it's volume with a charge of Q= 6e-5 C. The radius of the this sphere is R=10. Find the electric field created by a sphere of radius r=4.

Step 1: Cut up the sphere into shells.

step 2: We know that r<R.

Next find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r = 9e9 * \frac{6*10^{-5} C}{(10m)^3}*4m = 2160 N/C </math>

===Difficult===
A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge −e. (The actual charge density in the ground state is nonuniform.)

[[File:circle_outline.png|200px|thumb|left|Figure 4: This circle represents a hydrogen atom. Use the radius and total charge to find alpha ]]

For the uniform-density model, calculate the polarizability α of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud

(s « R in the diagram). (Use the following as necessary: R and ε0.)

Useful equations:

<math> \vec p = \alpha *E</math>

<math> \vec p = Q*s </math>

Assume an applied electric field of strength E. This electric field polarized the hydrogen atom. Now there is a spherical charge of radius s. Use the volume ratio, and then use the useful equations to find <math> \alpha </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{e}{s^2}\frac{\frac{4}{3} \pi s^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s </math>

<math>\alpha = \frac{e*s}{\frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s} = 4\pi \epsilon_0 R^3 </math>

==Connectedness==
It may be particularly useful to discuss real-life applications of a charged solid sphere. Two examples stem from the structure of an atom. The nucleus of an atom is packed very tightly so that we can consider the charge to be uniformly distributed. The electron cloud also can be viewed as a packed spherical region of charged. Of course the radii of these structures are very small; radii are about 10e-15 m and 10e-10 m for nuclei and electron clouds respectively!

==History==
ADD MORE ABOUT HISTORY HERE

== See also ==

===Further reading===

*http://www.physicsbook.gatech.edu/Point_Charge

*http://www.physicsbook.gatech.edu/Electric_Dipole

*http://www.physicsbook.gatech.edu/Gauss%27s_Flux_Theorem

===External Links===

==References==

*http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html

*https://www.physicsforums.com/threads/calculate-the-polarizability-a-lpha-of-atomic-hydrogen-in-terms-of-r.339994/

*Matter and Interactions, 4th Edition: 1-2

Charged Spherical Shell

2020-05-05T21:36:53Z

Sydney: /* Simple */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

===A Mathematical Model===

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

===A Computational Model===
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.

'''Observation Location Outside the Spherical Shell:'''

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

[[File:ShellOutside.JPG|400px|thumb|Figure 1: a gaussian surface outside of a spherical shell]]

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

'''Observation Location Inside the Hollow Portion of the Spherical Shell:'''

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because

[[File:ShellHollow2.JPG|400px|thumb|Figure 2: gaussian surface inside of a spherical shell]]

the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a<b

the charge enclosed by the gaussian surface is 0, <math>0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

'''Observation Location Within the Spherical Shell:'''

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.

[[File:ShellWithin.jpg|400px|thumb|Figure 3: gaussian surface inside spherical shell]]

If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Remember that the electric field is zero withing the hollow center of the shell:
<math>E = 0</math>.

</div></div>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

</div></div>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

</div></div>

'''Step Problem'''

A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

</div></div>

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

*[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

*[[Electric Field]] More general ideas about electric fields <br>

*[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

*Principles of Electrodynamics by Melvin Schwartz, ISBN: 9780486134673

===External links===

*http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

*http://www.hydroquebec.com/fields/corps-humain.html

*http://physicstasks.eu/1531/field-of-charged-spherical-shell

*https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

*Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Spherical Shell

2020-05-05T21:35:02Z

Sydney: /* Examples */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

===A Mathematical Model===

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

===A Computational Model===
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.

'''Observation Location Outside the Spherical Shell:'''

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

[[File:ShellOutside.JPG|400px|thumb|Figure 1: a gaussian surface outside of a spherical shell]]

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

'''Observation Location Inside the Hollow Portion of the Spherical Shell:'''

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because

[[File:ShellHollow2.JPG|400px|thumb|Figure 2: gaussian surface inside of a spherical shell]]

the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a<b

the charge enclosed by the gaussian surface is 0, <math>0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

'''Observation Location Within the Spherical Shell:'''

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.

[[File:ShellWithin.jpg|400px|thumb|Figure 3: gaussian surface inside spherical shell]]

If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

<math>E = 0</math>

</div></div>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

</div></div>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

</div></div>

'''Step Problem'''

A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

</div></div>

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

*[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

*[[Electric Field]] More general ideas about electric fields <br>

*[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

*Principles of Electrodynamics by Melvin Schwartz, ISBN: 9780486134673

===External links===

*http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

*http://www.hydroquebec.com/fields/corps-humain.html

*http://physicstasks.eu/1531/field-of-charged-spherical-shell

*https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

*Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Spherical Shell

2020-05-05T21:31:13Z

Sydney:

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

===A Mathematical Model===

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

===A Computational Model===
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.

'''Observation Location Outside the Spherical Shell:'''

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

[[File:ShellOutside.JPG|400px|thumb|Figure 1: a gaussian surface outside of a spherical shell]]

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

'''Observation Location Inside the Hollow Portion of the Spherical Shell:'''

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because

[[File:ShellHollow2.JPG|400px|thumb|Figure 2: gaussian surface inside of a spherical shell]]

the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a<b

the charge enclosed by the gaussian surface is 0, <math>0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

'''Observation Location Within the Spherical Shell:'''

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.

[[File:ShellWithin.jpg|400px|thumb|Figure 3: gaussian surface inside spherical shell]]

If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<math>E = 0</math>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

=== Step Problem===
A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

Solution:
(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

*[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

*[[Electric Field]] More general ideas about electric fields <br>

*[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

*Principles of Electrodynamics by Melvin Schwartz, ISBN: 9780486134673

===External links===

*http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

*http://www.hydroquebec.com/fields/corps-humain.html

*http://physicstasks.eu/1531/field-of-charged-spherical-shell

*https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

*Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Capacitor

2020-05-05T21:27:01Z

Sydney: /* History */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
<br>This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

===A Mathematical Model===

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
<br>The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
<br>

'''Derivation'''

[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
<br>Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
<br><br>Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
<br>If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
<br>

'''Gauss's Law'''

[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
<br>Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
<br>Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
<br>Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

===A Computational Model===

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
<br><br>
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capacitance is 3 F.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
<br><math>A = \frac{Cd}{\varepsilon_{0}}</math>
<br><math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
<br><br>

</div></div>

[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]

Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V</math>
<br><math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
<br>

</div></div>

===Middling===

[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
<br>To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
<br>To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.

</div></div>

[[File:qwertt.png|200px|thumb|Figure 6: Middling problem #2]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
<br>Since <math>C = \frac{Q}{V}</math>,
<br>the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
<br>and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
<br>The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>

</div></div>

===Difficult===

A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

[[File:zxcv.png|300px|thumb|right|Figure 7: Difficult problem]]
<br><br>
<br>First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
<br> Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
<br><math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
<br><math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
<br><br>Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
<br><math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

</div></div>

==Connectedness==
ADD MORE INFO ABOUT CONNECTEDNESS HERE

==History==
ADD MORE INFO ABOUT HISTORY HERE

==See also==

===Further reading===

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
<br><math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
<br><math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
<br><math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
<br><br>
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
<br><br>
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2020-05-05T21:26:32Z

Sydney: /* Connectedness */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
<br>This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

===A Mathematical Model===

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
<br>The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
<br>

'''Derivation'''

[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
<br>Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
<br><br>Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
<br>If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
<br>

'''Gauss's Law'''

[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
<br>Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
<br>Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
<br>Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

===A Computational Model===

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
<br><br>
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capacitance is 3 F.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
<br><math>A = \frac{Cd}{\varepsilon_{0}}</math>
<br><math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
<br><br>

</div></div>

[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]

Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V</math>
<br><math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
<br>

</div></div>

===Middling===

[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
<br>To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
<br>To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.

</div></div>

[[File:qwertt.png|200px|thumb|Figure 6: Middling problem #2]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
<br>Since <math>C = \frac{Q}{V}</math>,
<br>the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
<br>and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
<br>The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>

</div></div>

===Difficult===

A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

[[File:zxcv.png|300px|thumb|right|Figure 7: Difficult problem]]
<br><br>
<br>First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
<br> Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
<br><math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
<br><math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
<br><br>Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
<br><math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

</div></div>

==History==
ADD MORE INFO ABOUT HISTORY HERE

===Connectedness===
ADD MORE INFO ABOUT CONNECTEDNESS HERE

==See also==

===Further reading===

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
<br><math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
<br><math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
<br><math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
<br><br>
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
<br><br>
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2020-05-05T21:26:06Z

Sydney: /* Further reading */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
<br>This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

===A Mathematical Model===

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
<br>The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
<br>

'''Derivation'''

[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
<br>Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
<br><br>Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
<br>If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
<br>

'''Gauss's Law'''

[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
<br>Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
<br>Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
<br>Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

===A Computational Model===

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
<br><br>
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capacitance is 3 F.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
<br><math>A = \frac{Cd}{\varepsilon_{0}}</math>
<br><math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
<br><br>

</div></div>

[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]

Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V</math>
<br><math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
<br>

</div></div>

===Middling===

[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
<br>To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
<br>To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.

</div></div>

[[File:qwertt.png|200px|thumb|Figure 6: Middling problem #2]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
<br>Since <math>C = \frac{Q}{V}</math>,
<br>the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
<br>and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
<br>The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>

</div></div>

===Difficult===

A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

[[File:zxcv.png|300px|thumb|right|Figure 7: Difficult problem]]
<br><br>
<br>First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
<br> Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
<br><math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
<br><math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
<br><br>Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
<br><math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

</div></div>

==History==
ADD MORE INFO ABOUT HISTORY HERE

==Connectedness==
ADD MORE INFO ABOUT CONNECTEDNESS HERE

==See also==

===Further reading===

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
<br><math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
<br><math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
<br><math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
<br><br>
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
<br><br>
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2020-05-05T21:23:52Z

Sydney: /* Examples */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
<br>This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

===A Mathematical Model===

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
<br>The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
<br>

'''Derivation'''

[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
<br>Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
<br><br>Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
<br>If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
<br>

'''Gauss's Law'''

[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
<br>Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
<br>Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
<br>Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

===A Computational Model===

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
<br><br>
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capacitance is 3 F.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
<br><math>A = \frac{Cd}{\varepsilon_{0}}</math>
<br><math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
<br><br>

</div></div>

[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]

Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V</math>
<br><math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
<br>

</div></div>

===Middling===

[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
<br>To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
<br>To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.

</div></div>

[[File:qwertt.png|200px|thumb|Figure 6: Middling problem #2]]
Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
<br>Since <math>C = \frac{Q}{V}</math>,
<br>the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
<br>and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
<br>The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>

</div></div>

===Difficult===

A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

[[File:zxcv.png|300px|thumb|right|Figure 7: Difficult problem]]
<br><br>
<br>First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
<br> Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
<br><math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
<br><math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
<br><br>Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
<br><math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

</div></div>

==History==
ADD MORE INFO ABOUT HISTORY HERE

==Connectedness==
ADD MORE INFO ABOUT CONNECTEDNESS HERE

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
<br><math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
<br><math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
<br><math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
<br><br>
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
<br><br>
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2020-05-05T21:07:50Z

Sydney: /* A Computational Model */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
<br>This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

===A Mathematical Model===

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
<br>The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
<br>

'''Derivation'''

[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
<br>Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
<br><br>Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
<br>If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
<br>

'''Gauss's Law'''

[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
<br>Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
<br>Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
<br>Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

===A Computational Model===

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
<br><br>
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
<br>

<br><br>Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
<br><math>A = \frac{Cd}{\varepsilon_{0}}</math>
<br><math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
<br><br>
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]
<br><br>Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V</math>
<br><math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
<br>

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
<br>
[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
<br><br>The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
<br>To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
<br>To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
<br><br>
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png|200px|thumb|Figure 6: Middling problem #2]]
<br><br>SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
<br>Since <math>C = \frac{Q}{V}</math>,
<br>the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
<br>and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
<br>The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
<br>

===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
<br>
[[File:zxcv.png|300px|thumb|Figure 7: Difficult problem]]
<br><br>
<br>First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
<br> Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
<br><math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
<br><math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
<br><br>Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
<br><math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==
ADD MORE INFO ABOUT HISTORY HERE

==Connectedness==
ADD MORE INFO ABOUT CONNECTEDNESS HERE

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
<br><math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
<br><math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
<br><math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
<br><br>
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
<br><br>
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2020-05-05T21:07:22Z

Sydney: /* A Mathematical Model */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
<br>This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

===A Mathematical Model===

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
<br>The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
<br>

'''Derivation'''

[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
<br>Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
<br><br>Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
<br>If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
<br>

'''Gauss's Law'''

[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
<br>Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
<br>Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
<br>Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
<br><br>
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
<br>

<br><br>Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
<br><math>A = \frac{Cd}{\varepsilon_{0}}</math>
<br><math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
<br><br>
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]
<br><br>Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V</math>
<br><math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
<br>

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
<br>
[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
<br><br>The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
<br>To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
<br>To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
<br>Since <math>C = \frac{Q}{V}</math>,
<br><math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
<br><br>
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png|200px|thumb|Figure 6: Middling problem #2]]
<br><br>SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
<br>Since <math>C = \frac{Q}{V}</math>,
<br>the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
<br>and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
<br>The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
<br>

===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
<br>
[[File:zxcv.png|300px|thumb|Figure 7: Difficult problem]]
<br><br>
<br>First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
<br> Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
<br><math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
<br><math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
<br><br>Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
<br><math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==
ADD MORE INFO ABOUT HISTORY HERE

==Connectedness==
ADD MORE INFO ABOUT CONNECTEDNESS HERE

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
<br><math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
<br><math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
<br><math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
<br><br>
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
<br><br>
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Disk

2020-05-05T21:05:40Z

Sydney: /* Examples */

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

===A Mathematical Model===

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

===A Computational Model===

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Simple===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

</div></div>

===Middling===

[[File:circle_3.png|300px|thumb|Figure 3: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

</div></div>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

</div></div>

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

'''Charged Disks in Television'''

Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

'''Charged Disks in Chemical Engineering'''

The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

'''Industrial Applications'''

One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png|400px|center|thumb|Figure 5: a capacitor]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

*[[Capacitor]]

*[[Charged Ring]]

===Further reading===

*[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

===External links===

*[https://www.youtube.com/watch?v=B5UyXkFCg5s Video: Electric Field Due to a Charged Disk, Infinite Sheet of Charge, Parallel Plates - Physics Problems]

*[http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html Solved problems for the electric field of a ring, disk, and plane]

==References==

*[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

*[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

*[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

*Note: all images were created by author

[[Category:What category did you place this in?]]

Charged Disk

2020-05-05T21:00:42Z

Sydney: /* Connectedness */

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

===A Mathematical Model===

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

===A Computational Model===

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Simple===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 3: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

'''Charged Disks in Television'''

Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

'''Charged Disks in Chemical Engineering'''

The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

'''Industrial Applications'''

One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png|400px|center|thumb|Figure 5: a capacitor]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

*[[Capacitor]]

*[[Charged Ring]]

===Further reading===

*[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

===External links===

*[https://www.youtube.com/watch?v=B5UyXkFCg5s Video: Electric Field Due to a Charged Disk, Infinite Sheet of Charge, Parallel Plates - Physics Problems]

*[http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html Solved problems for the electric field of a ring, disk, and plane]

==References==

*[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

*[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

*[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

*Note: all images were created by author

[[Category:What category did you place this in?]]

Charged Disk

2020-05-05T21:00:10Z

Sydney: /* External links */

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

===A Mathematical Model===

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

===A Computational Model===

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Simple===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 3: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

'''Charged Disks in Television'''
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

'''Charged Disks in Chemical Engineering'''
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

'''Industrial Applications'''
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png|400px|center|thumb|Figure 5: a capacitor]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

*[[Capacitor]]

*[[Charged Ring]]

===Further reading===

*[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

===External links===

*[https://www.youtube.com/watch?v=B5UyXkFCg5s Video: Electric Field Due to a Charged Disk, Infinite Sheet of Charge, Parallel Plates - Physics Problems]

*[http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html Solved problems for the electric field of a ring, disk, and plane]

==References==

*[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

*[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

*[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

*Note: all images were created by author

[[Category:What category did you place this in?]]

Charged Disk

2020-05-05T20:55:07Z

Sydney: /* References */

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

===A Mathematical Model===

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

===A Computational Model===

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Simple===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 3: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

'''Charged Disks in Television'''
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

'''Charged Disks in Chemical Engineering'''
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

'''Industrial Applications'''
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png|400px|center|thumb|Figure 5: a capacitor]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

*[[Capacitor]]

*[[Charged Ring]]

===Further reading===

*[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

===External links===

*[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

*[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

*[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

==References==

*[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

*[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

*[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

*Note: all images were created by author

[[Category:What category did you place this in?]]

Charged Disk

2020-05-05T20:54:55Z

Sydney: /* External Links */

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

===A Mathematical Model===

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

===A Computational Model===

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Simple===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 3: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

'''Charged Disks in Television'''
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

'''Charged Disks in Chemical Engineering'''
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

'''Industrial Applications'''
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png|400px|center|thumb|Figure 5: a capacitor]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

*[[Capacitor]]

*[[Charged Ring]]

===Further reading===

*[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

===External links===

*[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

*[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

*[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

==References==

*Note: all images were created by author

[[Category:What category did you place this in?]]

Charged Disk

2020-05-05T20:54:26Z

Sydney: /* =A Mathematical Model */

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

===A Mathematical Model===

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

===A Computational Model===

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Simple===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 3: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

'''Charged Disks in Television'''
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

'''Charged Disks in Chemical Engineering'''
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

'''Industrial Applications'''
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png|400px|center|thumb|Figure 5: a capacitor]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

*[[Capacitor]]

*[[Charged Ring]]

===Further reading===

*[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

===External Links===

*[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

*[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

*[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

==References==

*Note: all images were created by author

[[Category:What category did you place this in?]]

Charged Disk

2020-05-05T20:54:03Z

Sydney:

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

===A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

===A Computational Model===

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Simple===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 3: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

'''Charged Disks in Television'''
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

'''Charged Disks in Chemical Engineering'''
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

'''Industrial Applications'''
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png|400px|center|thumb|Figure 5: a capacitor]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

*[[Capacitor]]

*[[Charged Ring]]

===Further reading===

*[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

===External Links===

*[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

*[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

*[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

==References==

*Note: all images were created by author

[[Category:What category did you place this in?]]

Charged Ring

2020-05-05T20:48:20Z

Sydney: /* Easy */

==The Main Idea==

Charges may be arranged in a variety of ways. When points are uniformly distributed a finite distance away from some point, which we define as the origin, we call this curve as a ring. We can position a [[Point Charge]] at every point upon this ring. To analyze this continuous arrangement of charges, we say that their individual contributions are infinitesimally small, and then sum each of these contributions together to arrive at the electric field produced by this charge distribution. We can compute the net electric field of this charge distribution with Coulomb's Law and by applying integration principles. The ring field can also be used to calculate the electric field of a [[Charged Disk]].

===A Mathematical Model===

This mathematical model is based on the individual [[Electric Field]] contributions of a number of point charges, each of which is defined by

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}</math>.

'''Do not forget this equation'''. This equation must be memorized as it isn't given on the formula sheet.

We also say that this vector field is a member of a linear space of vectors. This is to say that we can apply the [[Superposition Principle]] meaning that we can sum any number of these electric field vectors and obtain another vector which is the electric field vector contributed by those charges. When we continuously sum all of the vectors produced by these charges, we get the electric field produced by the entire arrangement of charges.

[[File:red_ring.png|400px|thumb|Figure 1: diagram of a charged ring]]

The diagram above shows the electric field due to one infinitesimal piece of the ring, <math>dq</math>. In order to avoid rigorous computations, we can see that the electric field of the charges cancels out in the vertical direction. Only the horizontal component will remain. For an observation location that is on the symmetry axis as the ring, the other two components will be zero. If the observation location were off axis, then it would be very different requiring more math. Since each piece, <math>dq</math>, contributes a <math>d\vec{E}</math> of <math>\frac{1}{4\pi\epsilon_{0}}\frac{dq}{|\vec{d}|^{2}}\hat{d}</math>, we can compute the horizontal component by multiplying the magnitude of <math>d\vec{E}</math> with <math>\cos{θ}</math>.

<math>|d\vec{E}_{z}| = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\cos{θ} = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\frac{z}{(r^2 + z^2)^\frac{1}{2}}</math>

Now we integrate to sum all the electric field contributions of each infinitesimal <math>dq</math>.

<math>\vec{E}_{z} = \frac{1}{4\pi\epsilon_{0}}\frac{z}{(r^2 + z^2)^\frac{3}{2}}\int_{}^{} dq = \frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(r^2 + z^2)^\frac{3}{2}}</math>

As a reminder, this is the equation given on the formula sheet. It '''only''' works when the location at which the field is being measured is along the <math>{z}</math> axis. However, the equation for it being off axis is not given on the equation sheet. That requires a separate and more difficult integration. Typically, the only way this would be asked on a test is to set up an equation that would be integrated that requires you to find a new <math>dQ</math>.

'''Rigorous Derivation'''

1. Define shape characteristics of the ring

:The ring has some finite charge.

:We see that this arrangement is circular, so a coordinate system with which we can define radial and angular coordinates would be useful. Naturally, this would be the polar coordinate system.

:We also see that all charge is uniformly distributed some finite distance R from the center of the ring. It would be useful to let the center of the ring be the origin of our coordinate axes.

:Here is a reasonable arrangement for this charged ring.

[[File:blue_graph.png|400px|thumb|Figure 2: defining a coordinate system about a charged ring]]

:Since all charge is concentrated upon the edge of the circle, we can consider our charge distribution to be invariant with respect to radial distance. However, we do see that our charge distribution is the function of theta.

2. Compute charge distribution function

:Let us call the charge distribution as <math display="inline">\sigma</math>

:We have a charge distributed on the edge of the circle, so

:<math> \sigma=\frac{Q}{2\pi} </math>

:where <math display="inline">Q</math> represents the charge of the ring.

:What this equation means is that the charge is uniformly distributed along each unit of angular length. In this way, it is called angular charge density.

3. Compute infinitesimal charge contribution

:Let us consider an infinitesimal section of the ring which contains exactly one point charge. The dimension of this section is given by <math display="inline">d\theta</math> which is the infinitesimal angular size. So, the infinitesimal charge contribution, <math display="inline">dQ</math>, is

:<math> dQ = \frac{Q}{2\pi}d\theta </math>

4. Compute infinitesimal electric field contribution

:Let us define some arbitrary location at which we are observing this ring of charge.

:<math> \vec{r} = x\hat{x}+y\hat{y} + z\hat{z} </math>

:in polar coordinates, we see this becomes

:<math> \vec{r} = Rcos(\theta)\hat{x} + Rsin(\theta)\hat{y} + z\hat{z} </math>.

:Recall that both sine and cosine are periodic with the period of <math>2\pi </math>. This will become important later.

:The magnitude of this vector is

:<math> |\vec{r}| = \sqrt{R^{2}+z^{2}} </math>

:owing to the usage of the Pythagorean trigonometric identity.

:Now, we have what we need to write the electric field vector contributed by each piece of the ring of charge. Let this vector field piece be <math display="inline">d\vec{E}</math>.

:<math> d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z}) </math>

5. Compute electric field vector

:So, now all that is left is to sum everything up. We are summing over the circumference of a circle, so our path is defined by <math display="inline">0\leq\theta\leq 2\pi </math>. Let us now set up our integral.

:<math>\vec{E}=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

:Since both sine and cosine are <math display="inline">2\pi</math> periodic functions, the <math display="inline">x</math> and <math display="inline">y</math> components of <math display="inline">\vec{E}</math> go to <math display="inline">\vec{0}</math>, which is very convenient. The result is

:<math> \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(z^{2}+R^{2})^{3/2}}\hat{z} </math>.

''Remark''

If the path we are interested in is over a half circle, third-circle or quarter-circle, or indeed any circular section, simply adjust the bounds of integration and the charge density to the ones required by the curve of interest.

===A Computational Model===

[https://trinket.io/glowscript/707d492e19 This simulation] shows the result of the computation for a ring composed of 2000 electrons. This is why the vector is pointing into the ring rather than out of the ring, which would happen for a ring composed of positively charged points.

==Examples==

===Simple===

1.) Find the function <math>E(x)</math> that represents the magnitude of the electric field along the center axis due to a uniformly charged ring of radius 0.04 m with total charge 8 C.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can use the derived formula for the electric field due to a charged ring. Only the horizontal components of the electric field will remain as the rest will cancel out.

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{(8 C)*x}{((0.04 m)^2 + x^2)^\frac{3}{2}}</math>

</div></div>

===Middling===

2.) A uniformly charged ring of radius 10.0 cm has a total charge of 91.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

a. 1 cm

b. 5 cm

c. 30 cm

d. 100 cm

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Use the derived formula, and convert the given distances into meters and the charge into Coulombs. Plug these into <math>x</math>

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{Q*x}{(x^2 + R^2)^\frac{3}{2}}</math>

a. 8.069e6 N/C

b. 29.3e6 N/C

c. 7.76e6 N/C

d. 8.06e5 N/C

</div></div>

===Difficult===

Find the electric field at a distance <math>z</math> along the center axis away from a uniformly charged semicircular ring of radius R and total charge Q.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can tweak the same integral for a uniformly charged ring by changing the limits of integration.

<math>\vec{E}=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(0\hat{x}+2R\hat{y}+{\pi}z\hat{z})</math>

</div></div>

==Connectedness==

Edited by AYESHA AHUJA Spring 2020

The idea of charge density is somewhat analogous to the idea of the mass density, which is useful in a variety of contexts, including the computation of the center of mass, the first and second moments of mass, which are useful in statics and rigid body dynamics. Instead of computing the uniform distribution of the mass of this ring, we are computing the uniform distribution of charge. This concept is also useful in visualizing what happens within a wire in a steady state circuit. The wire can be viewed to be a continuous length of rings of charge, which act as a channel through which electrons are transported, and that the electric field of these rings of charge pushes the electrons within the wire.

It is also important in the context of Maxwell's equations, specifically in Gauss's Law and in the Maxwell-Faraday equation, which are concerned with electron flux and induced fields, respectively.

==History==

Added by AYESHA AHUJA SPRING 2020

The first known time when charged rings were utilized to describe electromagnetism was during Faraday’s discovery of electrical induction. To conduct his experiments, Faraday used a ring made of iron that was attached to a battery. When this makeshift circuit was turned on it manipulated a compass needle by deflecting it. He showcased an induced current in the ring.

== See also ==

===Further reading===

*[[Charged Rod]]

*[[Charged Disk]]

*[[Charged Spherical Shell]]

*[[Charged Capacitor]]

===External links===

*https://www.youtube.com/watch?v=80mM3kSTZcE

*http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

*http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

==References==

*"Electric Field on the Axis of a Ring of Charge". University of Delaware Physics Library. Adapted from Stephen Kevan's lecture on Electric Fields and Charge Distribution. April 8, 1996. http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

*Chabay, R., & Sherwood, B. (2015). Matter and Interactions (4th ed., Vol. 2, pp. 597-599). Wiley.

*All images produced by the author

[[Category:Electric Fields]]

Charged Ring

2020-05-05T20:47:53Z

Sydney: /* See also */

==The Main Idea==

Charges may be arranged in a variety of ways. When points are uniformly distributed a finite distance away from some point, which we define as the origin, we call this curve as a ring. We can position a [[Point Charge]] at every point upon this ring. To analyze this continuous arrangement of charges, we say that their individual contributions are infinitesimally small, and then sum each of these contributions together to arrive at the electric field produced by this charge distribution. We can compute the net electric field of this charge distribution with Coulomb's Law and by applying integration principles. The ring field can also be used to calculate the electric field of a [[Charged Disk]].

===A Mathematical Model===

This mathematical model is based on the individual [[Electric Field]] contributions of a number of point charges, each of which is defined by

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}</math>.

'''Do not forget this equation'''. This equation must be memorized as it isn't given on the formula sheet.

We also say that this vector field is a member of a linear space of vectors. This is to say that we can apply the [[Superposition Principle]] meaning that we can sum any number of these electric field vectors and obtain another vector which is the electric field vector contributed by those charges. When we continuously sum all of the vectors produced by these charges, we get the electric field produced by the entire arrangement of charges.

[[File:red_ring.png|400px|thumb|Figure 1: diagram of a charged ring]]

The diagram above shows the electric field due to one infinitesimal piece of the ring, <math>dq</math>. In order to avoid rigorous computations, we can see that the electric field of the charges cancels out in the vertical direction. Only the horizontal component will remain. For an observation location that is on the symmetry axis as the ring, the other two components will be zero. If the observation location were off axis, then it would be very different requiring more math. Since each piece, <math>dq</math>, contributes a <math>d\vec{E}</math> of <math>\frac{1}{4\pi\epsilon_{0}}\frac{dq}{|\vec{d}|^{2}}\hat{d}</math>, we can compute the horizontal component by multiplying the magnitude of <math>d\vec{E}</math> with <math>\cos{θ}</math>.

<math>|d\vec{E}_{z}| = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\cos{θ} = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\frac{z}{(r^2 + z^2)^\frac{1}{2}}</math>

Now we integrate to sum all the electric field contributions of each infinitesimal <math>dq</math>.

<math>\vec{E}_{z} = \frac{1}{4\pi\epsilon_{0}}\frac{z}{(r^2 + z^2)^\frac{3}{2}}\int_{}^{} dq = \frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(r^2 + z^2)^\frac{3}{2}}</math>

As a reminder, this is the equation given on the formula sheet. It '''only''' works when the location at which the field is being measured is along the <math>{z}</math> axis. However, the equation for it being off axis is not given on the equation sheet. That requires a separate and more difficult integration. Typically, the only way this would be asked on a test is to set up an equation that would be integrated that requires you to find a new <math>dQ</math>.

'''Rigorous Derivation'''

1. Define shape characteristics of the ring

:The ring has some finite charge.

:We see that this arrangement is circular, so a coordinate system with which we can define radial and angular coordinates would be useful. Naturally, this would be the polar coordinate system.

:We also see that all charge is uniformly distributed some finite distance R from the center of the ring. It would be useful to let the center of the ring be the origin of our coordinate axes.

:Here is a reasonable arrangement for this charged ring.

[[File:blue_graph.png|400px|thumb|Figure 2: defining a coordinate system about a charged ring]]

:Since all charge is concentrated upon the edge of the circle, we can consider our charge distribution to be invariant with respect to radial distance. However, we do see that our charge distribution is the function of theta.

2. Compute charge distribution function

:Let us call the charge distribution as <math display="inline">\sigma</math>

:We have a charge distributed on the edge of the circle, so

:<math> \sigma=\frac{Q}{2\pi} </math>

:where <math display="inline">Q</math> represents the charge of the ring.

:What this equation means is that the charge is uniformly distributed along each unit of angular length. In this way, it is called angular charge density.

3. Compute infinitesimal charge contribution

:Let us consider an infinitesimal section of the ring which contains exactly one point charge. The dimension of this section is given by <math display="inline">d\theta</math> which is the infinitesimal angular size. So, the infinitesimal charge contribution, <math display="inline">dQ</math>, is

:<math> dQ = \frac{Q}{2\pi}d\theta </math>

4. Compute infinitesimal electric field contribution

:Let us define some arbitrary location at which we are observing this ring of charge.

:<math> \vec{r} = x\hat{x}+y\hat{y} + z\hat{z} </math>

:in polar coordinates, we see this becomes

:<math> \vec{r} = Rcos(\theta)\hat{x} + Rsin(\theta)\hat{y} + z\hat{z} </math>.

:Recall that both sine and cosine are periodic with the period of <math>2\pi </math>. This will become important later.

:The magnitude of this vector is

:<math> |\vec{r}| = \sqrt{R^{2}+z^{2}} </math>

:owing to the usage of the Pythagorean trigonometric identity.

:Now, we have what we need to write the electric field vector contributed by each piece of the ring of charge. Let this vector field piece be <math display="inline">d\vec{E}</math>.

:<math> d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z}) </math>

5. Compute electric field vector

:So, now all that is left is to sum everything up. We are summing over the circumference of a circle, so our path is defined by <math display="inline">0\leq\theta\leq 2\pi </math>. Let us now set up our integral.

:<math>\vec{E}=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

:Since both sine and cosine are <math display="inline">2\pi</math> periodic functions, the <math display="inline">x</math> and <math display="inline">y</math> components of <math display="inline">\vec{E}</math> go to <math display="inline">\vec{0}</math>, which is very convenient. The result is

:<math> \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(z^{2}+R^{2})^{3/2}}\hat{z} </math>.

''Remark''

If the path we are interested in is over a half circle, third-circle or quarter-circle, or indeed any circular section, simply adjust the bounds of integration and the charge density to the ones required by the curve of interest.

===A Computational Model===

[https://trinket.io/glowscript/707d492e19 This simulation] shows the result of the computation for a ring composed of 2000 electrons. This is why the vector is pointing into the ring rather than out of the ring, which would happen for a ring composed of positively charged points.

==Examples==

===Easy===

1.) Find the function <math>E(x)</math> that represents the magnitude of the electric field along the center axis due to a uniformly charged ring of radius 0.04 m with total charge 8 C.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can use the derived formula for the electric field due to a charged ring. Only the horizontal components of the electric field will remain as the rest will cancel out.

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{(8 C)*x}{((0.04 m)^2 + x^2)^\frac{3}{2}}</math>

</div></div>

===Middling===

2.) A uniformly charged ring of radius 10.0 cm has a total charge of 91.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

a. 1 cm

b. 5 cm

c. 30 cm

d. 100 cm

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Use the derived formula, and convert the given distances into meters and the charge into Coulombs. Plug these into <math>x</math>

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{Q*x}{(x^2 + R^2)^\frac{3}{2}}</math>

a. 8.069e6 N/C

b. 29.3e6 N/C

c. 7.76e6 N/C

d. 8.06e5 N/C

</div></div>

===Difficult===

Find the electric field at a distance <math>z</math> along the center axis away from a uniformly charged semicircular ring of radius R and total charge Q.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can tweak the same integral for a uniformly charged ring by changing the limits of integration.

<math>\vec{E}=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(0\hat{x}+2R\hat{y}+{\pi}z\hat{z})</math>

</div></div>

==Connectedness==

Edited by AYESHA AHUJA Spring 2020

The idea of charge density is somewhat analogous to the idea of the mass density, which is useful in a variety of contexts, including the computation of the center of mass, the first and second moments of mass, which are useful in statics and rigid body dynamics. Instead of computing the uniform distribution of the mass of this ring, we are computing the uniform distribution of charge. This concept is also useful in visualizing what happens within a wire in a steady state circuit. The wire can be viewed to be a continuous length of rings of charge, which act as a channel through which electrons are transported, and that the electric field of these rings of charge pushes the electrons within the wire.

It is also important in the context of Maxwell's equations, specifically in Gauss's Law and in the Maxwell-Faraday equation, which are concerned with electron flux and induced fields, respectively.

==History==

Added by AYESHA AHUJA SPRING 2020

The first known time when charged rings were utilized to describe electromagnetism was during Faraday’s discovery of electrical induction. To conduct his experiments, Faraday used a ring made of iron that was attached to a battery. When this makeshift circuit was turned on it manipulated a compass needle by deflecting it. He showcased an induced current in the ring.

== See also ==

===Further reading===

*[[Charged Rod]]

*[[Charged Disk]]

*[[Charged Spherical Shell]]

*[[Charged Capacitor]]

===External links===

*https://www.youtube.com/watch?v=80mM3kSTZcE

*http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

*http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

==References==

*"Electric Field on the Axis of a Ring of Charge". University of Delaware Physics Library. Adapted from Stephen Kevan's lecture on Electric Fields and Charge Distribution. April 8, 1996. http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

*Chabay, R., & Sherwood, B. (2015). Matter and Interactions (4th ed., Vol. 2, pp. 597-599). Wiley.

*All images produced by the author

[[Category:Electric Fields]]

Charged Ring

2020-05-05T20:47:15Z

Sydney: /* See also */

==The Main Idea==

Charges may be arranged in a variety of ways. When points are uniformly distributed a finite distance away from some point, which we define as the origin, we call this curve as a ring. We can position a [[Point Charge]] at every point upon this ring. To analyze this continuous arrangement of charges, we say that their individual contributions are infinitesimally small, and then sum each of these contributions together to arrive at the electric field produced by this charge distribution. We can compute the net electric field of this charge distribution with Coulomb's Law and by applying integration principles. The ring field can also be used to calculate the electric field of a [[Charged Disk]].

===A Mathematical Model===

This mathematical model is based on the individual [[Electric Field]] contributions of a number of point charges, each of which is defined by

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}</math>.

'''Do not forget this equation'''. This equation must be memorized as it isn't given on the formula sheet.

We also say that this vector field is a member of a linear space of vectors. This is to say that we can apply the [[Superposition Principle]] meaning that we can sum any number of these electric field vectors and obtain another vector which is the electric field vector contributed by those charges. When we continuously sum all of the vectors produced by these charges, we get the electric field produced by the entire arrangement of charges.

[[File:red_ring.png|400px|thumb|Figure 1: diagram of a charged ring]]

The diagram above shows the electric field due to one infinitesimal piece of the ring, <math>dq</math>. In order to avoid rigorous computations, we can see that the electric field of the charges cancels out in the vertical direction. Only the horizontal component will remain. For an observation location that is on the symmetry axis as the ring, the other two components will be zero. If the observation location were off axis, then it would be very different requiring more math. Since each piece, <math>dq</math>, contributes a <math>d\vec{E}</math> of <math>\frac{1}{4\pi\epsilon_{0}}\frac{dq}{|\vec{d}|^{2}}\hat{d}</math>, we can compute the horizontal component by multiplying the magnitude of <math>d\vec{E}</math> with <math>\cos{θ}</math>.

<math>|d\vec{E}_{z}| = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\cos{θ} = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\frac{z}{(r^2 + z^2)^\frac{1}{2}}</math>

Now we integrate to sum all the electric field contributions of each infinitesimal <math>dq</math>.

<math>\vec{E}_{z} = \frac{1}{4\pi\epsilon_{0}}\frac{z}{(r^2 + z^2)^\frac{3}{2}}\int_{}^{} dq = \frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(r^2 + z^2)^\frac{3}{2}}</math>

As a reminder, this is the equation given on the formula sheet. It '''only''' works when the location at which the field is being measured is along the <math>{z}</math> axis. However, the equation for it being off axis is not given on the equation sheet. That requires a separate and more difficult integration. Typically, the only way this would be asked on a test is to set up an equation that would be integrated that requires you to find a new <math>dQ</math>.

'''Rigorous Derivation'''

1. Define shape characteristics of the ring

:The ring has some finite charge.

:We see that this arrangement is circular, so a coordinate system with which we can define radial and angular coordinates would be useful. Naturally, this would be the polar coordinate system.

:We also see that all charge is uniformly distributed some finite distance R from the center of the ring. It would be useful to let the center of the ring be the origin of our coordinate axes.

:Here is a reasonable arrangement for this charged ring.

[[File:blue_graph.png|400px|thumb|Figure 2: defining a coordinate system about a charged ring]]

:Since all charge is concentrated upon the edge of the circle, we can consider our charge distribution to be invariant with respect to radial distance. However, we do see that our charge distribution is the function of theta.

2. Compute charge distribution function

:Let us call the charge distribution as <math display="inline">\sigma</math>

:We have a charge distributed on the edge of the circle, so

:<math> \sigma=\frac{Q}{2\pi} </math>

:where <math display="inline">Q</math> represents the charge of the ring.

:What this equation means is that the charge is uniformly distributed along each unit of angular length. In this way, it is called angular charge density.

3. Compute infinitesimal charge contribution

:Let us consider an infinitesimal section of the ring which contains exactly one point charge. The dimension of this section is given by <math display="inline">d\theta</math> which is the infinitesimal angular size. So, the infinitesimal charge contribution, <math display="inline">dQ</math>, is

:<math> dQ = \frac{Q}{2\pi}d\theta </math>

4. Compute infinitesimal electric field contribution

:Let us define some arbitrary location at which we are observing this ring of charge.

:<math> \vec{r} = x\hat{x}+y\hat{y} + z\hat{z} </math>

:in polar coordinates, we see this becomes

:<math> \vec{r} = Rcos(\theta)\hat{x} + Rsin(\theta)\hat{y} + z\hat{z} </math>.

:Recall that both sine and cosine are periodic with the period of <math>2\pi </math>. This will become important later.

:The magnitude of this vector is

:<math> |\vec{r}| = \sqrt{R^{2}+z^{2}} </math>

:owing to the usage of the Pythagorean trigonometric identity.

:Now, we have what we need to write the electric field vector contributed by each piece of the ring of charge. Let this vector field piece be <math display="inline">d\vec{E}</math>.

:<math> d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z}) </math>

5. Compute electric field vector

:So, now all that is left is to sum everything up. We are summing over the circumference of a circle, so our path is defined by <math display="inline">0\leq\theta\leq 2\pi </math>. Let us now set up our integral.

:<math>\vec{E}=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

:Since both sine and cosine are <math display="inline">2\pi</math> periodic functions, the <math display="inline">x</math> and <math display="inline">y</math> components of <math display="inline">\vec{E}</math> go to <math display="inline">\vec{0}</math>, which is very convenient. The result is

:<math> \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(z^{2}+R^{2})^{3/2}}\hat{z} </math>.

''Remark''

If the path we are interested in is over a half circle, third-circle or quarter-circle, or indeed any circular section, simply adjust the bounds of integration and the charge density to the ones required by the curve of interest.

===A Computational Model===

[https://trinket.io/glowscript/707d492e19 This simulation] shows the result of the computation for a ring composed of 2000 electrons. This is why the vector is pointing into the ring rather than out of the ring, which would happen for a ring composed of positively charged points.

==Examples==

===Easy===

1.) Find the function <math>E(x)</math> that represents the magnitude of the electric field along the center axis due to a uniformly charged ring of radius 0.04 m with total charge 8 C.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can use the derived formula for the electric field due to a charged ring. Only the horizontal components of the electric field will remain as the rest will cancel out.

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{(8 C)*x}{((0.04 m)^2 + x^2)^\frac{3}{2}}</math>

</div></div>

===Middling===

2.) A uniformly charged ring of radius 10.0 cm has a total charge of 91.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

a. 1 cm

b. 5 cm

c. 30 cm

d. 100 cm

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Use the derived formula, and convert the given distances into meters and the charge into Coulombs. Plug these into <math>x</math>

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{Q*x}{(x^2 + R^2)^\frac{3}{2}}</math>

a. 8.069e6 N/C

b. 29.3e6 N/C

c. 7.76e6 N/C

d. 8.06e5 N/C

</div></div>

===Difficult===

Find the electric field at a distance <math>z</math> along the center axis away from a uniformly charged semicircular ring of radius R and total charge Q.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can tweak the same integral for a uniformly charged ring by changing the limits of integration.

<math>\vec{E}=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(0\hat{x}+2R\hat{y}+{\pi}z\hat{z})</math>

</div></div>

==Connectedness==

Edited by AYESHA AHUJA Spring 2020

The idea of charge density is somewhat analogous to the idea of the mass density, which is useful in a variety of contexts, including the computation of the center of mass, the first and second moments of mass, which are useful in statics and rigid body dynamics. Instead of computing the uniform distribution of the mass of this ring, we are computing the uniform distribution of charge. This concept is also useful in visualizing what happens within a wire in a steady state circuit. The wire can be viewed to be a continuous length of rings of charge, which act as a channel through which electrons are transported, and that the electric field of these rings of charge pushes the electrons within the wire.

It is also important in the context of Maxwell's equations, specifically in Gauss's Law and in the Maxwell-Faraday equation, which are concerned with electron flux and induced fields, respectively.

==History==

Added by AYESHA AHUJA SPRING 2020

The first known time when charged rings were utilized to describe electromagnetism was during Faraday’s discovery of electrical induction. To conduct his experiments, Faraday used a ring made of iron that was attached to a battery. When this makeshift circuit was turned on it manipulated a compass needle by deflecting it. He showcased an induced current in the ring.

== See also ==

===Further Reading===

*[[Charged Rod]]

*[[Charged Disk]]

*[[Charged Spherical Shell]]

*[[Charged Capacitor]]

===External Links===

*https://www.youtube.com/watch?v=80mM3kSTZcE

*http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

*http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

==References==

*"Electric Field on the Axis of a Ring of Charge". University of Delaware Physics Library. Adapted from Stephen Kevan's lecture on Electric Fields and Charge Distribution. April 8, 1996. http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

*Chabay, R., & Sherwood, B. (2015). Matter and Interactions (4th ed., Vol. 2, pp. 597-599). Wiley.

*All images produced by the author

[[Category:Electric Fields]]

Charged Ring

2020-05-05T20:45:32Z

Sydney: /* External Links */

==The Main Idea==

Charges may be arranged in a variety of ways. When points are uniformly distributed a finite distance away from some point, which we define as the origin, we call this curve as a ring. We can position a [[Point Charge]] at every point upon this ring. To analyze this continuous arrangement of charges, we say that their individual contributions are infinitesimally small, and then sum each of these contributions together to arrive at the electric field produced by this charge distribution. We can compute the net electric field of this charge distribution with Coulomb's Law and by applying integration principles. The ring field can also be used to calculate the electric field of a [[Charged Disk]].

===A Mathematical Model===

This mathematical model is based on the individual [[Electric Field]] contributions of a number of point charges, each of which is defined by

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}</math>.

'''Do not forget this equation'''. This equation must be memorized as it isn't given on the formula sheet.

We also say that this vector field is a member of a linear space of vectors. This is to say that we can apply the [[Superposition Principle]] meaning that we can sum any number of these electric field vectors and obtain another vector which is the electric field vector contributed by those charges. When we continuously sum all of the vectors produced by these charges, we get the electric field produced by the entire arrangement of charges.

[[File:red_ring.png|400px|thumb|Figure 1: diagram of a charged ring]]

The diagram above shows the electric field due to one infinitesimal piece of the ring, <math>dq</math>. In order to avoid rigorous computations, we can see that the electric field of the charges cancels out in the vertical direction. Only the horizontal component will remain. For an observation location that is on the symmetry axis as the ring, the other two components will be zero. If the observation location were off axis, then it would be very different requiring more math. Since each piece, <math>dq</math>, contributes a <math>d\vec{E}</math> of <math>\frac{1}{4\pi\epsilon_{0}}\frac{dq}{|\vec{d}|^{2}}\hat{d}</math>, we can compute the horizontal component by multiplying the magnitude of <math>d\vec{E}</math> with <math>\cos{θ}</math>.

<math>|d\vec{E}_{z}| = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\cos{θ} = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\frac{z}{(r^2 + z^2)^\frac{1}{2}}</math>

Now we integrate to sum all the electric field contributions of each infinitesimal <math>dq</math>.

<math>\vec{E}_{z} = \frac{1}{4\pi\epsilon_{0}}\frac{z}{(r^2 + z^2)^\frac{3}{2}}\int_{}^{} dq = \frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(r^2 + z^2)^\frac{3}{2}}</math>

As a reminder, this is the equation given on the formula sheet. It '''only''' works when the location at which the field is being measured is along the <math>{z}</math> axis. However, the equation for it being off axis is not given on the equation sheet. That requires a separate and more difficult integration. Typically, the only way this would be asked on a test is to set up an equation that would be integrated that requires you to find a new <math>dQ</math>.

'''Rigorous Derivation'''

1. Define shape characteristics of the ring

:The ring has some finite charge.

:We see that this arrangement is circular, so a coordinate system with which we can define radial and angular coordinates would be useful. Naturally, this would be the polar coordinate system.

:We also see that all charge is uniformly distributed some finite distance R from the center of the ring. It would be useful to let the center of the ring be the origin of our coordinate axes.

:Here is a reasonable arrangement for this charged ring.

[[File:blue_graph.png|400px|thumb|Figure 2: defining a coordinate system about a charged ring]]

:Since all charge is concentrated upon the edge of the circle, we can consider our charge distribution to be invariant with respect to radial distance. However, we do see that our charge distribution is the function of theta.

2. Compute charge distribution function

:Let us call the charge distribution as <math display="inline">\sigma</math>

:We have a charge distributed on the edge of the circle, so

:<math> \sigma=\frac{Q}{2\pi} </math>

:where <math display="inline">Q</math> represents the charge of the ring.

:What this equation means is that the charge is uniformly distributed along each unit of angular length. In this way, it is called angular charge density.

3. Compute infinitesimal charge contribution

:Let us consider an infinitesimal section of the ring which contains exactly one point charge. The dimension of this section is given by <math display="inline">d\theta</math> which is the infinitesimal angular size. So, the infinitesimal charge contribution, <math display="inline">dQ</math>, is

:<math> dQ = \frac{Q}{2\pi}d\theta </math>

4. Compute infinitesimal electric field contribution

:Let us define some arbitrary location at which we are observing this ring of charge.

:<math> \vec{r} = x\hat{x}+y\hat{y} + z\hat{z} </math>

:in polar coordinates, we see this becomes

:<math> \vec{r} = Rcos(\theta)\hat{x} + Rsin(\theta)\hat{y} + z\hat{z} </math>.

:Recall that both sine and cosine are periodic with the period of <math>2\pi </math>. This will become important later.

:The magnitude of this vector is

:<math> |\vec{r}| = \sqrt{R^{2}+z^{2}} </math>

:owing to the usage of the Pythagorean trigonometric identity.

:Now, we have what we need to write the electric field vector contributed by each piece of the ring of charge. Let this vector field piece be <math display="inline">d\vec{E}</math>.

:<math> d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z}) </math>

5. Compute electric field vector

:So, now all that is left is to sum everything up. We are summing over the circumference of a circle, so our path is defined by <math display="inline">0\leq\theta\leq 2\pi </math>. Let us now set up our integral.

:<math>\vec{E}=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

:Since both sine and cosine are <math display="inline">2\pi</math> periodic functions, the <math display="inline">x</math> and <math display="inline">y</math> components of <math display="inline">\vec{E}</math> go to <math display="inline">\vec{0}</math>, which is very convenient. The result is

:<math> \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(z^{2}+R^{2})^{3/2}}\hat{z} </math>.

''Remark''

If the path we are interested in is over a half circle, third-circle or quarter-circle, or indeed any circular section, simply adjust the bounds of integration and the charge density to the ones required by the curve of interest.

===A Computational Model===

[https://trinket.io/glowscript/707d492e19 This simulation] shows the result of the computation for a ring composed of 2000 electrons. This is why the vector is pointing into the ring rather than out of the ring, which would happen for a ring composed of positively charged points.

==Examples==

===Easy===

1.) Find the function <math>E(x)</math> that represents the magnitude of the electric field along the center axis due to a uniformly charged ring of radius 0.04 m with total charge 8 C.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can use the derived formula for the electric field due to a charged ring. Only the horizontal components of the electric field will remain as the rest will cancel out.

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{(8 C)*x}{((0.04 m)^2 + x^2)^\frac{3}{2}}</math>

</div></div>

===Middling===

2.) A uniformly charged ring of radius 10.0 cm has a total charge of 91.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

a. 1 cm

b. 5 cm

c. 30 cm

d. 100 cm

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Use the derived formula, and convert the given distances into meters and the charge into Coulombs. Plug these into <math>x</math>

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{Q*x}{(x^2 + R^2)^\frac{3}{2}}</math>

a. 8.069e6 N/C

b. 29.3e6 N/C

c. 7.76e6 N/C

d. 8.06e5 N/C

</div></div>

===Difficult===

Find the electric field at a distance <math>z</math> along the center axis away from a uniformly charged semicircular ring of radius R and total charge Q.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can tweak the same integral for a uniformly charged ring by changing the limits of integration.

<math>\vec{E}=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(0\hat{x}+2R\hat{y}+{\pi}z\hat{z})</math>

</div></div>

==Connectedness==

Edited by AYESHA AHUJA Spring 2020

The idea of charge density is somewhat analogous to the idea of the mass density, which is useful in a variety of contexts, including the computation of the center of mass, the first and second moments of mass, which are useful in statics and rigid body dynamics. Instead of computing the uniform distribution of the mass of this ring, we are computing the uniform distribution of charge. This concept is also useful in visualizing what happens within a wire in a steady state circuit. The wire can be viewed to be a continuous length of rings of charge, which act as a channel through which electrons are transported, and that the electric field of these rings of charge pushes the electrons within the wire.

It is also important in the context of Maxwell's equations, specifically in Gauss's Law and in the Maxwell-Faraday equation, which are concerned with electron flux and induced fields, respectively.

==History==

Added by AYESHA AHUJA SPRING 2020

The first known time when charged rings were utilized to describe electromagnetism was during Faraday’s discovery of electrical induction. To conduct his experiments, Faraday used a ring made of iron that was attached to a battery. When this makeshift circuit was turned on it manipulated a compass needle by deflecting it. He showcased an induced current in the ring.

== See also ==

[[Charged Rod]]

[[Charged Disk]]

[[Charged Spherical Shell]]

[[Charged Capacitor]]

===References===

"Electric Field on the Axis of a Ring of Charge". University of Delaware Physics Library. Adapted from Stephen Kevan's lecture on Electric Fields and Charge Distribution. April 8, 1996. http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

Chabay, R., & Sherwood, B. (2015). Matter and Interactions (4th ed., Vol. 2, pp. 597-599). Wiley.

All images produced by the author

===External Links===

https://www.youtube.com/watch?v=80mM3kSTZcE

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

[[Category:Electric Fields]]

Charged Ring

2020-05-05T20:44:47Z

Sydney: /* References */

==The Main Idea==

Charges may be arranged in a variety of ways. When points are uniformly distributed a finite distance away from some point, which we define as the origin, we call this curve as a ring. We can position a [[Point Charge]] at every point upon this ring. To analyze this continuous arrangement of charges, we say that their individual contributions are infinitesimally small, and then sum each of these contributions together to arrive at the electric field produced by this charge distribution. We can compute the net electric field of this charge distribution with Coulomb's Law and by applying integration principles. The ring field can also be used to calculate the electric field of a [[Charged Disk]].

===A Mathematical Model===

This mathematical model is based on the individual [[Electric Field]] contributions of a number of point charges, each of which is defined by

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}</math>.

'''Do not forget this equation'''. This equation must be memorized as it isn't given on the formula sheet.

We also say that this vector field is a member of a linear space of vectors. This is to say that we can apply the [[Superposition Principle]] meaning that we can sum any number of these electric field vectors and obtain another vector which is the electric field vector contributed by those charges. When we continuously sum all of the vectors produced by these charges, we get the electric field produced by the entire arrangement of charges.

[[File:red_ring.png|400px|thumb|Figure 1: diagram of a charged ring]]

The diagram above shows the electric field due to one infinitesimal piece of the ring, <math>dq</math>. In order to avoid rigorous computations, we can see that the electric field of the charges cancels out in the vertical direction. Only the horizontal component will remain. For an observation location that is on the symmetry axis as the ring, the other two components will be zero. If the observation location were off axis, then it would be very different requiring more math. Since each piece, <math>dq</math>, contributes a <math>d\vec{E}</math> of <math>\frac{1}{4\pi\epsilon_{0}}\frac{dq}{|\vec{d}|^{2}}\hat{d}</math>, we can compute the horizontal component by multiplying the magnitude of <math>d\vec{E}</math> with <math>\cos{θ}</math>.

<math>|d\vec{E}_{z}| = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\cos{θ} = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\frac{z}{(r^2 + z^2)^\frac{1}{2}}</math>

Now we integrate to sum all the electric field contributions of each infinitesimal <math>dq</math>.

<math>\vec{E}_{z} = \frac{1}{4\pi\epsilon_{0}}\frac{z}{(r^2 + z^2)^\frac{3}{2}}\int_{}^{} dq = \frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(r^2 + z^2)^\frac{3}{2}}</math>

As a reminder, this is the equation given on the formula sheet. It '''only''' works when the location at which the field is being measured is along the <math>{z}</math> axis. However, the equation for it being off axis is not given on the equation sheet. That requires a separate and more difficult integration. Typically, the only way this would be asked on a test is to set up an equation that would be integrated that requires you to find a new <math>dQ</math>.

'''Rigorous Derivation'''

1. Define shape characteristics of the ring

:The ring has some finite charge.

:We see that this arrangement is circular, so a coordinate system with which we can define radial and angular coordinates would be useful. Naturally, this would be the polar coordinate system.

:We also see that all charge is uniformly distributed some finite distance R from the center of the ring. It would be useful to let the center of the ring be the origin of our coordinate axes.

:Here is a reasonable arrangement for this charged ring.

[[File:blue_graph.png|400px|thumb|Figure 2: defining a coordinate system about a charged ring]]

:Since all charge is concentrated upon the edge of the circle, we can consider our charge distribution to be invariant with respect to radial distance. However, we do see that our charge distribution is the function of theta.

2. Compute charge distribution function

:Let us call the charge distribution as <math display="inline">\sigma</math>

:We have a charge distributed on the edge of the circle, so

:<math> \sigma=\frac{Q}{2\pi} </math>

:where <math display="inline">Q</math> represents the charge of the ring.

:What this equation means is that the charge is uniformly distributed along each unit of angular length. In this way, it is called angular charge density.

3. Compute infinitesimal charge contribution

:Let us consider an infinitesimal section of the ring which contains exactly one point charge. The dimension of this section is given by <math display="inline">d\theta</math> which is the infinitesimal angular size. So, the infinitesimal charge contribution, <math display="inline">dQ</math>, is

:<math> dQ = \frac{Q}{2\pi}d\theta </math>

4. Compute infinitesimal electric field contribution

:Let us define some arbitrary location at which we are observing this ring of charge.

:<math> \vec{r} = x\hat{x}+y\hat{y} + z\hat{z} </math>

:in polar coordinates, we see this becomes

:<math> \vec{r} = Rcos(\theta)\hat{x} + Rsin(\theta)\hat{y} + z\hat{z} </math>.

:Recall that both sine and cosine are periodic with the period of <math>2\pi </math>. This will become important later.

:The magnitude of this vector is

:<math> |\vec{r}| = \sqrt{R^{2}+z^{2}} </math>

:owing to the usage of the Pythagorean trigonometric identity.

:Now, we have what we need to write the electric field vector contributed by each piece of the ring of charge. Let this vector field piece be <math display="inline">d\vec{E}</math>.

:<math> d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z}) </math>

5. Compute electric field vector

:So, now all that is left is to sum everything up. We are summing over the circumference of a circle, so our path is defined by <math display="inline">0\leq\theta\leq 2\pi </math>. Let us now set up our integral.

:<math>\vec{E}=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

:Since both sine and cosine are <math display="inline">2\pi</math> periodic functions, the <math display="inline">x</math> and <math display="inline">y</math> components of <math display="inline">\vec{E}</math> go to <math display="inline">\vec{0}</math>, which is very convenient. The result is

:<math> \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(z^{2}+R^{2})^{3/2}}\hat{z} </math>.

''Remark''

If the path we are interested in is over a half circle, third-circle or quarter-circle, or indeed any circular section, simply adjust the bounds of integration and the charge density to the ones required by the curve of interest.

===A Computational Model===

[https://trinket.io/glowscript/707d492e19 This simulation] shows the result of the computation for a ring composed of 2000 electrons. This is why the vector is pointing into the ring rather than out of the ring, which would happen for a ring composed of positively charged points.

==Examples==

===Easy===

1.) Find the function <math>E(x)</math> that represents the magnitude of the electric field along the center axis due to a uniformly charged ring of radius 0.04 m with total charge 8 C.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can use the derived formula for the electric field due to a charged ring. Only the horizontal components of the electric field will remain as the rest will cancel out.

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{(8 C)*x}{((0.04 m)^2 + x^2)^\frac{3}{2}}</math>

</div></div>

===Middling===

2.) A uniformly charged ring of radius 10.0 cm has a total charge of 91.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

a. 1 cm

b. 5 cm

c. 30 cm

d. 100 cm

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Use the derived formula, and convert the given distances into meters and the charge into Coulombs. Plug these into <math>x</math>

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{Q*x}{(x^2 + R^2)^\frac{3}{2}}</math>

a. 8.069e6 N/C

b. 29.3e6 N/C

c. 7.76e6 N/C

d. 8.06e5 N/C

</div></div>

===Difficult===

Find the electric field at a distance <math>z</math> along the center axis away from a uniformly charged semicircular ring of radius R and total charge Q.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can tweak the same integral for a uniformly charged ring by changing the limits of integration.

<math>\vec{E}=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(0\hat{x}+2R\hat{y}+{\pi}z\hat{z})</math>

</div></div>

==Connectedness==

Edited by AYESHA AHUJA Spring 2020

The idea of charge density is somewhat analogous to the idea of the mass density, which is useful in a variety of contexts, including the computation of the center of mass, the first and second moments of mass, which are useful in statics and rigid body dynamics. Instead of computing the uniform distribution of the mass of this ring, we are computing the uniform distribution of charge. This concept is also useful in visualizing what happens within a wire in a steady state circuit. The wire can be viewed to be a continuous length of rings of charge, which act as a channel through which electrons are transported, and that the electric field of these rings of charge pushes the electrons within the wire.

It is also important in the context of Maxwell's equations, specifically in Gauss's Law and in the Maxwell-Faraday equation, which are concerned with electron flux and induced fields, respectively.

==History==

Added by AYESHA AHUJA SPRING 2020

The first known time when charged rings were utilized to describe electromagnetism was during Faraday’s discovery of electrical induction. To conduct his experiments, Faraday used a ring made of iron that was attached to a battery. When this makeshift circuit was turned on it manipulated a compass needle by deflecting it. He showcased an induced current in the ring.

== See also ==

[[Charged Rod]]

[[Charged Disk]]

[[Charged Spherical Shell]]

[[Charged Capacitor]]

===References===

"Electric Field on the Axis of a Ring of Charge". University of Delaware Physics Library. Adapted from Stephen Kevan's lecture on Electric Fields and Charge Distribution. April 8, 1996. http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

Chabay, R., & Sherwood, B. (2015). Matter and Interactions (4th ed., Vol. 2, pp. 597-599). Wiley.

All images produced by the author

==External Links==

https://www.youtube.com/watch?v=80mM3kSTZcE

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

[[Category:Electric Fields]]

Charged Ring

2020-05-05T20:43:06Z

Sydney: /* History */

==The Main Idea==

Charges may be arranged in a variety of ways. When points are uniformly distributed a finite distance away from some point, which we define as the origin, we call this curve as a ring. We can position a [[Point Charge]] at every point upon this ring. To analyze this continuous arrangement of charges, we say that their individual contributions are infinitesimally small, and then sum each of these contributions together to arrive at the electric field produced by this charge distribution. We can compute the net electric field of this charge distribution with Coulomb's Law and by applying integration principles. The ring field can also be used to calculate the electric field of a [[Charged Disk]].

===A Mathematical Model===

This mathematical model is based on the individual [[Electric Field]] contributions of a number of point charges, each of which is defined by

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}</math>.

'''Do not forget this equation'''. This equation must be memorized as it isn't given on the formula sheet.

We also say that this vector field is a member of a linear space of vectors. This is to say that we can apply the [[Superposition Principle]] meaning that we can sum any number of these electric field vectors and obtain another vector which is the electric field vector contributed by those charges. When we continuously sum all of the vectors produced by these charges, we get the electric field produced by the entire arrangement of charges.

[[File:red_ring.png|400px|thumb|Figure 1: diagram of a charged ring]]

The diagram above shows the electric field due to one infinitesimal piece of the ring, <math>dq</math>. In order to avoid rigorous computations, we can see that the electric field of the charges cancels out in the vertical direction. Only the horizontal component will remain. For an observation location that is on the symmetry axis as the ring, the other two components will be zero. If the observation location were off axis, then it would be very different requiring more math. Since each piece, <math>dq</math>, contributes a <math>d\vec{E}</math> of <math>\frac{1}{4\pi\epsilon_{0}}\frac{dq}{|\vec{d}|^{2}}\hat{d}</math>, we can compute the horizontal component by multiplying the magnitude of <math>d\vec{E}</math> with <math>\cos{θ}</math>.

<math>|d\vec{E}_{z}| = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\cos{θ} = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\frac{z}{(r^2 + z^2)^\frac{1}{2}}</math>

Now we integrate to sum all the electric field contributions of each infinitesimal <math>dq</math>.

<math>\vec{E}_{z} = \frac{1}{4\pi\epsilon_{0}}\frac{z}{(r^2 + z^2)^\frac{3}{2}}\int_{}^{} dq = \frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(r^2 + z^2)^\frac{3}{2}}</math>

As a reminder, this is the equation given on the formula sheet. It '''only''' works when the location at which the field is being measured is along the <math>{z}</math> axis. However, the equation for it being off axis is not given on the equation sheet. That requires a separate and more difficult integration. Typically, the only way this would be asked on a test is to set up an equation that would be integrated that requires you to find a new <math>dQ</math>.

'''Rigorous Derivation'''

1. Define shape characteristics of the ring

:The ring has some finite charge.

:We see that this arrangement is circular, so a coordinate system with which we can define radial and angular coordinates would be useful. Naturally, this would be the polar coordinate system.

:We also see that all charge is uniformly distributed some finite distance R from the center of the ring. It would be useful to let the center of the ring be the origin of our coordinate axes.

:Here is a reasonable arrangement for this charged ring.

[[File:blue_graph.png|400px|thumb|Figure 2: defining a coordinate system about a charged ring]]

:Since all charge is concentrated upon the edge of the circle, we can consider our charge distribution to be invariant with respect to radial distance. However, we do see that our charge distribution is the function of theta.

2. Compute charge distribution function

:Let us call the charge distribution as <math display="inline">\sigma</math>

:We have a charge distributed on the edge of the circle, so

:<math> \sigma=\frac{Q}{2\pi} </math>

:where <math display="inline">Q</math> represents the charge of the ring.

:What this equation means is that the charge is uniformly distributed along each unit of angular length. In this way, it is called angular charge density.

3. Compute infinitesimal charge contribution

:Let us consider an infinitesimal section of the ring which contains exactly one point charge. The dimension of this section is given by <math display="inline">d\theta</math> which is the infinitesimal angular size. So, the infinitesimal charge contribution, <math display="inline">dQ</math>, is

:<math> dQ = \frac{Q}{2\pi}d\theta </math>

4. Compute infinitesimal electric field contribution

:Let us define some arbitrary location at which we are observing this ring of charge.

:<math> \vec{r} = x\hat{x}+y\hat{y} + z\hat{z} </math>

:in polar coordinates, we see this becomes

:<math> \vec{r} = Rcos(\theta)\hat{x} + Rsin(\theta)\hat{y} + z\hat{z} </math>.

:Recall that both sine and cosine are periodic with the period of <math>2\pi </math>. This will become important later.

:The magnitude of this vector is

:<math> |\vec{r}| = \sqrt{R^{2}+z^{2}} </math>

:owing to the usage of the Pythagorean trigonometric identity.

:Now, we have what we need to write the electric field vector contributed by each piece of the ring of charge. Let this vector field piece be <math display="inline">d\vec{E}</math>.

:<math> d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z}) </math>

5. Compute electric field vector

:So, now all that is left is to sum everything up. We are summing over the circumference of a circle, so our path is defined by <math display="inline">0\leq\theta\leq 2\pi </math>. Let us now set up our integral.

:<math>\vec{E}=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

:Since both sine and cosine are <math display="inline">2\pi</math> periodic functions, the <math display="inline">x</math> and <math display="inline">y</math> components of <math display="inline">\vec{E}</math> go to <math display="inline">\vec{0}</math>, which is very convenient. The result is

:<math> \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(z^{2}+R^{2})^{3/2}}\hat{z} </math>.

''Remark''

If the path we are interested in is over a half circle, third-circle or quarter-circle, or indeed any circular section, simply adjust the bounds of integration and the charge density to the ones required by the curve of interest.

===A Computational Model===

[https://trinket.io/glowscript/707d492e19 This simulation] shows the result of the computation for a ring composed of 2000 electrons. This is why the vector is pointing into the ring rather than out of the ring, which would happen for a ring composed of positively charged points.

==Examples==

===Easy===

1.) Find the function <math>E(x)</math> that represents the magnitude of the electric field along the center axis due to a uniformly charged ring of radius 0.04 m with total charge 8 C.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can use the derived formula for the electric field due to a charged ring. Only the horizontal components of the electric field will remain as the rest will cancel out.

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{(8 C)*x}{((0.04 m)^2 + x^2)^\frac{3}{2}}</math>

</div></div>

===Middling===

2.) A uniformly charged ring of radius 10.0 cm has a total charge of 91.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

a. 1 cm

b. 5 cm

c. 30 cm

d. 100 cm

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Use the derived formula, and convert the given distances into meters and the charge into Coulombs. Plug these into <math>x</math>

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{Q*x}{(x^2 + R^2)^\frac{3}{2}}</math>

a. 8.069e6 N/C

b. 29.3e6 N/C

c. 7.76e6 N/C

d. 8.06e5 N/C

</div></div>

===Difficult===

Find the electric field at a distance <math>z</math> along the center axis away from a uniformly charged semicircular ring of radius R and total charge Q.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can tweak the same integral for a uniformly charged ring by changing the limits of integration.

<math>\vec{E}=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(0\hat{x}+2R\hat{y}+{\pi}z\hat{z})</math>

</div></div>

==Connectedness==

Edited by AYESHA AHUJA Spring 2020

The idea of charge density is somewhat analogous to the idea of the mass density, which is useful in a variety of contexts, including the computation of the center of mass, the first and second moments of mass, which are useful in statics and rigid body dynamics. Instead of computing the uniform distribution of the mass of this ring, we are computing the uniform distribution of charge. This concept is also useful in visualizing what happens within a wire in a steady state circuit. The wire can be viewed to be a continuous length of rings of charge, which act as a channel through which electrons are transported, and that the electric field of these rings of charge pushes the electrons within the wire.

It is also important in the context of Maxwell's equations, specifically in Gauss's Law and in the Maxwell-Faraday equation, which are concerned with electron flux and induced fields, respectively.

==History==

Added by AYESHA AHUJA SPRING 2020

The first known time when charged rings were utilized to describe electromagnetism was during Faraday’s discovery of electrical induction. To conduct his experiments, Faraday used a ring made of iron that was attached to a battery. When this makeshift circuit was turned on it manipulated a compass needle by deflecting it. He showcased an induced current in the ring.

== See also ==

[[Charged Rod]]

[[Charged Disk]]

[[Charged Spherical Shell]]

[[Charged Capacitor]]

==References==

"Electric Field on the Axis of a Ring of Charge". University of Delaware Physics Library. Adapted from Stephen Kevan's lecture on Electric Fields and Charge Distribution. April 8, 1996. http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

Chabay, R., & Sherwood, B. (2015). Matter and Interactions (4th ed., Vol. 2, pp. 597-599). Wiley.

All images produced by the author

==External Links==

https://www.youtube.com/watch?v=80mM3kSTZcE

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

[[Category:Electric Fields]]

Charged Ring

2020-05-05T20:42:23Z

Sydney: /* Connectedness */

==The Main Idea==

Charges may be arranged in a variety of ways. When points are uniformly distributed a finite distance away from some point, which we define as the origin, we call this curve as a ring. We can position a [[Point Charge]] at every point upon this ring. To analyze this continuous arrangement of charges, we say that their individual contributions are infinitesimally small, and then sum each of these contributions together to arrive at the electric field produced by this charge distribution. We can compute the net electric field of this charge distribution with Coulomb's Law and by applying integration principles. The ring field can also be used to calculate the electric field of a [[Charged Disk]].

===A Mathematical Model===

This mathematical model is based on the individual [[Electric Field]] contributions of a number of point charges, each of which is defined by

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}</math>.

'''Do not forget this equation'''. This equation must be memorized as it isn't given on the formula sheet.

We also say that this vector field is a member of a linear space of vectors. This is to say that we can apply the [[Superposition Principle]] meaning that we can sum any number of these electric field vectors and obtain another vector which is the electric field vector contributed by those charges. When we continuously sum all of the vectors produced by these charges, we get the electric field produced by the entire arrangement of charges.

[[File:red_ring.png|400px|thumb|Figure 1: diagram of a charged ring]]

The diagram above shows the electric field due to one infinitesimal piece of the ring, <math>dq</math>. In order to avoid rigorous computations, we can see that the electric field of the charges cancels out in the vertical direction. Only the horizontal component will remain. For an observation location that is on the symmetry axis as the ring, the other two components will be zero. If the observation location were off axis, then it would be very different requiring more math. Since each piece, <math>dq</math>, contributes a <math>d\vec{E}</math> of <math>\frac{1}{4\pi\epsilon_{0}}\frac{dq}{|\vec{d}|^{2}}\hat{d}</math>, we can compute the horizontal component by multiplying the magnitude of <math>d\vec{E}</math> with <math>\cos{θ}</math>.

<math>|d\vec{E}_{z}| = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\cos{θ} = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\frac{z}{(r^2 + z^2)^\frac{1}{2}}</math>

Now we integrate to sum all the electric field contributions of each infinitesimal <math>dq</math>.

<math>\vec{E}_{z} = \frac{1}{4\pi\epsilon_{0}}\frac{z}{(r^2 + z^2)^\frac{3}{2}}\int_{}^{} dq = \frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(r^2 + z^2)^\frac{3}{2}}</math>

As a reminder, this is the equation given on the formula sheet. It '''only''' works when the location at which the field is being measured is along the <math>{z}</math> axis. However, the equation for it being off axis is not given on the equation sheet. That requires a separate and more difficult integration. Typically, the only way this would be asked on a test is to set up an equation that would be integrated that requires you to find a new <math>dQ</math>.

'''Rigorous Derivation'''

1. Define shape characteristics of the ring

:The ring has some finite charge.

:We see that this arrangement is circular, so a coordinate system with which we can define radial and angular coordinates would be useful. Naturally, this would be the polar coordinate system.

:We also see that all charge is uniformly distributed some finite distance R from the center of the ring. It would be useful to let the center of the ring be the origin of our coordinate axes.

:Here is a reasonable arrangement for this charged ring.

[[File:blue_graph.png|400px|thumb|Figure 2: defining a coordinate system about a charged ring]]

:Since all charge is concentrated upon the edge of the circle, we can consider our charge distribution to be invariant with respect to radial distance. However, we do see that our charge distribution is the function of theta.

2. Compute charge distribution function

:Let us call the charge distribution as <math display="inline">\sigma</math>

:We have a charge distributed on the edge of the circle, so

:<math> \sigma=\frac{Q}{2\pi} </math>

:where <math display="inline">Q</math> represents the charge of the ring.

:What this equation means is that the charge is uniformly distributed along each unit of angular length. In this way, it is called angular charge density.

3. Compute infinitesimal charge contribution

:Let us consider an infinitesimal section of the ring which contains exactly one point charge. The dimension of this section is given by <math display="inline">d\theta</math> which is the infinitesimal angular size. So, the infinitesimal charge contribution, <math display="inline">dQ</math>, is

:<math> dQ = \frac{Q}{2\pi}d\theta </math>

4. Compute infinitesimal electric field contribution

:Let us define some arbitrary location at which we are observing this ring of charge.

:<math> \vec{r} = x\hat{x}+y\hat{y} + z\hat{z} </math>

:in polar coordinates, we see this becomes

:<math> \vec{r} = Rcos(\theta)\hat{x} + Rsin(\theta)\hat{y} + z\hat{z} </math>.

:Recall that both sine and cosine are periodic with the period of <math>2\pi </math>. This will become important later.

:The magnitude of this vector is

:<math> |\vec{r}| = \sqrt{R^{2}+z^{2}} </math>

:owing to the usage of the Pythagorean trigonometric identity.

:Now, we have what we need to write the electric field vector contributed by each piece of the ring of charge. Let this vector field piece be <math display="inline">d\vec{E}</math>.

:<math> d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z}) </math>

5. Compute electric field vector

:So, now all that is left is to sum everything up. We are summing over the circumference of a circle, so our path is defined by <math display="inline">0\leq\theta\leq 2\pi </math>. Let us now set up our integral.

:<math>\vec{E}=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

:Since both sine and cosine are <math display="inline">2\pi</math> periodic functions, the <math display="inline">x</math> and <math display="inline">y</math> components of <math display="inline">\vec{E}</math> go to <math display="inline">\vec{0}</math>, which is very convenient. The result is

:<math> \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(z^{2}+R^{2})^{3/2}}\hat{z} </math>.

''Remark''

If the path we are interested in is over a half circle, third-circle or quarter-circle, or indeed any circular section, simply adjust the bounds of integration and the charge density to the ones required by the curve of interest.

===A Computational Model===

[https://trinket.io/glowscript/707d492e19 This simulation] shows the result of the computation for a ring composed of 2000 electrons. This is why the vector is pointing into the ring rather than out of the ring, which would happen for a ring composed of positively charged points.

==Examples==

===Easy===

1.) Find the function <math>E(x)</math> that represents the magnitude of the electric field along the center axis due to a uniformly charged ring of radius 0.04 m with total charge 8 C.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can use the derived formula for the electric field due to a charged ring. Only the horizontal components of the electric field will remain as the rest will cancel out.

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{(8 C)*x}{((0.04 m)^2 + x^2)^\frac{3}{2}}</math>

</div></div>

===Middling===

2.) A uniformly charged ring of radius 10.0 cm has a total charge of 91.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

a. 1 cm

b. 5 cm

c. 30 cm

d. 100 cm

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Use the derived formula, and convert the given distances into meters and the charge into Coulombs. Plug these into <math>x</math>

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{Q*x}{(x^2 + R^2)^\frac{3}{2}}</math>

a. 8.069e6 N/C

b. 29.3e6 N/C

c. 7.76e6 N/C

d. 8.06e5 N/C

</div></div>

===Difficult===

Find the electric field at a distance <math>z</math> along the center axis away from a uniformly charged semicircular ring of radius R and total charge Q.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can tweak the same integral for a uniformly charged ring by changing the limits of integration.

<math>\vec{E}=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(0\hat{x}+2R\hat{y}+{\pi}z\hat{z})</math>

</div></div>

==History==

Added by AYESHA AHUJA SPRING 2020

The first known time when charged rings were utilized to describe electromagnetism was during Faraday’s discovery of electrical induction. To conduct his experiments, Faraday used a ring made of iron that was attached to a battery. When this makeshift circuit was turned on it manipulated a compass needle by deflecting it. He showcased an induced current in the ring.

===Connectedness===

Edited by AYESHA AHUJA Spring 2020

The idea of charge density is somewhat analogous to the idea of the mass density, which is useful in a variety of contexts, including the computation of the center of mass, the first and second moments of mass, which are useful in statics and rigid body dynamics. Instead of computing the uniform distribution of the mass of this ring, we are computing the uniform distribution of charge. This concept is also useful in visualizing what happens within a wire in a steady state circuit. The wire can be viewed to be a continuous length of rings of charge, which act as a channel through which electrons are transported, and that the electric field of these rings of charge pushes the electrons within the wire.

It is also important in the context of Maxwell's equations, specifically in Gauss's Law and in the Maxwell-Faraday equation, which are concerned with electron flux and induced fields, respectively.

== See also ==

[[Charged Rod]]

[[Charged Disk]]

[[Charged Spherical Shell]]

[[Charged Capacitor]]

==References==

"Electric Field on the Axis of a Ring of Charge". University of Delaware Physics Library. Adapted from Stephen Kevan's lecture on Electric Fields and Charge Distribution. April 8, 1996. http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

Chabay, R., & Sherwood, B. (2015). Matter and Interactions (4th ed., Vol. 2, pp. 597-599). Wiley.

All images produced by the author

==External Links==

https://www.youtube.com/watch?v=80mM3kSTZcE

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

[[Category:Electric Fields]]

Charged Ring

2020-05-05T20:41:30Z

Sydney: /* Examples */

==The Main Idea==

Charges may be arranged in a variety of ways. When points are uniformly distributed a finite distance away from some point, which we define as the origin, we call this curve as a ring. We can position a [[Point Charge]] at every point upon this ring. To analyze this continuous arrangement of charges, we say that their individual contributions are infinitesimally small, and then sum each of these contributions together to arrive at the electric field produced by this charge distribution. We can compute the net electric field of this charge distribution with Coulomb's Law and by applying integration principles. The ring field can also be used to calculate the electric field of a [[Charged Disk]].

===A Mathematical Model===

This mathematical model is based on the individual [[Electric Field]] contributions of a number of point charges, each of which is defined by

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}</math>.

'''Do not forget this equation'''. This equation must be memorized as it isn't given on the formula sheet.

We also say that this vector field is a member of a linear space of vectors. This is to say that we can apply the [[Superposition Principle]] meaning that we can sum any number of these electric field vectors and obtain another vector which is the electric field vector contributed by those charges. When we continuously sum all of the vectors produced by these charges, we get the electric field produced by the entire arrangement of charges.

[[File:red_ring.png|400px|thumb|Figure 1: diagram of a charged ring]]

The diagram above shows the electric field due to one infinitesimal piece of the ring, <math>dq</math>. In order to avoid rigorous computations, we can see that the electric field of the charges cancels out in the vertical direction. Only the horizontal component will remain. For an observation location that is on the symmetry axis as the ring, the other two components will be zero. If the observation location were off axis, then it would be very different requiring more math. Since each piece, <math>dq</math>, contributes a <math>d\vec{E}</math> of <math>\frac{1}{4\pi\epsilon_{0}}\frac{dq}{|\vec{d}|^{2}}\hat{d}</math>, we can compute the horizontal component by multiplying the magnitude of <math>d\vec{E}</math> with <math>\cos{θ}</math>.

<math>|d\vec{E}_{z}| = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\cos{θ} = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\frac{z}{(r^2 + z^2)^\frac{1}{2}}</math>

Now we integrate to sum all the electric field contributions of each infinitesimal <math>dq</math>.

<math>\vec{E}_{z} = \frac{1}{4\pi\epsilon_{0}}\frac{z}{(r^2 + z^2)^\frac{3}{2}}\int_{}^{} dq = \frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(r^2 + z^2)^\frac{3}{2}}</math>

As a reminder, this is the equation given on the formula sheet. It '''only''' works when the location at which the field is being measured is along the <math>{z}</math> axis. However, the equation for it being off axis is not given on the equation sheet. That requires a separate and more difficult integration. Typically, the only way this would be asked on a test is to set up an equation that would be integrated that requires you to find a new <math>dQ</math>.

'''Rigorous Derivation'''

1. Define shape characteristics of the ring

:The ring has some finite charge.

:We see that this arrangement is circular, so a coordinate system with which we can define radial and angular coordinates would be useful. Naturally, this would be the polar coordinate system.

:We also see that all charge is uniformly distributed some finite distance R from the center of the ring. It would be useful to let the center of the ring be the origin of our coordinate axes.

:Here is a reasonable arrangement for this charged ring.

[[File:blue_graph.png|400px|thumb|Figure 2: defining a coordinate system about a charged ring]]

:Since all charge is concentrated upon the edge of the circle, we can consider our charge distribution to be invariant with respect to radial distance. However, we do see that our charge distribution is the function of theta.

2. Compute charge distribution function

:Let us call the charge distribution as <math display="inline">\sigma</math>

:We have a charge distributed on the edge of the circle, so

:<math> \sigma=\frac{Q}{2\pi} </math>

:where <math display="inline">Q</math> represents the charge of the ring.

:What this equation means is that the charge is uniformly distributed along each unit of angular length. In this way, it is called angular charge density.

3. Compute infinitesimal charge contribution

:Let us consider an infinitesimal section of the ring which contains exactly one point charge. The dimension of this section is given by <math display="inline">d\theta</math> which is the infinitesimal angular size. So, the infinitesimal charge contribution, <math display="inline">dQ</math>, is

:<math> dQ = \frac{Q}{2\pi}d\theta </math>

4. Compute infinitesimal electric field contribution

:Let us define some arbitrary location at which we are observing this ring of charge.

:<math> \vec{r} = x\hat{x}+y\hat{y} + z\hat{z} </math>

:in polar coordinates, we see this becomes

:<math> \vec{r} = Rcos(\theta)\hat{x} + Rsin(\theta)\hat{y} + z\hat{z} </math>.

:Recall that both sine and cosine are periodic with the period of <math>2\pi </math>. This will become important later.

:The magnitude of this vector is

:<math> |\vec{r}| = \sqrt{R^{2}+z^{2}} </math>

:owing to the usage of the Pythagorean trigonometric identity.

:Now, we have what we need to write the electric field vector contributed by each piece of the ring of charge. Let this vector field piece be <math display="inline">d\vec{E}</math>.

:<math> d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z}) </math>

5. Compute electric field vector

:So, now all that is left is to sum everything up. We are summing over the circumference of a circle, so our path is defined by <math display="inline">0\leq\theta\leq 2\pi </math>. Let us now set up our integral.

:<math>\vec{E}=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

:Since both sine and cosine are <math display="inline">2\pi</math> periodic functions, the <math display="inline">x</math> and <math display="inline">y</math> components of <math display="inline">\vec{E}</math> go to <math display="inline">\vec{0}</math>, which is very convenient. The result is

:<math> \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(z^{2}+R^{2})^{3/2}}\hat{z} </math>.

''Remark''

If the path we are interested in is over a half circle, third-circle or quarter-circle, or indeed any circular section, simply adjust the bounds of integration and the charge density to the ones required by the curve of interest.

===A Computational Model===

[https://trinket.io/glowscript/707d492e19 This simulation] shows the result of the computation for a ring composed of 2000 electrons. This is why the vector is pointing into the ring rather than out of the ring, which would happen for a ring composed of positively charged points.

==Examples==

===Easy===

1.) Find the function <math>E(x)</math> that represents the magnitude of the electric field along the center axis due to a uniformly charged ring of radius 0.04 m with total charge 8 C.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can use the derived formula for the electric field due to a charged ring. Only the horizontal components of the electric field will remain as the rest will cancel out.

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{(8 C)*x}{((0.04 m)^2 + x^2)^\frac{3}{2}}</math>

</div></div>

===Middling===

2.) A uniformly charged ring of radius 10.0 cm has a total charge of 91.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

a. 1 cm

b. 5 cm

c. 30 cm

d. 100 cm

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

Use the derived formula, and convert the given distances into meters and the charge into Coulombs. Plug these into <math>x</math>

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{Q*x}{(x^2 + R^2)^\frac{3}{2}}</math>

a. 8.069e6 N/C

b. 29.3e6 N/C

c. 7.76e6 N/C

d. 8.06e5 N/C

</div></div>

===Difficult===

Find the electric field at a distance <math>z</math> along the center axis away from a uniformly charged semicircular ring of radius R and total charge Q.

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can tweak the same integral for a uniformly charged ring by changing the limits of integration.

<math>\vec{E}=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(0\hat{x}+2R\hat{y}+{\pi}z\hat{z})</math>

</div></div>

==History==

Added by AYESHA AHUJA SPRING 2020

The first known time when charged rings were utilized to describe electromagnetism was during Faraday’s discovery of electrical induction. To conduct his experiments, Faraday used a ring made of iron that was attached to a battery. When this makeshift circuit was turned on it manipulated a compass needle by deflecting it. He showcased an induced current in the ring.

==Connectedness==

Edited by AYESHA AHUJA Spring 2020

The idea of charge density is somewhat analogous to the idea of the mass density, which is useful in a variety of contexts, including the computation of the center of mass, the first and second moments of mass, which are useful in statics and rigid body dynamics. Instead of computing the uniform distribution of the mass of this ring, we are computing the uniform distribution of charge. This concept is also useful in visualizing what happens within a wire in a steady state circuit. The wire can be viewed to be a continuous length of rings of charge, which act as a channel through which electrons are transported, and that the electric field of these rings of charge pushes the electrons within the wire.

It is also important in the context of Maxwell's equations, specifically in Gauss's Law and in the Maxwell-Faraday equation, which are concerned with electron flux and induced fields, respectively.

== See also ==

[[Charged Rod]]

[[Charged Disk]]

[[Charged Spherical Shell]]

[[Charged Capacitor]]

==References==

"Electric Field on the Axis of a Ring of Charge". University of Delaware Physics Library. Adapted from Stephen Kevan's lecture on Electric Fields and Charge Distribution. April 8, 1996. http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

Chabay, R., & Sherwood, B. (2015). Matter and Interactions (4th ed., Vol. 2, pp. 597-599). Wiley.

All images produced by the author

==External Links==

https://www.youtube.com/watch?v=80mM3kSTZcE

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

[[Category:Electric Fields]]

Charged Ring

2020-05-05T20:37:45Z

Sydney: /* Easy */

==The Main Idea==

Charges may be arranged in a variety of ways. When points are uniformly distributed a finite distance away from some point, which we define as the origin, we call this curve as a ring. We can position a [[Point Charge]] at every point upon this ring. To analyze this continuous arrangement of charges, we say that their individual contributions are infinitesimally small, and then sum each of these contributions together to arrive at the electric field produced by this charge distribution. We can compute the net electric field of this charge distribution with Coulomb's Law and by applying integration principles. The ring field can also be used to calculate the electric field of a [[Charged Disk]].

===A Mathematical Model===

This mathematical model is based on the individual [[Electric Field]] contributions of a number of point charges, each of which is defined by

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}</math>.

'''Do not forget this equation'''. This equation must be memorized as it isn't given on the formula sheet.

We also say that this vector field is a member of a linear space of vectors. This is to say that we can apply the [[Superposition Principle]] meaning that we can sum any number of these electric field vectors and obtain another vector which is the electric field vector contributed by those charges. When we continuously sum all of the vectors produced by these charges, we get the electric field produced by the entire arrangement of charges.

[[File:red_ring.png|400px|thumb|Figure 1: diagram of a charged ring]]

The diagram above shows the electric field due to one infinitesimal piece of the ring, <math>dq</math>. In order to avoid rigorous computations, we can see that the electric field of the charges cancels out in the vertical direction. Only the horizontal component will remain. For an observation location that is on the symmetry axis as the ring, the other two components will be zero. If the observation location were off axis, then it would be very different requiring more math. Since each piece, <math>dq</math>, contributes a <math>d\vec{E}</math> of <math>\frac{1}{4\pi\epsilon_{0}}\frac{dq}{|\vec{d}|^{2}}\hat{d}</math>, we can compute the horizontal component by multiplying the magnitude of <math>d\vec{E}</math> with <math>\cos{θ}</math>.

<math>|d\vec{E}_{z}| = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\cos{θ} = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\frac{z}{(r^2 + z^2)^\frac{1}{2}}</math>

Now we integrate to sum all the electric field contributions of each infinitesimal <math>dq</math>.

<math>\vec{E}_{z} = \frac{1}{4\pi\epsilon_{0}}\frac{z}{(r^2 + z^2)^\frac{3}{2}}\int_{}^{} dq = \frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(r^2 + z^2)^\frac{3}{2}}</math>

As a reminder, this is the equation given on the formula sheet. It '''only''' works when the location at which the field is being measured is along the <math>{z}</math> axis. However, the equation for it being off axis is not given on the equation sheet. That requires a separate and more difficult integration. Typically, the only way this would be asked on a test is to set up an equation that would be integrated that requires you to find a new <math>dQ</math>.

'''Rigorous Derivation'''

1. Define shape characteristics of the ring

:The ring has some finite charge.

:We see that this arrangement is circular, so a coordinate system with which we can define radial and angular coordinates would be useful. Naturally, this would be the polar coordinate system.

:We also see that all charge is uniformly distributed some finite distance R from the center of the ring. It would be useful to let the center of the ring be the origin of our coordinate axes.

:Here is a reasonable arrangement for this charged ring.

[[File:blue_graph.png|400px|thumb|Figure 2: defining a coordinate system about a charged ring]]

:Since all charge is concentrated upon the edge of the circle, we can consider our charge distribution to be invariant with respect to radial distance. However, we do see that our charge distribution is the function of theta.

2. Compute charge distribution function

:Let us call the charge distribution as <math display="inline">\sigma</math>

:We have a charge distributed on the edge of the circle, so

:<math> \sigma=\frac{Q}{2\pi} </math>

:where <math display="inline">Q</math> represents the charge of the ring.

:What this equation means is that the charge is uniformly distributed along each unit of angular length. In this way, it is called angular charge density.

3. Compute infinitesimal charge contribution

:Let us consider an infinitesimal section of the ring which contains exactly one point charge. The dimension of this section is given by <math display="inline">d\theta</math> which is the infinitesimal angular size. So, the infinitesimal charge contribution, <math display="inline">dQ</math>, is

:<math> dQ = \frac{Q}{2\pi}d\theta </math>

4. Compute infinitesimal electric field contribution

:Let us define some arbitrary location at which we are observing this ring of charge.

:<math> \vec{r} = x\hat{x}+y\hat{y} + z\hat{z} </math>

:in polar coordinates, we see this becomes

:<math> \vec{r} = Rcos(\theta)\hat{x} + Rsin(\theta)\hat{y} + z\hat{z} </math>.

:Recall that both sine and cosine are periodic with the period of <math>2\pi </math>. This will become important later.

:The magnitude of this vector is

:<math> |\vec{r}| = \sqrt{R^{2}+z^{2}} </math>

:owing to the usage of the Pythagorean trigonometric identity.

:Now, we have what we need to write the electric field vector contributed by each piece of the ring of charge. Let this vector field piece be <math display="inline">d\vec{E}</math>.

:<math> d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z}) </math>

5. Compute electric field vector

:So, now all that is left is to sum everything up. We are summing over the circumference of a circle, so our path is defined by <math display="inline">0\leq\theta\leq 2\pi </math>. Let us now set up our integral.

:<math>\vec{E}=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

:Since both sine and cosine are <math display="inline">2\pi</math> periodic functions, the <math display="inline">x</math> and <math display="inline">y</math> components of <math display="inline">\vec{E}</math> go to <math display="inline">\vec{0}</math>, which is very convenient. The result is

:<math> \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(z^{2}+R^{2})^{3/2}}\hat{z} </math>.

''Remark''

If the path we are interested in is over a half circle, third-circle or quarter-circle, or indeed any circular section, simply adjust the bounds of integration and the charge density to the ones required by the curve of interest.

===A Computational Model===

[https://trinket.io/glowscript/707d492e19 This simulation] shows the result of the computation for a ring composed of 2000 electrons. This is why the vector is pointing into the ring rather than out of the ring, which would happen for a ring composed of positively charged points.

==Examples==

===Easy===

1.) Find the function <math>E(x)</math> that represents the magnitude of the electric field along the center axis due to a uniformly charged ring of radius 0.04 m with total charge 8 C.

'''Solution'''

We can use the derived formula for the electric field due to a charged ring. Only the horizontal components of the electric field will remain as the rest will cancel out.

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{(8 C)*x}{((0.04 m)^2 + x^2)^\frac{3}{2}}</math>

===Middling===

2.) A uniformly charged ring of radius 10.0 cm has a total charge of 91.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

a. 1 cm

b. 5 cm

c. 30 cm

d. 100 cm

''Solution''

Use the derived formula, and convert the given distances into meters and the charge into Coulombs. Plug these into <math>x</math>

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{Q*x}{(x^2 + R^2)^\frac{3}{2}}</math>

a. 8.069e6 N/C

b. 29.3e6 N/C

c. 7.76e6 N/C

d. 8.06e5 N/C

===Difficult===

Find the electric field at a distance <math>z</math> along the center axis away from a uniformly charged semicircular ring of radius R and total charge Q.

''Solution''

We can tweak the same integral for a uniformly charged ring by changing the limits of integration.

<math>\vec{E}=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(0\hat{x}+2R\hat{y}+{\pi}z\hat{z})</math>

==History==

Added by AYESHA AHUJA SPRING 2020

The first known time when charged rings were utilized to describe electromagnetism was during Faraday’s discovery of electrical induction. To conduct his experiments, Faraday used a ring made of iron that was attached to a battery. When this makeshift circuit was turned on it manipulated a compass needle by deflecting it. He showcased an induced current in the ring.

==Connectedness==

Edited by AYESHA AHUJA Spring 2020

The idea of charge density is somewhat analogous to the idea of the mass density, which is useful in a variety of contexts, including the computation of the center of mass, the first and second moments of mass, which are useful in statics and rigid body dynamics. Instead of computing the uniform distribution of the mass of this ring, we are computing the uniform distribution of charge. This concept is also useful in visualizing what happens within a wire in a steady state circuit. The wire can be viewed to be a continuous length of rings of charge, which act as a channel through which electrons are transported, and that the electric field of these rings of charge pushes the electrons within the wire.

It is also important in the context of Maxwell's equations, specifically in Gauss's Law and in the Maxwell-Faraday equation, which are concerned with electron flux and induced fields, respectively.

== See also ==

[[Charged Rod]]

[[Charged Disk]]

[[Charged Spherical Shell]]

[[Charged Capacitor]]

==References==

"Electric Field on the Axis of a Ring of Charge". University of Delaware Physics Library. Adapted from Stephen Kevan's lecture on Electric Fields and Charge Distribution. April 8, 1996. http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

Chabay, R., & Sherwood, B. (2015). Matter and Interactions (4th ed., Vol. 2, pp. 597-599). Wiley.

All images produced by the author

==External Links==

https://www.youtube.com/watch?v=80mM3kSTZcE

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

[[Category:Electric Fields]]

Charged Ring

2020-05-05T20:37:21Z

Sydney: /* A Mathematical Model */

==The Main Idea==

Charges may be arranged in a variety of ways. When points are uniformly distributed a finite distance away from some point, which we define as the origin, we call this curve as a ring. We can position a [[Point Charge]] at every point upon this ring. To analyze this continuous arrangement of charges, we say that their individual contributions are infinitesimally small, and then sum each of these contributions together to arrive at the electric field produced by this charge distribution. We can compute the net electric field of this charge distribution with Coulomb's Law and by applying integration principles. The ring field can also be used to calculate the electric field of a [[Charged Disk]].

===A Mathematical Model===

This mathematical model is based on the individual [[Electric Field]] contributions of a number of point charges, each of which is defined by

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}</math>.

'''Do not forget this equation'''. This equation must be memorized as it isn't given on the formula sheet.

We also say that this vector field is a member of a linear space of vectors. This is to say that we can apply the [[Superposition Principle]] meaning that we can sum any number of these electric field vectors and obtain another vector which is the electric field vector contributed by those charges. When we continuously sum all of the vectors produced by these charges, we get the electric field produced by the entire arrangement of charges.

[[File:red_ring.png|400px|thumb|Figure 1: diagram of a charged ring]]

The diagram above shows the electric field due to one infinitesimal piece of the ring, <math>dq</math>. In order to avoid rigorous computations, we can see that the electric field of the charges cancels out in the vertical direction. Only the horizontal component will remain. For an observation location that is on the symmetry axis as the ring, the other two components will be zero. If the observation location were off axis, then it would be very different requiring more math. Since each piece, <math>dq</math>, contributes a <math>d\vec{E}</math> of <math>\frac{1}{4\pi\epsilon_{0}}\frac{dq}{|\vec{d}|^{2}}\hat{d}</math>, we can compute the horizontal component by multiplying the magnitude of <math>d\vec{E}</math> with <math>\cos{θ}</math>.

<math>|d\vec{E}_{z}| = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\cos{θ} = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\frac{z}{(r^2 + z^2)^\frac{1}{2}}</math>

Now we integrate to sum all the electric field contributions of each infinitesimal <math>dq</math>.

<math>\vec{E}_{z} = \frac{1}{4\pi\epsilon_{0}}\frac{z}{(r^2 + z^2)^\frac{3}{2}}\int_{}^{} dq = \frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(r^2 + z^2)^\frac{3}{2}}</math>

As a reminder, this is the equation given on the formula sheet. It '''only''' works when the location at which the field is being measured is along the <math>{z}</math> axis. However, the equation for it being off axis is not given on the equation sheet. That requires a separate and more difficult integration. Typically, the only way this would be asked on a test is to set up an equation that would be integrated that requires you to find a new <math>dQ</math>.

'''Rigorous Derivation'''

1. Define shape characteristics of the ring

:The ring has some finite charge.

:We see that this arrangement is circular, so a coordinate system with which we can define radial and angular coordinates would be useful. Naturally, this would be the polar coordinate system.

:We also see that all charge is uniformly distributed some finite distance R from the center of the ring. It would be useful to let the center of the ring be the origin of our coordinate axes.

:Here is a reasonable arrangement for this charged ring.

[[File:blue_graph.png|400px|thumb|Figure 2: defining a coordinate system about a charged ring]]

:Since all charge is concentrated upon the edge of the circle, we can consider our charge distribution to be invariant with respect to radial distance. However, we do see that our charge distribution is the function of theta.

2. Compute charge distribution function

:Let us call the charge distribution as <math display="inline">\sigma</math>

:We have a charge distributed on the edge of the circle, so

:<math> \sigma=\frac{Q}{2\pi} </math>

:where <math display="inline">Q</math> represents the charge of the ring.

:What this equation means is that the charge is uniformly distributed along each unit of angular length. In this way, it is called angular charge density.

3. Compute infinitesimal charge contribution

:Let us consider an infinitesimal section of the ring which contains exactly one point charge. The dimension of this section is given by <math display="inline">d\theta</math> which is the infinitesimal angular size. So, the infinitesimal charge contribution, <math display="inline">dQ</math>, is

:<math> dQ = \frac{Q}{2\pi}d\theta </math>

4. Compute infinitesimal electric field contribution

:Let us define some arbitrary location at which we are observing this ring of charge.

:<math> \vec{r} = x\hat{x}+y\hat{y} + z\hat{z} </math>

:in polar coordinates, we see this becomes

:<math> \vec{r} = Rcos(\theta)\hat{x} + Rsin(\theta)\hat{y} + z\hat{z} </math>.

:Recall that both sine and cosine are periodic with the period of <math>2\pi </math>. This will become important later.

:The magnitude of this vector is

:<math> |\vec{r}| = \sqrt{R^{2}+z^{2}} </math>

:owing to the usage of the Pythagorean trigonometric identity.

:Now, we have what we need to write the electric field vector contributed by each piece of the ring of charge. Let this vector field piece be <math display="inline">d\vec{E}</math>.

:<math> d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z}) </math>

5. Compute electric field vector

:So, now all that is left is to sum everything up. We are summing over the circumference of a circle, so our path is defined by <math display="inline">0\leq\theta\leq 2\pi </math>. Let us now set up our integral.

:<math>\vec{E}=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

:Since both sine and cosine are <math display="inline">2\pi</math> periodic functions, the <math display="inline">x</math> and <math display="inline">y</math> components of <math display="inline">\vec{E}</math> go to <math display="inline">\vec{0}</math>, which is very convenient. The result is

:<math> \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(z^{2}+R^{2})^{3/2}}\hat{z} </math>.

''Remark''

If the path we are interested in is over a half circle, third-circle or quarter-circle, or indeed any circular section, simply adjust the bounds of integration and the charge density to the ones required by the curve of interest.

===A Computational Model===

[https://trinket.io/glowscript/707d492e19 This simulation] shows the result of the computation for a ring composed of 2000 electrons. This is why the vector is pointing into the ring rather than out of the ring, which would happen for a ring composed of positively charged points.

==Examples==

===Easy===

1.) Find the function <math>E(x)</math> that represents the magnitude of the electric field along the center axis due to a uniformly charged ring of radius 0.04 m with total charge 8 C.

''Solution''

We can use the derived formula for the electric field due to a charged ring. Only the horizontal components of the electric field will remain as the rest will cancel out.

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{(8 C)*x}{((0.04 m)^2 + x^2)^\frac{3}{2}}</math>

===Middling===

2.) A uniformly charged ring of radius 10.0 cm has a total charge of 91.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

a. 1 cm

b. 5 cm

c. 30 cm

d. 100 cm

''Solution''

Use the derived formula, and convert the given distances into meters and the charge into Coulombs. Plug these into <math>x</math>

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{Q*x}{(x^2 + R^2)^\frac{3}{2}}</math>

a. 8.069e6 N/C

b. 29.3e6 N/C

c. 7.76e6 N/C

d. 8.06e5 N/C

===Difficult===

Find the electric field at a distance <math>z</math> along the center axis away from a uniformly charged semicircular ring of radius R and total charge Q.

''Solution''

We can tweak the same integral for a uniformly charged ring by changing the limits of integration.

<math>\vec{E}=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(0\hat{x}+2R\hat{y}+{\pi}z\hat{z})</math>

==History==

Added by AYESHA AHUJA SPRING 2020

The first known time when charged rings were utilized to describe electromagnetism was during Faraday’s discovery of electrical induction. To conduct his experiments, Faraday used a ring made of iron that was attached to a battery. When this makeshift circuit was turned on it manipulated a compass needle by deflecting it. He showcased an induced current in the ring.

==Connectedness==

Edited by AYESHA AHUJA Spring 2020

The idea of charge density is somewhat analogous to the idea of the mass density, which is useful in a variety of contexts, including the computation of the center of mass, the first and second moments of mass, which are useful in statics and rigid body dynamics. Instead of computing the uniform distribution of the mass of this ring, we are computing the uniform distribution of charge. This concept is also useful in visualizing what happens within a wire in a steady state circuit. The wire can be viewed to be a continuous length of rings of charge, which act as a channel through which electrons are transported, and that the electric field of these rings of charge pushes the electrons within the wire.

It is also important in the context of Maxwell's equations, specifically in Gauss's Law and in the Maxwell-Faraday equation, which are concerned with electron flux and induced fields, respectively.

== See also ==

[[Charged Rod]]

[[Charged Disk]]

[[Charged Spherical Shell]]

[[Charged Capacitor]]

==References==

"Electric Field on the Axis of a Ring of Charge". University of Delaware Physics Library. Adapted from Stephen Kevan's lecture on Electric Fields and Charge Distribution. April 8, 1996. http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

Chabay, R., & Sherwood, B. (2015). Matter and Interactions (4th ed., Vol. 2, pp. 597-599). Wiley.

All images produced by the author

==External Links==

https://www.youtube.com/watch?v=80mM3kSTZcE

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

[[Category:Electric Fields]]

Charged Ring

2020-05-05T20:34:56Z

Sydney:

==The Main Idea==

Charges may be arranged in a variety of ways. When points are uniformly distributed a finite distance away from some point, which we define as the origin, we call this curve as a ring. We can position a [[Point Charge]] at every point upon this ring. To analyze this continuous arrangement of charges, we say that their individual contributions are infinitesimally small, and then sum each of these contributions together to arrive at the electric field produced by this charge distribution. We can compute the net electric field of this charge distribution with Coulomb's Law and by applying integration principles. The ring field can also be used to calculate the electric field of a [[Charged Disk]].

===A Mathematical Model===

This mathematical model is based on the individual [[Electric Field]] contributions of a number of point charges, each of which is defined by

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}</math>.

'''Do not forget this equation'''. This equation must be memorized as it isn't given on the formula sheet.

We also say that this vector field is a member of a linear space of vectors. This is to say that we can apply the [[Superposition Principle]] meaning that we can sum any number of these electric field vectors and obtain another vector which is the electric field vector contributed by those charges. When we continuously sum all of the vectors produced by these charges, we get the electric field produced by the entire arrangement of charges.

[[File:red_ring.png|400px|thumb|Figure 1: diagram of a charged ring]]

The diagram above shows the electric field due to one infinitesimal piece of the ring, <math>dq</math>. In order to avoid rigorous computations, we can see that the electric field of the charges cancels out in the vertical direction. Only the horizontal component will remain. For an observation location that is on the symmetry axis as the ring, the other two components will be zero. If the observation location were off axis, then it would be very different requiring more math. Since each piece, <math>dq</math>, contributes a <math>d\vec{E}</math> of <math>\frac{1}{4\pi\epsilon_{0}}\frac{dq}{|\vec{d}|^{2}}\hat{d}</math>, we can compute the horizontal component by multiplying the magnitude of <math>d\vec{E}</math> with <math>\cos{θ}</math>.

<math>|d\vec{E}_{z}| = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\cos{θ} = \frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^2 + z^2}\frac{z}{(r^2 + z^2)^\frac{1}{2}}</math>

Now we integrate to sum all the electric field contributions of each infinitesimal <math>dq</math>.

<math>\vec{E}_{z} = \frac{1}{4\pi\epsilon_{0}}\frac{z}{(r^2 + z^2)^\frac{3}{2}}\int_{}^{} dq = \frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(r^2 + z^2)^\frac{3}{2}}</math>

As a reminder, this is the equation given on the formula sheet. It '''only''' works when the location at which the field is being measured is along the <math>{z}</math> axis. However, the equation for it being off axis is not given on the equation sheet. That requires a separate and more difficult integration. Typically, the only way this would be asked on a test is to set up an equation that would be integrated that requires you to find a new <math>dQ</math>.

'''Rigorous Derivation'''

1. Define shape characteristics of the ring

:The ring has some finite charge.

:We see that this arrangement is circular, so a coordinate system with which we can define radial and angular coordinates would be useful. Naturally, this would be the polar coordinate system.

We also see that all charge is uniformly distributed some finite distance R from the center of the ring. It would be useful to let the center of the ring be the origin of our coordinate axes.

Here is a reasonable arrangement for this charged ring.

[[File:blue_graph.png|400px|thumb|Figure 2: defining a coordinate system about a charged ring]]

Since all charge is concentrated upon the edge of the circle, we can consider our charge distribution to be invariant with respect to radial distance. However, we do see that our charge distribution is the function of theta.

#Compute charge distribution function

Let us call the charge distribution as <math display="inline">\sigma</math>

We have a charge distributed on the edge of the circle, so

<math> \sigma=\frac{Q}{2\pi} </math>

where <math display="inline">Q</math> represents the charge of the ring.

What this equation means is that the charge is uniformly distributed along each unit of angular length. In this way, it is called angular charge density.

#Compute infinitesimal charge contribution

Let us consider an infinitesimal section of the ring which contains exactly one point charge. The dimension of this section is given by <math display="inline">d\theta</math> which is the infinitesimal angular size. So, the infinitesimal charge contribution, <math display="inline">dQ</math>, is

<math> dQ = \frac{Q}{2\pi}d\theta </math>

#Compute infinitesimal electric field contribution

Let us define some arbitrary location at which we are observing this ring of charge.

<math> \vec{r} = x\hat{x}+y\hat{y} + z\hat{z} </math>

in polar coordinates, we see this becomes

<math> \vec{r} = Rcos(\theta)\hat{x} + Rsin(\theta)\hat{y} + z\hat{z} </math>.

Recall that both sine and cosine are periodic with the period of <math>2\pi </math>. This will become important later.

The magnitude of this vector is

<math> |\vec{r}| = \sqrt{R^{2}+z^{2}} </math>

owing to the usage of the Pythagorean trigonometric identity.

Now, we have what we need to write the electric field vector contributed by each piece of the ring of charge. Let this vector field piece be <math display="inline">d\vec{E}</math>.

<math> d\vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r}=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z}) </math>

#Compute electric field vector

So, now all that is left is to sum everything up. We are summing over the circumference of a circle, so our path is defined by <math display="inline">0\leq\theta\leq 2\pi </math>. Let us now set up our integral.

<math>\vec{E}=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{2\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

Since both sine and cosine are <math display="inline">2\pi</math> periodic functions, the <math display="inline">x</math> and <math display="inline">y</math> components of <math display="inline">\vec{E}</math> go to <math display="inline">\vec{0}</math>, which is very convenient. The result is

<math> \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{Qz}{(z^{2}+R^{2})^{3/2}}\hat{z} </math>.

#Remark

If the path we are interested in is over a half circle, third-circle or quarter-circle, or indeed any circular section, simply adjust the bounds of integration and the charge density to the ones required by the curve of interest.
===A Computational Model===

[https://trinket.io/glowscript/707d492e19 This simulation] shows the result of the computation for a ring composed of 2000 electrons. This is why the vector is pointing into the ring rather than out of the ring, which would happen for a ring composed of positively charged points.

==Examples==

===Easy===

1.) Find the function <math>E(x)</math> that represents the magnitude of the electric field along the center axis due to a uniformly charged ring of radius 0.04 m with total charge 8 C.

''Solution''

We can use the derived formula for the electric field due to a charged ring. Only the horizontal components of the electric field will remain as the rest will cancel out.

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{(8 C)*x}{((0.04 m)^2 + x^2)^\frac{3}{2}}</math>

===Middling===

2.) A uniformly charged ring of radius 10.0 cm has a total charge of 91.0 µC. Find the electric field on the axis of the ring at the following distances from the center of the ring.

a. 1 cm

b. 5 cm

c. 30 cm

d. 100 cm

''Solution''

Use the derived formula, and convert the given distances into meters and the charge into Coulombs. Plug these into <math>x</math>

<math>E(x) = \frac{1}{4\pi\epsilon_{0}}\frac{Q*x}{(x^2 + R^2)^\frac{3}{2}}</math>

a. 8.069e6 N/C

b. 29.3e6 N/C

c. 7.76e6 N/C

d. 8.06e5 N/C

===Difficult===

Find the electric field at a distance <math>z</math> along the center axis away from a uniformly charged semicircular ring of radius R and total charge Q.

''Solution''

We can tweak the same integral for a uniformly charged ring by changing the limits of integration.

<math>\vec{E}=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{q}{|\vec{r}|^{2}}\hat{r} d\theta=\int_{0}^{\pi}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(Rcos(\theta)\hat{x}+Rsin(\theta)\hat{y}+z\hat{z})d\theta</math>

<math>\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{\pi(R^{2}+z^{2})^{3/2}}(0\hat{x}+2R\hat{y}+{\pi}z\hat{z})</math>

==History==

Added by AYESHA AHUJA SPRING 2020

The first known time when charged rings were utilized to describe electromagnetism was during Faraday’s discovery of electrical induction. To conduct his experiments, Faraday used a ring made of iron that was attached to a battery. When this makeshift circuit was turned on it manipulated a compass needle by deflecting it. He showcased an induced current in the ring.

==Connectedness==

Edited by AYESHA AHUJA Spring 2020

The idea of charge density is somewhat analogous to the idea of the mass density, which is useful in a variety of contexts, including the computation of the center of mass, the first and second moments of mass, which are useful in statics and rigid body dynamics. Instead of computing the uniform distribution of the mass of this ring, we are computing the uniform distribution of charge. This concept is also useful in visualizing what happens within a wire in a steady state circuit. The wire can be viewed to be a continuous length of rings of charge, which act as a channel through which electrons are transported, and that the electric field of these rings of charge pushes the electrons within the wire.

It is also important in the context of Maxwell's equations, specifically in Gauss's Law and in the Maxwell-Faraday equation, which are concerned with electron flux and induced fields, respectively.

== See also ==

[[Charged Rod]]

[[Charged Disk]]

[[Charged Spherical Shell]]

[[Charged Capacitor]]

==References==

"Electric Field on the Axis of a Ring of Charge". University of Delaware Physics Library. Adapted from Stephen Kevan's lecture on Electric Fields and Charge Distribution. April 8, 1996. http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

Chabay, R., & Sherwood, B. (2015). Matter and Interactions (4th ed., Vol. 2, pp. 597-599). Wiley.

All images produced by the author

==External Links==

https://www.youtube.com/watch?v=80mM3kSTZcE

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

[[Category:Electric Fields]]

Field of a Charged Rod

2020-05-05T20:25:16Z

Sydney: /* Connectedness */

== The Main Idea ==

Previously, we've learned about the electric field of a point particle. Often, when analyzing physical systems, it is the case that we're unable to analyze each individual particle that composes an object and need to therefore generalize collections of particles into shapes (in this case, a rod) whereby the mathematics corresponding to electric field calculations can be simplified. This can essentially be done by adding up the contributions to the electric field made by parts of an object, approximating each part of an object as a point charge.

Some objects, such as rods, can be modeled as a uniformly charged object in order to calculate the electric field at some observation location. The following wiki page provides an overview of electric fields created by uniformly charged thin rods, briefly presenting its inception, some specific cases, and proof of concept experiments. The case of a uniformly charged thin rod is a fundamental example of electric field patterns and calculations within physics. Its implications can be applied to other charged objects, such as rings, disks, and spheres.

In practical situations, objects have charges spread all over their surface. When going about calculating the electric fields of these objects, we can either use one of two processes: numerical summation or integration, or dividing an object into many pieces and summing the individual pieces' electric field contributions. As with point charges, the direction of the field is determined by the sign of the object's charge (positive-points away, negative-points toward) and the size of the field is determined by the observation distance and the magnitude of the object's charge.

The process of finding the electric field due to charge distributed over an object has four steps:

1. Divide the charged object into small pieces. Make a diagram and draw the electric field <math>\Delta \vec{E}</math> contributed by one of the pieces.

2. Choose an origin and the axes. Write an algebraic expression for the electric field <math>\Delta \vec{E}</math> due to one piece.

3. Add up the contributions of all pieces, either numerically or symbolically.

4. Check that the result is physically correct.

=== A Mathematical Model ===
The process of calculating a uniformly charged rod's electric field is tedious, but breaking the process down into several steps makes the task at hand easier. Consider a uniformly charged thin rod of length <math>L</math> and positive charge <math>Q</math> centered on and lying along the x-axis. The rod is being observed from above at a point on the y-axis.

'''Step 1: Divide the Distribution into Pieces; Draw <math>\Delta \vec{E}</math>'''

Imagine dividing the rod into a series of very thin slices, each with the same charge <math>\Delta Q</math>. This charge <math>\Delta Q</math> is a small part of the overall charge. Picture it as a point charge. Each slice contributes its own electric field, <math>\Delta E</math>. Summing all these individual slices of <math>E</math> gives you the total electric field of the rod. This process is the same as taking an integral, as each thickness approaches 0 and the the number of slices approaches infinity. Note that in this example, the variable that is changing for each slice is its x-coordinate.

'''Step 2: Write an Expression for the Electric Field Due to One Piece'''

The second step is to write a mathematical expression for the field <math>\Delta E</math> contributed by a single slice of the rod. We use the formula of the electric field for a point charge because we are imagining each slice as a point charge. First, determine <math>r</math>, the vector pointing from the source to the observation location. For our example, this is <math> r = obs - source = <0,y,0> - < x,0,0> = <-x,y,0></math>. Now use this to calculate the magnitude and direction of <math>r</math>. So <math>|\vec{r}| = \sqrt{(-x)^2 + y^2} = \sqrt{x^2 + y^2}</math> and <math>\hat{r} = \frac{\vec{r}}{\hat{r}} = \frac{< -x,y,0>}{\sqrt{x^2 + y^2}} </math>. <math> \hat{r}</math> is the vector portion of the expression for the field. The scalar portion is <math> \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{|\vec{r}|^2}</math>. Thus the expression for one slice of the rod is:
<math> \Delta \vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{(\sqrt{x^2+y^2})^{3/2}} \cdot < -x,y,0> </math>.

'''Determining <math>\Delta Q</math> and the integration variable'''

In the first step, we determined that the changing variable for this rod was its x-coordinate. This means the integration variable is <math> dx</math>. We need to put this integration variable into our expression for the electric field. More specifically, we need to express <math>\Delta Q</math> in terms of the integration variable. Recall that the rod is uniformly charged, so the charge on any single slice of it is: <math>
\Delta Q = (\frac{\Delta x}{L})\cdot Q</math>. This quantity can also be expressed in terms of the charge density.

'''Expression for <math> \Delta \vec{E}</math>

Substitute the expression for the integration variable into the formula for the electric field of one slice. Separating the equation into separate x and y components, we get <math> \Delta \vec{E_x} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-x}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math> and <math> \Delta \vec{E_y} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math>. Note that we have replaced <math> \Delta x </math> with <math> dx</math> in preparation for integration.

'''Step 3: Add Up the Contributions of All the Pieces'''

The third step is to sum all of our slices. We can go about this in two ways. One way is with numerical summation, or separating the object into a finite number of small pieces, calculating the individual contributing electric fields, and then summing them. Another, more precise method is to integrate. Most of the work of finding the field of a uniformly charged object is setting up this integral. If you have reached the correct expression to integrate, the rest is simple math. The bounds for integration are the coordinates of the start and stop of the rod. In this example the bounds are from <math>-L/2</math> to <math>+L/2</math>. So the expression is <math> \int\limits_{-L/2}^{L/2}\ \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx. </math> Solving this gives the final expression <math> E_y = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{x} \cdot \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. Note that the field parallel to the x axis is zero. This can be observed due to the symmetry of the problem.
This equation can be written more generally as <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \cdot \frac{1}{(\sqrt{r^2+ (L/2)^2})} </math> where r represents the distance from the rod to the observation location.

'''Step 4:Checking the Result'''

Finally, the fourth step is to check the result. The units should be the same as the units of the expression for the electric field for a single point particle <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} </math>.
Our answer has the right units, since <math> \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. has the same units of <math>\frac{Q}{r^2}</math>
Is the direction qualitatively correct? We have the electric field pointing straight away from the midpoint of the rod, which is correct, given the symmetry of the situation. The vertical component of the electric field should indeed be zero.

=== A Computational Model ===

'''Finding the Electric Field from a Rod with Code'''

Here is some code that you can run which shows the electric field vector
at a given distance from the rod along its length. The rod is shown as a
series of green balls to help emphasize that when using the numerical
integrations mentioned on this page, you are measuring the field produced
by discrete parts of the rod being analyzed.

Notice the edge-effects of the electric field of the rod. For reasons
discussed above, if we used the long rod approximation (L>>d), these
effects would be negligible.

[http://www.glowscript.org/#/user/yoderlukas/folder/Public/program/ElectricFieldAlongRodLength Click Here to Run the Code]

==Examples==

Although this is not a very difficult topic, some reasonably difficult conceptual questions can be asked about it.

===Simple===

[[Image:LukasYoderNo.jpg|400px|center|thumb|Figure1: Problem 1]]

===Middling===

[[Image:LukasYoderMaybe.jpg|400px|center|thumb|Figure 2: Problem 2]]

===Difficult===

[[Image:LukasYoderYes.jpg|400px|center|thumb|Figure 3: Problem 3]]

== Connectedness ==
The following two DIY experiments are shown to better understand and visualize the physical consequences of electric fields.

'''Charged Rod and Aluminum Can'''

In our first example we set up an experiment using two charged rods placed to the left and right of an aluminum can, distanced by a length d. If one of the rods is positively charged and the other is negatively charged, what will the can do? Because the positively charged rod induces a negative charge on the left side of the can, creating an attractive force between the rod and the can, and the negatively charged rod induces an equal positive charge on the right side of the can, which creates an attractive force between the can and that rod, the net force on the can is zero. Thus the can will stay still. The setup is depicted in the image below.

[[File:plusq.png|300px|center|thumb|Figure 4: apparatus diagram with +q]]

Next, lets consider a setup but with both rods having equal positive charge, as shown in the image below. What will the can do in this situation?

[[File:minusq.png|300px|center|thumb|Figure 5: apparatus diagram with -q]]

Again the can will also stay still, but this time it is because the polarization force between two objects is always attractive.

So in what scenario will the can move, and what time of movement will the can exhibit? Considering the first setup, imagine this time we initially touch the negatively charged rod and the can for a brief moment. Holding the rods at equal distance on either side of the can, the can will now roll toward the positively charged rod. This is because the can acquires a net negative charge after being touched, so it is then attracted to the positively charged rod.

'''Charged Rod and Pith Ball'''

Another DIY experiment to visualize the effects of an electric field are shown in the video embedded below, which depicts the interaction between an initially neutral pith ball hanging on a string from a stand, and a charged rod.
[http://www.youtube.com/watch?v=aeiqw81kGio Interaction between a Charged Rod and Pith Ball]

== History ==

Physicists and scientists make use of electric fields and charged objects all the time. Many times, we may need to know which objects are contributing how much charge in certain areas. Charged objects may attract or repel (depending on the signs of their charge), so we often need to know how objects will interact with each other based on their charges. The phenomenon of this interaction, or electric force between charged particles, was finally confirmed and stated as a law in 1785 by French physicist Charles-Augustin de Coulomb, hence "Coulomb's Law."

== See also ==

The equation for the electric field of a charged rod was derived from the equation for an electric field of a charged particle. See the article "[[Electric Field]]" for more information.

=== Further Reading ===

The page on electric fields: [[Electric Field]]

=== External Links ===

http://online.cctt.org/physicslab/content/phyapc/lessonnotes/Efields/EchargedRods.asp

https://pages.uncc.edu/phys2102/online-lectures/chapter-02-electric-field/2-4-electric-field-of-charge-distributions/example-1-electric-field-of-a-charged-rod-along-its-axis/

http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ContinuousChargedRod.xml

== References ==

https://www.youtube.com/watch?v=BBWd0zUe0mI

(For the above reference, I chose to follow the textbook's method in not defining the charge distribution and assuming it was constant, though this was helpful in figuring out a better way to introduce it.)

Chabay and Sherwood: Matter and Interactions, Fourth Edition, Chapter 15

All figures created by author
[[Category: Electric Field]]

Field of a Charged Rod

2020-05-05T20:24:35Z

Sydney: /* A Computational Model */

== The Main Idea ==

Previously, we've learned about the electric field of a point particle. Often, when analyzing physical systems, it is the case that we're unable to analyze each individual particle that composes an object and need to therefore generalize collections of particles into shapes (in this case, a rod) whereby the mathematics corresponding to electric field calculations can be simplified. This can essentially be done by adding up the contributions to the electric field made by parts of an object, approximating each part of an object as a point charge.

Some objects, such as rods, can be modeled as a uniformly charged object in order to calculate the electric field at some observation location. The following wiki page provides an overview of electric fields created by uniformly charged thin rods, briefly presenting its inception, some specific cases, and proof of concept experiments. The case of a uniformly charged thin rod is a fundamental example of electric field patterns and calculations within physics. Its implications can be applied to other charged objects, such as rings, disks, and spheres.

In practical situations, objects have charges spread all over their surface. When going about calculating the electric fields of these objects, we can either use one of two processes: numerical summation or integration, or dividing an object into many pieces and summing the individual pieces' electric field contributions. As with point charges, the direction of the field is determined by the sign of the object's charge (positive-points away, negative-points toward) and the size of the field is determined by the observation distance and the magnitude of the object's charge.

The process of finding the electric field due to charge distributed over an object has four steps:

1. Divide the charged object into small pieces. Make a diagram and draw the electric field <math>\Delta \vec{E}</math> contributed by one of the pieces.

2. Choose an origin and the axes. Write an algebraic expression for the electric field <math>\Delta \vec{E}</math> due to one piece.

3. Add up the contributions of all pieces, either numerically or symbolically.

4. Check that the result is physically correct.

=== A Mathematical Model ===
The process of calculating a uniformly charged rod's electric field is tedious, but breaking the process down into several steps makes the task at hand easier. Consider a uniformly charged thin rod of length <math>L</math> and positive charge <math>Q</math> centered on and lying along the x-axis. The rod is being observed from above at a point on the y-axis.

'''Step 1: Divide the Distribution into Pieces; Draw <math>\Delta \vec{E}</math>'''

Imagine dividing the rod into a series of very thin slices, each with the same charge <math>\Delta Q</math>. This charge <math>\Delta Q</math> is a small part of the overall charge. Picture it as a point charge. Each slice contributes its own electric field, <math>\Delta E</math>. Summing all these individual slices of <math>E</math> gives you the total electric field of the rod. This process is the same as taking an integral, as each thickness approaches 0 and the the number of slices approaches infinity. Note that in this example, the variable that is changing for each slice is its x-coordinate.

'''Step 2: Write an Expression for the Electric Field Due to One Piece'''

The second step is to write a mathematical expression for the field <math>\Delta E</math> contributed by a single slice of the rod. We use the formula of the electric field for a point charge because we are imagining each slice as a point charge. First, determine <math>r</math>, the vector pointing from the source to the observation location. For our example, this is <math> r = obs - source = <0,y,0> - < x,0,0> = <-x,y,0></math>. Now use this to calculate the magnitude and direction of <math>r</math>. So <math>|\vec{r}| = \sqrt{(-x)^2 + y^2} = \sqrt{x^2 + y^2}</math> and <math>\hat{r} = \frac{\vec{r}}{\hat{r}} = \frac{< -x,y,0>}{\sqrt{x^2 + y^2}} </math>. <math> \hat{r}</math> is the vector portion of the expression for the field. The scalar portion is <math> \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{|\vec{r}|^2}</math>. Thus the expression for one slice of the rod is:
<math> \Delta \vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{(\sqrt{x^2+y^2})^{3/2}} \cdot < -x,y,0> </math>.

'''Determining <math>\Delta Q</math> and the integration variable'''

In the first step, we determined that the changing variable for this rod was its x-coordinate. This means the integration variable is <math> dx</math>. We need to put this integration variable into our expression for the electric field. More specifically, we need to express <math>\Delta Q</math> in terms of the integration variable. Recall that the rod is uniformly charged, so the charge on any single slice of it is: <math>
\Delta Q = (\frac{\Delta x}{L})\cdot Q</math>. This quantity can also be expressed in terms of the charge density.

'''Expression for <math> \Delta \vec{E}</math>

Substitute the expression for the integration variable into the formula for the electric field of one slice. Separating the equation into separate x and y components, we get <math> \Delta \vec{E_x} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-x}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math> and <math> \Delta \vec{E_y} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math>. Note that we have replaced <math> \Delta x </math> with <math> dx</math> in preparation for integration.

'''Step 3: Add Up the Contributions of All the Pieces'''

The third step is to sum all of our slices. We can go about this in two ways. One way is with numerical summation, or separating the object into a finite number of small pieces, calculating the individual contributing electric fields, and then summing them. Another, more precise method is to integrate. Most of the work of finding the field of a uniformly charged object is setting up this integral. If you have reached the correct expression to integrate, the rest is simple math. The bounds for integration are the coordinates of the start and stop of the rod. In this example the bounds are from <math>-L/2</math> to <math>+L/2</math>. So the expression is <math> \int\limits_{-L/2}^{L/2}\ \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx. </math> Solving this gives the final expression <math> E_y = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{x} \cdot \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. Note that the field parallel to the x axis is zero. This can be observed due to the symmetry of the problem.
This equation can be written more generally as <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \cdot \frac{1}{(\sqrt{r^2+ (L/2)^2})} </math> where r represents the distance from the rod to the observation location.

'''Step 4:Checking the Result'''

Finally, the fourth step is to check the result. The units should be the same as the units of the expression for the electric field for a single point particle <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} </math>.
Our answer has the right units, since <math> \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. has the same units of <math>\frac{Q}{r^2}</math>
Is the direction qualitatively correct? We have the electric field pointing straight away from the midpoint of the rod, which is correct, given the symmetry of the situation. The vertical component of the electric field should indeed be zero.

=== A Computational Model ===

'''Finding the Electric Field from a Rod with Code'''

Here is some code that you can run which shows the electric field vector
at a given distance from the rod along its length. The rod is shown as a
series of green balls to help emphasize that when using the numerical
integrations mentioned on this page, you are measuring the field produced
by discrete parts of the rod being analyzed.

Notice the edge-effects of the electric field of the rod. For reasons
discussed above, if we used the long rod approximation (L>>d), these
effects would be negligible.

[http://www.glowscript.org/#/user/yoderlukas/folder/Public/program/ElectricFieldAlongRodLength Click Here to Run the Code]

==Examples==

Although this is not a very difficult topic, some reasonably difficult conceptual questions can be asked about it.

===Simple===

[[Image:LukasYoderNo.jpg|400px|center|thumb|Figure1: Problem 1]]

===Middling===

[[Image:LukasYoderMaybe.jpg|400px|center|thumb|Figure 2: Problem 2]]

===Difficult===

[[Image:LukasYoderYes.jpg|400px|center|thumb|Figure 3: Problem 3]]

== Connectedness ==
The following two DIY experiments are shown to better understand and visualize the physical consequences of electric fields.

===Charged Rod and Aluminum Can===

In our first example we set up an experiment using two charged rods placed to the left and right of an aluminum can, distanced by a length d. If one of the rods is positively charged and the other is negatively charged, what will the can do? Because the positively charged rod induces a negative charge on the left side of the can, creating an attractive force between the rod and the can, and the negatively charged rod induces an equal positive charge on the right side of the can, which creates an attractive force between the can and that rod, the net force on the can is zero. Thus the can will stay still. The setup is depicted in the image below.

[[File:plusq.png|300px|center|thumb|Figure 4: apparatus diagram with +q]]

Next, lets consider a setup but with both rods having equal positive charge, as shown in the image below. What will the can do in this situation?

[[File:minusq.png|300px|center|thumb|Figure 5: apparatus diagram with -q]]

Again the can will also stay still, but this time it is because the polarization force between two objects is always attractive.

So in what scenario will the can move, and what time of movement will the can exhibit? Considering the first setup, imagine this time we initially touch the negatively charged rod and the can for a brief moment. Holding the rods at equal distance on either side of the can, the can will now roll toward the positively charged rod. This is because the can acquires a net negative charge after being touched, so it is then attracted to the positively charged rod.

===Charged Rod and Pith Ball===

Another DIY experiment to visualize the effects of an electric field are shown in the video embedded below, which depicts the interaction between an initially neutral pith ball hanging on a string from a stand, and a charged rod.
[http://www.youtube.com/watch?v=aeiqw81kGio Interaction between a Charged Rod and Pith Ball]

== History ==

Physicists and scientists make use of electric fields and charged objects all the time. Many times, we may need to know which objects are contributing how much charge in certain areas. Charged objects may attract or repel (depending on the signs of their charge), so we often need to know how objects will interact with each other based on their charges. The phenomenon of this interaction, or electric force between charged particles, was finally confirmed and stated as a law in 1785 by French physicist Charles-Augustin de Coulomb, hence "Coulomb's Law."

== See also ==

The equation for the electric field of a charged rod was derived from the equation for an electric field of a charged particle. See the article "[[Electric Field]]" for more information.

=== Further Reading ===

The page on electric fields: [[Electric Field]]

=== External Links ===

http://online.cctt.org/physicslab/content/phyapc/lessonnotes/Efields/EchargedRods.asp

https://pages.uncc.edu/phys2102/online-lectures/chapter-02-electric-field/2-4-electric-field-of-charge-distributions/example-1-electric-field-of-a-charged-rod-along-its-axis/

http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ContinuousChargedRod.xml

== References ==

https://www.youtube.com/watch?v=BBWd0zUe0mI

(For the above reference, I chose to follow the textbook's method in not defining the charge distribution and assuming it was constant, though this was helpful in figuring out a better way to introduce it.)

Chabay and Sherwood: Matter and Interactions, Fourth Edition, Chapter 15

All figures created by author
[[Category: Electric Field]]

Field of a Charged Rod

2020-05-05T20:24:18Z

Sydney: /* A Mathematical Model */

== The Main Idea ==

Previously, we've learned about the electric field of a point particle. Often, when analyzing physical systems, it is the case that we're unable to analyze each individual particle that composes an object and need to therefore generalize collections of particles into shapes (in this case, a rod) whereby the mathematics corresponding to electric field calculations can be simplified. This can essentially be done by adding up the contributions to the electric field made by parts of an object, approximating each part of an object as a point charge.

Some objects, such as rods, can be modeled as a uniformly charged object in order to calculate the electric field at some observation location. The following wiki page provides an overview of electric fields created by uniformly charged thin rods, briefly presenting its inception, some specific cases, and proof of concept experiments. The case of a uniformly charged thin rod is a fundamental example of electric field patterns and calculations within physics. Its implications can be applied to other charged objects, such as rings, disks, and spheres.

In practical situations, objects have charges spread all over their surface. When going about calculating the electric fields of these objects, we can either use one of two processes: numerical summation or integration, or dividing an object into many pieces and summing the individual pieces' electric field contributions. As with point charges, the direction of the field is determined by the sign of the object's charge (positive-points away, negative-points toward) and the size of the field is determined by the observation distance and the magnitude of the object's charge.

The process of finding the electric field due to charge distributed over an object has four steps:

1. Divide the charged object into small pieces. Make a diagram and draw the electric field <math>\Delta \vec{E}</math> contributed by one of the pieces.

2. Choose an origin and the axes. Write an algebraic expression for the electric field <math>\Delta \vec{E}</math> due to one piece.

3. Add up the contributions of all pieces, either numerically or symbolically.

4. Check that the result is physically correct.

=== A Mathematical Model ===
The process of calculating a uniformly charged rod's electric field is tedious, but breaking the process down into several steps makes the task at hand easier. Consider a uniformly charged thin rod of length <math>L</math> and positive charge <math>Q</math> centered on and lying along the x-axis. The rod is being observed from above at a point on the y-axis.

'''Step 1: Divide the Distribution into Pieces; Draw <math>\Delta \vec{E}</math>'''

Imagine dividing the rod into a series of very thin slices, each with the same charge <math>\Delta Q</math>. This charge <math>\Delta Q</math> is a small part of the overall charge. Picture it as a point charge. Each slice contributes its own electric field, <math>\Delta E</math>. Summing all these individual slices of <math>E</math> gives you the total electric field of the rod. This process is the same as taking an integral, as each thickness approaches 0 and the the number of slices approaches infinity. Note that in this example, the variable that is changing for each slice is its x-coordinate.

'''Step 2: Write an Expression for the Electric Field Due to One Piece'''

The second step is to write a mathematical expression for the field <math>\Delta E</math> contributed by a single slice of the rod. We use the formula of the electric field for a point charge because we are imagining each slice as a point charge. First, determine <math>r</math>, the vector pointing from the source to the observation location. For our example, this is <math> r = obs - source = <0,y,0> - < x,0,0> = <-x,y,0></math>. Now use this to calculate the magnitude and direction of <math>r</math>. So <math>|\vec{r}| = \sqrt{(-x)^2 + y^2} = \sqrt{x^2 + y^2}</math> and <math>\hat{r} = \frac{\vec{r}}{\hat{r}} = \frac{< -x,y,0>}{\sqrt{x^2 + y^2}} </math>. <math> \hat{r}</math> is the vector portion of the expression for the field. The scalar portion is <math> \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{|\vec{r}|^2}</math>. Thus the expression for one slice of the rod is:
<math> \Delta \vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{(\sqrt{x^2+y^2})^{3/2}} \cdot < -x,y,0> </math>.

'''Determining <math>\Delta Q</math> and the integration variable'''

In the first step, we determined that the changing variable for this rod was its x-coordinate. This means the integration variable is <math> dx</math>. We need to put this integration variable into our expression for the electric field. More specifically, we need to express <math>\Delta Q</math> in terms of the integration variable. Recall that the rod is uniformly charged, so the charge on any single slice of it is: <math>
\Delta Q = (\frac{\Delta x}{L})\cdot Q</math>. This quantity can also be expressed in terms of the charge density.

'''Expression for <math> \Delta \vec{E}</math>

Substitute the expression for the integration variable into the formula for the electric field of one slice. Separating the equation into separate x and y components, we get <math> \Delta \vec{E_x} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-x}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math> and <math> \Delta \vec{E_y} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math>. Note that we have replaced <math> \Delta x </math> with <math> dx</math> in preparation for integration.

'''Step 3: Add Up the Contributions of All the Pieces'''

The third step is to sum all of our slices. We can go about this in two ways. One way is with numerical summation, or separating the object into a finite number of small pieces, calculating the individual contributing electric fields, and then summing them. Another, more precise method is to integrate. Most of the work of finding the field of a uniformly charged object is setting up this integral. If you have reached the correct expression to integrate, the rest is simple math. The bounds for integration are the coordinates of the start and stop of the rod. In this example the bounds are from <math>-L/2</math> to <math>+L/2</math>. So the expression is <math> \int\limits_{-L/2}^{L/2}\ \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx. </math> Solving this gives the final expression <math> E_y = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{x} \cdot \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. Note that the field parallel to the x axis is zero. This can be observed due to the symmetry of the problem.
This equation can be written more generally as <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \cdot \frac{1}{(\sqrt{r^2+ (L/2)^2})} </math> where r represents the distance from the rod to the observation location.

'''Step 4:Checking the Result'''

Finally, the fourth step is to check the result. The units should be the same as the units of the expression for the electric field for a single point particle <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} </math>.
Our answer has the right units, since <math> \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. has the same units of <math>\frac{Q}{r^2}</math>
Is the direction qualitatively correct? We have the electric field pointing straight away from the midpoint of the rod, which is correct, given the symmetry of the situation. The vertical component of the electric field should indeed be zero.

== A Computational Model ==

'''Finding the Electric Field from a Rod with Code'''

Here is some code that you can run which shows the electric field vector
at a given distance from the rod along its length. The rod is shown as a
series of green balls to help emphasize that when using the numerical
integrations mentioned on this page, you are measuring the field produced
by discrete parts of the rod being analyzed.

Notice the edge-effects of the electric field of the rod. For reasons
discussed above, if we used the long rod approximation (L>>d), these
effects would be negligible.

[http://www.glowscript.org/#/user/yoderlukas/folder/Public/program/ElectricFieldAlongRodLength Click Here to Run the Code]

==Examples==

Although this is not a very difficult topic, some reasonably difficult conceptual questions can be asked about it.

===Simple===

[[Image:LukasYoderNo.jpg|400px|center|thumb|Figure1: Problem 1]]

===Middling===

[[Image:LukasYoderMaybe.jpg|400px|center|thumb|Figure 2: Problem 2]]

===Difficult===

[[Image:LukasYoderYes.jpg|400px|center|thumb|Figure 3: Problem 3]]

== Connectedness ==
The following two DIY experiments are shown to better understand and visualize the physical consequences of electric fields.

===Charged Rod and Aluminum Can===

In our first example we set up an experiment using two charged rods placed to the left and right of an aluminum can, distanced by a length d. If one of the rods is positively charged and the other is negatively charged, what will the can do? Because the positively charged rod induces a negative charge on the left side of the can, creating an attractive force between the rod and the can, and the negatively charged rod induces an equal positive charge on the right side of the can, which creates an attractive force between the can and that rod, the net force on the can is zero. Thus the can will stay still. The setup is depicted in the image below.

[[File:plusq.png|300px|center|thumb|Figure 4: apparatus diagram with +q]]

Next, lets consider a setup but with both rods having equal positive charge, as shown in the image below. What will the can do in this situation?

[[File:minusq.png|300px|center|thumb|Figure 5: apparatus diagram with -q]]

Again the can will also stay still, but this time it is because the polarization force between two objects is always attractive.

So in what scenario will the can move, and what time of movement will the can exhibit? Considering the first setup, imagine this time we initially touch the negatively charged rod and the can for a brief moment. Holding the rods at equal distance on either side of the can, the can will now roll toward the positively charged rod. This is because the can acquires a net negative charge after being touched, so it is then attracted to the positively charged rod.

===Charged Rod and Pith Ball===

Another DIY experiment to visualize the effects of an electric field are shown in the video embedded below, which depicts the interaction between an initially neutral pith ball hanging on a string from a stand, and a charged rod.
[http://www.youtube.com/watch?v=aeiqw81kGio Interaction between a Charged Rod and Pith Ball]

== History ==

Physicists and scientists make use of electric fields and charged objects all the time. Many times, we may need to know which objects are contributing how much charge in certain areas. Charged objects may attract or repel (depending on the signs of their charge), so we often need to know how objects will interact with each other based on their charges. The phenomenon of this interaction, or electric force between charged particles, was finally confirmed and stated as a law in 1785 by French physicist Charles-Augustin de Coulomb, hence "Coulomb's Law."

== See also ==

The equation for the electric field of a charged rod was derived from the equation for an electric field of a charged particle. See the article "[[Electric Field]]" for more information.

=== Further Reading ===

The page on electric fields: [[Electric Field]]

=== External Links ===

http://online.cctt.org/physicslab/content/phyapc/lessonnotes/Efields/EchargedRods.asp

https://pages.uncc.edu/phys2102/online-lectures/chapter-02-electric-field/2-4-electric-field-of-charge-distributions/example-1-electric-field-of-a-charged-rod-along-its-axis/

http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ContinuousChargedRod.xml

== References ==

https://www.youtube.com/watch?v=BBWd0zUe0mI

(For the above reference, I chose to follow the textbook's method in not defining the charge distribution and assuming it was constant, though this was helpful in figuring out a better way to introduce it.)

Chabay and Sherwood: Matter and Interactions, Fourth Edition, Chapter 15

All figures created by author
[[Category: Electric Field]]

Field of a Charged Rod

2020-05-05T20:23:55Z

Sydney: /* A Computational Model */

== The Main Idea ==

Previously, we've learned about the electric field of a point particle. Often, when analyzing physical systems, it is the case that we're unable to analyze each individual particle that composes an object and need to therefore generalize collections of particles into shapes (in this case, a rod) whereby the mathematics corresponding to electric field calculations can be simplified. This can essentially be done by adding up the contributions to the electric field made by parts of an object, approximating each part of an object as a point charge.

Some objects, such as rods, can be modeled as a uniformly charged object in order to calculate the electric field at some observation location. The following wiki page provides an overview of electric fields created by uniformly charged thin rods, briefly presenting its inception, some specific cases, and proof of concept experiments. The case of a uniformly charged thin rod is a fundamental example of electric field patterns and calculations within physics. Its implications can be applied to other charged objects, such as rings, disks, and spheres.

In practical situations, objects have charges spread all over their surface. When going about calculating the electric fields of these objects, we can either use one of two processes: numerical summation or integration, or dividing an object into many pieces and summing the individual pieces' electric field contributions. As with point charges, the direction of the field is determined by the sign of the object's charge (positive-points away, negative-points toward) and the size of the field is determined by the observation distance and the magnitude of the object's charge.

The process of finding the electric field due to charge distributed over an object has four steps:

1. Divide the charged object into small pieces. Make a diagram and draw the electric field <math>\Delta \vec{E}</math> contributed by one of the pieces.

2. Choose an origin and the axes. Write an algebraic expression for the electric field <math>\Delta \vec{E}</math> due to one piece.

3. Add up the contributions of all pieces, either numerically or symbolically.

4. Check that the result is physically correct.

== A Mathematical Model ==
The process of calculating a uniformly charged rod's electric field is tedious, but breaking the process down into several steps makes the task at hand easier. Consider a uniformly charged thin rod of length <math>L</math> and positive charge <math>Q</math> centered on and lying along the x-axis. The rod is being observed from above at a point on the y-axis.

'''Step 1: Divide the Distribution into Pieces; Draw <math>\Delta \vec{E}</math>'''

Imagine dividing the rod into a series of very thin slices, each with the same charge <math>\Delta Q</math>. This charge <math>\Delta Q</math> is a small part of the overall charge. Picture it as a point charge. Each slice contributes its own electric field, <math>\Delta E</math>. Summing all these individual slices of <math>E</math> gives you the total electric field of the rod. This process is the same as taking an integral, as each thickness approaches 0 and the the number of slices approaches infinity. Note that in this example, the variable that is changing for each slice is its x-coordinate.

'''Step 2: Write an Expression for the Electric Field Due to One Piece'''

The second step is to write a mathematical expression for the field <math>\Delta E</math> contributed by a single slice of the rod. We use the formula of the electric field for a point charge because we are imagining each slice as a point charge. First, determine <math>r</math>, the vector pointing from the source to the observation location. For our example, this is <math> r = obs - source = <0,y,0> - < x,0,0> = <-x,y,0></math>. Now use this to calculate the magnitude and direction of <math>r</math>. So <math>|\vec{r}| = \sqrt{(-x)^2 + y^2} = \sqrt{x^2 + y^2}</math> and <math>\hat{r} = \frac{\vec{r}}{\hat{r}} = \frac{< -x,y,0>}{\sqrt{x^2 + y^2}} </math>. <math> \hat{r}</math> is the vector portion of the expression for the field. The scalar portion is <math> \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{|\vec{r}|^2}</math>. Thus the expression for one slice of the rod is:
<math> \Delta \vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{(\sqrt{x^2+y^2})^{3/2}} \cdot < -x,y,0> </math>.

'''Determining <math>\Delta Q</math> and the integration variable'''

In the first step, we determined that the changing variable for this rod was its x-coordinate. This means the integration variable is <math> dx</math>. We need to put this integration variable into our expression for the electric field. More specifically, we need to express <math>\Delta Q</math> in terms of the integration variable. Recall that the rod is uniformly charged, so the charge on any single slice of it is: <math>
\Delta Q = (\frac{\Delta x}{L})\cdot Q</math>. This quantity can also be expressed in terms of the charge density.

'''Expression for <math> \Delta \vec{E}</math>

Substitute the expression for the integration variable into the formula for the electric field of one slice. Separating the equation into separate x and y components, we get <math> \Delta \vec{E_x} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-x}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math> and <math> \Delta \vec{E_y} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math>. Note that we have replaced <math> \Delta x </math> with <math> dx</math> in preparation for integration.

'''Step 3: Add Up the Contributions of All the Pieces'''

The third step is to sum all of our slices. We can go about this in two ways. One way is with numerical summation, or separating the object into a finite number of small pieces, calculating the individual contributing electric fields, and then summing them. Another, more precise method is to integrate. Most of the work of finding the field of a uniformly charged object is setting up this integral. If you have reached the correct expression to integrate, the rest is simple math. The bounds for integration are the coordinates of the start and stop of the rod. In this example the bounds are from <math>-L/2</math> to <math>+L/2</math>. So the expression is <math> \int\limits_{-L/2}^{L/2}\ \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx. </math> Solving this gives the final expression <math> E_y = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{x} \cdot \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. Note that the field parallel to the x axis is zero. This can be observed due to the symmetry of the problem.
This equation can be written more generally as <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \cdot \frac{1}{(\sqrt{r^2+ (L/2)^2})} </math> where r represents the distance from the rod to the observation location.

'''Step 4:Checking the Result'''

Finally, the fourth step is to check the result. The units should be the same as the units of the expression for the electric field for a single point particle <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} </math>.
Our answer has the right units, since <math> \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. has the same units of <math>\frac{Q}{r^2}</math>
Is the direction qualitatively correct? We have the electric field pointing straight away from the midpoint of the rod, which is correct, given the symmetry of the situation. The vertical component of the electric field should indeed be zero.

== A Computational Model ==

'''Finding the Electric Field from a Rod with Code'''

Here is some code that you can run which shows the electric field vector
at a given distance from the rod along its length. The rod is shown as a
series of green balls to help emphasize that when using the numerical
integrations mentioned on this page, you are measuring the field produced
by discrete parts of the rod being analyzed.

Notice the edge-effects of the electric field of the rod. For reasons
discussed above, if we used the long rod approximation (L>>d), these
effects would be negligible.

[http://www.glowscript.org/#/user/yoderlukas/folder/Public/program/ElectricFieldAlongRodLength Click Here to Run the Code]

==Examples==

Although this is not a very difficult topic, some reasonably difficult conceptual questions can be asked about it.

===Simple===

[[Image:LukasYoderNo.jpg|400px|center|thumb|Figure1: Problem 1]]

===Middling===

[[Image:LukasYoderMaybe.jpg|400px|center|thumb|Figure 2: Problem 2]]

===Difficult===

[[Image:LukasYoderYes.jpg|400px|center|thumb|Figure 3: Problem 3]]

== Connectedness ==
The following two DIY experiments are shown to better understand and visualize the physical consequences of electric fields.

===Charged Rod and Aluminum Can===

In our first example we set up an experiment using two charged rods placed to the left and right of an aluminum can, distanced by a length d. If one of the rods is positively charged and the other is negatively charged, what will the can do? Because the positively charged rod induces a negative charge on the left side of the can, creating an attractive force between the rod and the can, and the negatively charged rod induces an equal positive charge on the right side of the can, which creates an attractive force between the can and that rod, the net force on the can is zero. Thus the can will stay still. The setup is depicted in the image below.

[[File:plusq.png|300px|center|thumb|Figure 4: apparatus diagram with +q]]

Next, lets consider a setup but with both rods having equal positive charge, as shown in the image below. What will the can do in this situation?

[[File:minusq.png|300px|center|thumb|Figure 5: apparatus diagram with -q]]

Again the can will also stay still, but this time it is because the polarization force between two objects is always attractive.

So in what scenario will the can move, and what time of movement will the can exhibit? Considering the first setup, imagine this time we initially touch the negatively charged rod and the can for a brief moment. Holding the rods at equal distance on either side of the can, the can will now roll toward the positively charged rod. This is because the can acquires a net negative charge after being touched, so it is then attracted to the positively charged rod.

===Charged Rod and Pith Ball===

Another DIY experiment to visualize the effects of an electric field are shown in the video embedded below, which depicts the interaction between an initially neutral pith ball hanging on a string from a stand, and a charged rod.
[http://www.youtube.com/watch?v=aeiqw81kGio Interaction between a Charged Rod and Pith Ball]

== History ==

Physicists and scientists make use of electric fields and charged objects all the time. Many times, we may need to know which objects are contributing how much charge in certain areas. Charged objects may attract or repel (depending on the signs of their charge), so we often need to know how objects will interact with each other based on their charges. The phenomenon of this interaction, or electric force between charged particles, was finally confirmed and stated as a law in 1785 by French physicist Charles-Augustin de Coulomb, hence "Coulomb's Law."

== See also ==

The equation for the electric field of a charged rod was derived from the equation for an electric field of a charged particle. See the article "[[Electric Field]]" for more information.

=== Further Reading ===

The page on electric fields: [[Electric Field]]

=== External Links ===

http://online.cctt.org/physicslab/content/phyapc/lessonnotes/Efields/EchargedRods.asp

https://pages.uncc.edu/phys2102/online-lectures/chapter-02-electric-field/2-4-electric-field-of-charge-distributions/example-1-electric-field-of-a-charged-rod-along-its-axis/

http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ContinuousChargedRod.xml

== References ==

https://www.youtube.com/watch?v=BBWd0zUe0mI

(For the above reference, I chose to follow the textbook's method in not defining the charge distribution and assuming it was constant, though this was helpful in figuring out a better way to introduce it.)

Chabay and Sherwood: Matter and Interactions, Fourth Edition, Chapter 15

All figures created by author
[[Category: Electric Field]]

Conductors

2020-05-04T18:04:56Z

Sydney: /* Middling */

Claimed by Sydney Triplett Spring 2020.

Conductor are materials that allow electric current to travel with little resistance throughout. This is related to the structure of the atoms of the conductor. In this section, we will look at what a conductor is, why it is this way, and the applications.

==The Main Idea==

Conductors are defined as a material that allows for charged particles to move easily throughout. Charges placed on the surface of a conductor will not simply stay in one spot or only spread over the surface, but will immediately spread evenly throughout the conductor- given there are no interfering forces. If the conductor is in an electric field, for example, it will cause the (negative) charges to move in the opposite direction of the field.

Electric current flows by the net movement of electric charge. This can be by electrons, ions, or other charged particles.

Conductors allow for easy movement of charged particles because of the structure of the atoms. The outermost electrons in atoms that make up conductors are only loosely bound and allow for more interaction with other particles. These electrons that are free to move about the conductor are called the "sea of electrons". The sea of electrons is therefore able to move in response to other charges that are put on the conductor or in response to an electric field that the conductor is within.

[[File:Cross section and length image.gif|right|180 px]] There are some factors that can change the conductance of a conductor. Shape and size, for example, affect the conductance of an object. A thicker(larger cross sectional area A as shown in the diagram) piece will be a better conductor than a thinner piece of the same material and other dimensions. This is the same concept that a thicker piece of wire allows for greater current flow. The larger cross sectional area allows for more flow of charge carriers. A shorter conductor will also conduct better since it has less resistance than a longer piece. Conductance itself can also change conductivity. In actively conducting electric current, the conductor heats up. This is secretly the third factor affecting conductance: temperature. Changes in temperature can cause the same object to have a different conductance under otherwise identical conditions. The most well known example of this is glass. Glass is more of an insulator at typical to cold temperatures, but becomes a good conductor at higher temperatures. Generally, metals are better conductors at cooler temperatures. This is because an increase in temperature is an increase in energy, specifically for electrons.

[[File:Conductor chart.gif|600 px]]

===A Mathematical Model===

Ohm's Law of J = σE can be used to model the relationship of conductivity to electric current density where J is electric current density, σ is conductivity of the material, and E is electric field. This is a generalized form of the well known V = IR.

σ is larger for better conductors like metals and saltwater. For "perfect" conductors, σ approaches infinity. In this case, E would be zero since the current density J cannot be infinity.

Materials are generally divided into three categories based on σ:
*Lossless Materials: σ = 0
*Lossy Materials: σ > 0
*Conductors: σ >> 0

Below is a breakdown of how conductivity is calculated. This could be considered a formula for conductivity, but it would be more accurate to think of it as a definition.

[[File:Electric conductivity equation.jpg|400 px]]

Conductivity can also be explained as the inverse of resistivity. σ = 1/ρ where ρ is resistivity.

===A Computational Model===

[[File:Conductor computational model.PNG|right|400 px]]

[http://www.physics-chemistry-interactive-flash-animation.com/electricity_electromagnetism_interactive/electric_conductors_insulators.htm This simple interactive] is a great way to see which real objects are made of conducting or insulating material. Try to guess which objects will allow for flow of electricity before you test with the interactive.

https://phet.colorado.edu/en/simulation/semiconductor

==Examples==

===Simple===
Material A has a resitivity of <math> 5.90 \cdot 10^{-8} Ω \cdot m </math> and Material B has a conductivity of <math> 1.00 \cdot 10^7 S/m </math>. They are the same size and temperature. Which is a better conductor?

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

We can change Material A's resitivity to conductivity with the formula σ = 1/ρ. σ = <math>\frac{1}{(5.90 \cdot 10^{-8} Ω \cdot m)} = 1.69 \cdot 10^7 S/m</math>. Since Material A has a greater conductivity, it is a better conductor.

These numbers are the real conductivities of Zinc(Material A) and Iron(Material B).

</div></div>

===Middling===
A negatively charged iron block is placed in a region where there is an electric field downward (in the -y direction). What will be the charge distribution of the iron block in this field? (Problem 47 from Matter and Interactions, page 583)

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

[[File:Conductor question diagram.jpg|300 px]]

Remember that in the direction of an electric field is traditionally the direction of positive charge movement. Since the iron block is a conductor, it has electrons that are free to move and will travel opposite the direction of the electric field. This will leave an excess of positive charge at the bottom of the block and an excess of negative charge at the top of the block, as shown in the diagram.

</div></div>

===Difficult===
question

<div class="toccolours mw-collapsible mw-collapsed" style="width:800px; overflow:auto;">
<div style="font-weight:bold;line-height:1.6;">Solution</div>
<div class="mw-collapsible-content">

solution

</div></div>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Modern research on electricity and conductors starts in the 1700s. Many different scientists contributed to the research that led to the understanding and use of conductors. Stephan Gray was one of the first of these, first studying the idea of electricity and then conductors. In Gray's time, the general consensus was that "electric virtue" was a quality that some materials could attain and others could not. Some materials, like glass, could acquire electric virtue by friction, while others, like metal, could be given electric virtue by contact with a charged object. Gray tested this theory with many different types of material, even with a child(who did work as a conductor).

Dufay began research on the same topics of charge transfer and conductance just after Gray. He lengthened the list of objects that could be given electric virtue by friction. Dufay also named the growing categories. "Electrical bodies" were what we call insulators and "non-electric bodies" were what we know as conductors. While this seems backwards from the way we think about insulators and conductors, it comes from the idea that electric virtue was intrinsic to insulators because charge could be induced on these simply by friction, while conductors can only come to have a charge by contact with a charged insulator.

Only when Benjamin Franklin came around some later did the ideology and vocabulary make a big switch. Franklin suggested that electricity is not created by electrical bodies through friction, but that it is a fluid shared by all bodies and can pass between them. Franklin also caused the shift in language from non-electric bodies to conductors and electric bodies to non-conductors.

There is some evidence that ancient Egyptians also used electricity.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

https://www.youtube.com/watch?v=BY8ZPobU8B0

==References==

*https://en.wikipedia.org/wiki/Electrical_conductor

*https://www.thoughtco.com/examples-of-electrical-conductors-and-insulators-608315

*https://www.rpi.edu/dept/phys/ScIT/InformationProcessing/semicond/sc_glossary/scglossary.htm

*http://maxwells-equations.com/materials/conductivity.php

*https://en.wikipedia.org/wiki/Ohm%27s_law

*https://www.youtube.com/watch?v=BY8ZPobU8B0

*https://www.physicsclassroom.com/class/estatics/Lesson-1/Conductors-and-Insulators

*http://histoires-de-sciences.over-blog.fr/2018/04/history-of-electricity.the-discovery-of-conductors-and-insulators-by-gray-dufay-and-franklin.html

*Benjamin, Park. A History of Electricity (The Intellectual Rise of Electricity) from Antiquity to the Days of Benjamin Franklin. J. Wiley & Sons, 1898. Google Books, https://books.google.com/books?hl=en&lr=&id=K2dDAAAAIAAJ&oi=fnd&pg=PR1&dq=ancient egypt electricity&ots=edMffcocC0&sig=zFX9kUz2FKoPcPTCf8YVId2AjhQ#v=onepage&q&f=false.

*https://www.thoughtco.com/table-of-electrical-resistivity-conductivity-608499

[[Category:Which Category did you place this in?]]

File:Conductor question diagram.jpg

2020-05-04T17:58:52Z

Sydney: this is from Matter and Interactions by Chabay and Sherwood, page 583

this is from Matter and Interactions by Chabay and Sherwood, page 583

Electrostatic Discharge

2020-05-04T17:25:11Z

Sydney: /* References */

Edited by Jonathan Ledet (Spring 2017)

Electrostatic Discharge (ESD) is the swift transfer of charges between objects at different potentials. ESD commonly occurs after the accumulation of static electricity on a material and can have devastating effects on solid-state electronics.

==The Main Idea==

All materials (both insulators and conductors) are defined on the Triboelectric series, which essentially rates that material's affinity for electrons. When an object lower on the series is touched by an object higher on the series, the lower object will acquire a negative charge. When the objects are separated, their charges are equal and opposite- just waiting for the opportunity to discharge. When the objects finally do discharge, a high voltage spark can form, turning the bridging air to plasma or frying circuits. This process is referred to as tribocharging.

You may have seen a demonstration in class involving a glass rod and a piece of cloth, or rubbed a balloon on your hair and observed its wacky behavior. Both of these are examples of the Triboelectric effect. Additionally, although friction greatly increases the magnitude of the effect, it is not required for the effect to occur. Only an initial contact and subsequent separation are required; friction only amplifies the effect due to the increased contact and separation of molecules on the microscopic level.

Electrostatic Discharge can be caused by an electrical breakdown, a short circuit, and--most commonly--Tribocharging. ESD is measured using an electrostatic voltmeter.

Sparks and lightning are visible Electrostatic Discharge events, but only represent part of the threat of ESD. For example, such a high voltage can be very damaging to the delicate pathways and components on circuit boards.

ESD events, like a spark from a human hand, allow current to travel to the ground through electronic devices, burning holes in integrated circuits and dealing heat damage to the circuit board. This can happen when working barehanded (without an electrostatic wrist strap) with circuit boards and other sensitive electronic equipment, when negatively charged synthetic materials are on or near sensitive electronic equipment, or due to the fast movement of air near electronic equipment.

There are some preventative methods taken to discourage Electrostatic Discharge. During the process in which electronic components are assembled most manufacturers implement Electrostatic Discharge Protected Areas (EPA). These areas are specially designed to prevent the build up of charge on the components, workers, and all other conductive materials. To protect against Electrostatic Discharge during transit, antistatic bags act as Faraday Cages to protect sensitive devices.

===The Physics Principals and Visual Aids===

Triboelectric Series
*Celluloid
*Sulfer
*Rubber
*Copper, Brass
*Amber
*Wood
*Cotton
*Human Skin
*Silk
*Cat Fur
*Wool
*Glass
*Rabbit Fur
*Asbestos

The objects near the top (of the list above) will tend to gain negative charges, while those below them will gain positive charges. The law of Conservation of Charge is followed.

[[File:5255360854_b0db025afb.jpg]]

First, a tube made out of plastic is charged when rubbed with synthetic fur (top pictures). The tube, which is now charged, is brought close to neutral paper bits on the table (bottom left). You can see that the tube and paper now attract each other and that this attraction is strong enough to lift the pieces of paper off the table.

==Examples==

===Simple===

===Middling===

===Difficult===

==Connectedness==
The concept of Electrostatic Discharge (ESD) can be applied to many practical scenarios; however, one of the most notable instances of ESD is in the electronics manufacturing industry. Due to the fact that charges can accumulate relatively easily in electronics manufacturing, companies often have to take ESD precautions by making employees wear special clothing and by designing their workbenches and flooring out of special material which prevents electrostatic build up.

Electrostatic Discharge is connected to chemical engineering in that there is a great deal of equipment in plants that has the potential to acquire electrostatic charge. As a result, it is vital to safety to be knowledgeable and cautious when dealing with such equipment in order to reduce work related injuries.

Electrostatic Discharge also has connections to Aerospace Engineering by how potentially dangerous it is to not guard against ESD in environments in which many different electronic devices are integrated into air and spacecraft. Aerospace research agencies, such as NASA, JAXA, or RKA, and corporation, such as Lockheed Martin, have implemented ESD protected areas and grounded workbenches, mandated the use of protective equipment, introduced protocol in which charge generating materials cannot be worn in protected areas, and have audits and inspections to make sure important aerospace electronics stay protected from ESD.

Additionally, ESD poses a threat in the operation of air and spacecraft due to its volatile nature and possible interaction with fuel. When fueling aircraft, it is mandated procedure to ground the air frame to the fuel truck in order to prevent a spark from jumping between the metal surfaces.

One interesting way that ESD can be used is to create sparks. When the dielectric between two oppositely charged sources is damaged or if the electric field due to the build up charge exceeds the dielectric, then ESD can occur by means of a spark through the air or other dielectric medium.

==History==
The phenomena of electrostatic discharge (ESD) has been known for a very long time, dating all the way back to the ancient Greeks. While ESD was mostly considered a non-factor throughout most of history, due to the growth of solid state electronics in the 1950's companies and researchers have had to account for and study in greater detail the phenomena of ESD. As the prevalence of electronics increased people began to notice the ESD could have very negative effects on certain components, causing them to short-circuit or malfunction. The 1960s and 70s were characterized by companies discovering methods and techniques to test for ESD, some of which included the Human Body Discharge Model and the Horizontal Coupling Plate. In the 1980's the release of the IBM personal computer saw an increased need for materials with resistance to ESD and thus the focus shifted towards maximizing the ability of electronic components to avoid ESD and subsequent malfunctioning. Since the 1980's and well into modern times, companies have been working to hone and refine this process in order to maximize the efficacy of their circuit components by increasing their ability to resist ESD when in use.

== See also ==
*[[Charge Transfer]]

*[[Charge Motion in Metals]]

*[[Polarization]]

*[https://en.wikipedia.org/wiki/Michael_Faraday Micheal Faraday]

===Further reading===

*Electro Static Discharge: Understand, Simulate, and Fix ESD Problems, 3rd Edition

===External links===

*http://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Friction

==References==

*Aaq.auburn.edu,. 'Summary | Academy Of Aerospace Quality'. N.p., 2015. Web. 5 Dec. 2015. (http://aaq.auburn.edu/node/277)

*Hoolihan, Daniel D. "A Brief History of Electrostatic Discharge Testing of Electronic Products Read More: Http://incompliancemag.com/article/a-brief-history-of-electrostatic-discharge-testing-of-electronic-products/#ixzz4R5MRIIIQ Follow Us: @incompliancemag on Twitter | Incompliancemag on Facebook." INCompliance. ECM Consulting, 01 Mar. 2014. Web. 25 Nov. 2016.

*Michaels, Ken, AYMAN ZAHER, and Stan Herron. 'Electrostatic Discharge: Causes, Effects, And Solutions'. Ecmweb.com. N.p., 2013. Web. 6 Dec. 2015. (http://ecmweb.com/content/electrostatic-discharge-causes-effects-and-solutions)

*Physicsclassroom.com,. 'Charging By Friction'. N.p., 2015. Web. 6 Dec. 2015. (http://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Friction)

[[Category: Fields]]

Electrostatic Discharge

2020-05-04T17:24:39Z

Sydney: /* See also */

Edited by Jonathan Ledet (Spring 2017)

Electrostatic Discharge (ESD) is the swift transfer of charges between objects at different potentials. ESD commonly occurs after the accumulation of static electricity on a material and can have devastating effects on solid-state electronics.

==The Main Idea==

All materials (both insulators and conductors) are defined on the Triboelectric series, which essentially rates that material's affinity for electrons. When an object lower on the series is touched by an object higher on the series, the lower object will acquire a negative charge. When the objects are separated, their charges are equal and opposite- just waiting for the opportunity to discharge. When the objects finally do discharge, a high voltage spark can form, turning the bridging air to plasma or frying circuits. This process is referred to as tribocharging.

You may have seen a demonstration in class involving a glass rod and a piece of cloth, or rubbed a balloon on your hair and observed its wacky behavior. Both of these are examples of the Triboelectric effect. Additionally, although friction greatly increases the magnitude of the effect, it is not required for the effect to occur. Only an initial contact and subsequent separation are required; friction only amplifies the effect due to the increased contact and separation of molecules on the microscopic level.

Electrostatic Discharge can be caused by an electrical breakdown, a short circuit, and--most commonly--Tribocharging. ESD is measured using an electrostatic voltmeter.

Sparks and lightning are visible Electrostatic Discharge events, but only represent part of the threat of ESD. For example, such a high voltage can be very damaging to the delicate pathways and components on circuit boards.

ESD events, like a spark from a human hand, allow current to travel to the ground through electronic devices, burning holes in integrated circuits and dealing heat damage to the circuit board. This can happen when working barehanded (without an electrostatic wrist strap) with circuit boards and other sensitive electronic equipment, when negatively charged synthetic materials are on or near sensitive electronic equipment, or due to the fast movement of air near electronic equipment.

There are some preventative methods taken to discourage Electrostatic Discharge. During the process in which electronic components are assembled most manufacturers implement Electrostatic Discharge Protected Areas (EPA). These areas are specially designed to prevent the build up of charge on the components, workers, and all other conductive materials. To protect against Electrostatic Discharge during transit, antistatic bags act as Faraday Cages to protect sensitive devices.

===The Physics Principals and Visual Aids===

Triboelectric Series
*Celluloid
*Sulfer
*Rubber
*Copper, Brass
*Amber
*Wood
*Cotton
*Human Skin
*Silk
*Cat Fur
*Wool
*Glass
*Rabbit Fur
*Asbestos

The objects near the top (of the list above) will tend to gain negative charges, while those below them will gain positive charges. The law of Conservation of Charge is followed.

[[File:5255360854_b0db025afb.jpg]]

First, a tube made out of plastic is charged when rubbed with synthetic fur (top pictures). The tube, which is now charged, is brought close to neutral paper bits on the table (bottom left). You can see that the tube and paper now attract each other and that this attraction is strong enough to lift the pieces of paper off the table.

==Examples==

===Simple===

===Middling===

===Difficult===

==Connectedness==
The concept of Electrostatic Discharge (ESD) can be applied to many practical scenarios; however, one of the most notable instances of ESD is in the electronics manufacturing industry. Due to the fact that charges can accumulate relatively easily in electronics manufacturing, companies often have to take ESD precautions by making employees wear special clothing and by designing their workbenches and flooring out of special material which prevents electrostatic build up.

Electrostatic Discharge is connected to chemical engineering in that there is a great deal of equipment in plants that has the potential to acquire electrostatic charge. As a result, it is vital to safety to be knowledgeable and cautious when dealing with such equipment in order to reduce work related injuries.

Electrostatic Discharge also has connections to Aerospace Engineering by how potentially dangerous it is to not guard against ESD in environments in which many different electronic devices are integrated into air and spacecraft. Aerospace research agencies, such as NASA, JAXA, or RKA, and corporation, such as Lockheed Martin, have implemented ESD protected areas and grounded workbenches, mandated the use of protective equipment, introduced protocol in which charge generating materials cannot be worn in protected areas, and have audits and inspections to make sure important aerospace electronics stay protected from ESD.

Additionally, ESD poses a threat in the operation of air and spacecraft due to its volatile nature and possible interaction with fuel. When fueling aircraft, it is mandated procedure to ground the air frame to the fuel truck in order to prevent a spark from jumping between the metal surfaces.

One interesting way that ESD can be used is to create sparks. When the dielectric between two oppositely charged sources is damaged or if the electric field due to the build up charge exceeds the dielectric, then ESD can occur by means of a spark through the air or other dielectric medium.

==History==
The phenomena of electrostatic discharge (ESD) has been known for a very long time, dating all the way back to the ancient Greeks. While ESD was mostly considered a non-factor throughout most of history, due to the growth of solid state electronics in the 1950's companies and researchers have had to account for and study in greater detail the phenomena of ESD. As the prevalence of electronics increased people began to notice the ESD could have very negative effects on certain components, causing them to short-circuit or malfunction. The 1960s and 70s were characterized by companies discovering methods and techniques to test for ESD, some of which included the Human Body Discharge Model and the Horizontal Coupling Plate. In the 1980's the release of the IBM personal computer saw an increased need for materials with resistance to ESD and thus the focus shifted towards maximizing the ability of electronic components to avoid ESD and subsequent malfunctioning. Since the 1980's and well into modern times, companies have been working to hone and refine this process in order to maximize the efficacy of their circuit components by increasing their ability to resist ESD when in use.

== See also ==
*[[Charge Transfer]]

*[[Charge Motion in Metals]]

*[[Polarization]]

*[https://en.wikipedia.org/wiki/Michael_Faraday Micheal Faraday]

===Further reading===

*Electro Static Discharge: Understand, Simulate, and Fix ESD Problems, 3rd Edition

===External links===

*http://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Friction

==References==

Aaq.auburn.edu,. 'Summary | Academy Of Aerospace Quality'. N.p., 2015. Web. 5 Dec. 2015. (http://aaq.auburn.edu/node/277)

Hoolihan, Daniel D. "A Brief History of Electrostatic Discharge Testing of Electronic Products Read More: Http://incompliancemag.com/article/a-brief-history-of-electrostatic-discharge-testing-of-electronic-products/#ixzz4R5MRIIIQ Follow Us: @incompliancemag on Twitter | Incompliancemag on Facebook." INCompliance. ECM Consulting, 01 Mar. 2014. Web. 25 Nov. 2016.

Michaels, Ken, AYMAN ZAHER, and Stan Herron. 'Electrostatic Discharge: Causes, Effects, And Solutions'. Ecmweb.com. N.p., 2013. Web. 6 Dec. 2015. (http://ecmweb.com/content/electrostatic-discharge-causes-effects-and-solutions)

Physicsclassroom.com,. 'Charging By Friction'. N.p., 2015. Web. 6 Dec. 2015. (http://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Friction)

[[Category: Fields]]

Charged Conductor and Charged Insulator

2020-05-04T00:20:09Z

Sydney: /* The Main Idea */

Claimed by Amanda Barber Spring 2018

Okay – so we know that '''charged particles''' exist out there (shout out to those positrons and electrons for being “lit” when they annihilate) and that objects can either be negatively charged, positively charged, or neutral depending on the ratio of charged particles that object has. For example, if I happen to observe some random thing in nature that has 5 protons and 4 electrons – that thing has a net charge of +1. But what we haven’t really explored '''''how something''''' (like a '''conductor''' or an '''insulator''') '''''becomes positively, negatively, or neutrally charged''''' or '''''what happens when something becomes charged'''''. Guess what? That’s what this wiki is about! And by the time you’ve finished reading this, you should have a better understanding of how one can become “charged up” (in the context of E&M physics, not in the context of 6 Gawd Drizzy Drake preparing to release the hottest diss track of the decade in response to “twitter-finger” beef initiated by former rapper Meek Mill in the Summer of 2015).

==The Main Idea==

Okay, so remember in that short description above when I bolded and italicized the phrase “how something can become positively, negatively, or neutrally charged”? The processes by which this can occur are called '''charging''' and '''discharging'''. The definitions are simple: charging – how an object becomes positively or negatively charged; discharging – how a positively or negatively charged object becomes neutral. What happens to an object as a result of charging or discharging depends on the nature of that object and whether or not the object is a conductor or an insulator. All materials are made of atoms that contain positive and negative charges (protons and electrons). While all materials contain these two basic units, the charge distribution patterns change depending on the microscopic behavior of the atom's movement in an electric field. These differences have created two distinct classes of materials: conductors and insulators. This article will explore the differences between a charged conductor and a charged insulator. But before we go further into the specifics regarding conductors and insulators, let us further discuss the means by which charging can occur: '''conduction''' and '''induction'''. Both processes involve objects becoming charged by exchanging, gaining, or losing mobile charged particles. In most examples, objects will become charged by gaining or losing electrons, but that is not always the case – other mobile charged particles can contribute to an objects net charge, such as mobile potassium or sodium ions.

'''Charging by Conduction'''

Okay, so y’all remember that example in lecture where the professor kept pulling tape apart? And the lab that followed up afterwards? That’s what charging by induction is all about! (No worries if you don’t remember or if you haven’t gotten to that part of the course yet – I’m not gonna leave you hanging and just assume you know and/or remember what that was all about. I gotcha back homie! Checkout this link: https://www.youtube.com/watch?v=O7siDnbEuko

[[File:Charging tape.png|thumb|Charges on strips of tape as they are pulled apart from http://p3server.pa.msu.edu/coursewiki/doku.php?id=184_notes:charging_discharging]]

As illustrated in the pulling tape exercise, charging by conduction is all about '''making contact in order to transfer charged particles'''. When the two strips of tape are quickly ripped apart, mobile charged particles (electrons) are transferred from one strip of tape to the other – leaving one strip positively charged and the other strip negatively charged. What mobile charged particles are transferred (such as electrons) and where those mobile charged particles go depends on chemical makeup of the materials involved in the process of charging by conduction. For example, when one is charging by conduction with plastic and wool – the plastic gains electrons to become negatively charged. But when one is charging by conduction using plastic and silk, the plastic loses electrons to become positively charged.

'''Charging by Induction'''

Charging by induction is all about '''using a charged object to separate mobile charged particles between two other neutral objects in contact with one another'''. First, the two neutral objects are in physical contact with each other outside the presence of the charged object. Next, the two objects are subject to the presence of a charged object, causing mobile charged particles to separate among the two connected objects, resulting in excess charge building up on the opposing surfaces of the two connected objects. Then, the two connected objects are separated while still subject to the presence of the charged object. Finally, the separated objects removed from the presence of the charged object and the excess charges distribute across the entire surfaces of each object, resulting in what were initially two neutral objects now being charged.

[[File:Charging induction.png|thumb|Charging two neutral conductors by induction from http://p3server.pa.msu.edu/coursewiki/doku.php?id=184_notes:charging_discharging]]

Here's a video that might also help explain charging by induction: https://www.youtube.com/watch?v=cMM6hZiWnig

[[File:Inductiontransfer.gif]]

'''Discharging'''

Discharging is the opposite of charging: the process by which charged objects lose their excess charge and become neutral. There are a number of ways in which charged objects can discharge. For example, charged tape can be discharged by water in the surrounding environment (over time) or by using your finger and touching the tape. When we touch charged tape, our bodies act as electrical grounds. An electrical ground is huge pool of charged particles that remains neutral when small amounts of charged particles are added or taken away

[[File:05172.png]]

'''Charging Insulators'''

[[File:Insulator and conductor.png|thumb|Charging an insulator vs charging a conductor from http://p3server.pa.msu.edu/coursewiki/doku.php?id=184_notes:charging_discharging]]

Insulators can only be charged by conduction. When an insulator is charged by conduction, the charged particles remain at the initial point of contact throughout time until the insulator is discharged.

When a charged object enters the immediate region of an insulator, the result is polarization of the atoms and molecules within the insulator as seen in the example to the right (only while in the presence of that charged object). Since insulators impede the movement of charged particles, two insulators cannot be charged by induction in the presence of a charged object.

Here’s a neat little simulator to see how an insulator (a balloon) can become charged by conduction and see what happens afterwards:
https://phet.colorado.edu/en/simulation/balloons-and-static-electricity

'''Charging Conductors'''

When conductors are charged by conduction, the excess charges distribute evenly across the entire surface of the conductor over time since conductors enable charged particles to move freely. While the inside electric field remains zero. This all happens because of Coulomb's Law. It insists that charges lay as far away from each other as atom-ly (you get it? like human-ly) possible. If you charge a conductor, the conductor becomes charged with whatever type of charged was used. i.e. i charge a metal ball with a negative charge, thus the ball becomes negatively charged.

Here's a fun thing. You ever heard of the Van de Graaff? That metal ball thing that shocks you? It is an electrostatic generator that uses a belt of sorts to accumulate charge. It creates very big electric potentials so as you go to touch it, it discharges onto you, shocking you.

Remember how I kept using the word “object” when describing the neutral things needed in explaining charging by induction? Yea, so those objects are always going to be conductors. So charging by induction is always charging conductors by induction.

[[File:Inschargedist.gif|center|375]]

===A Mathematical Model===

===A Computational Model===

==Examples==
[[File:Screen Shot 2018-04-08 at 4.16.06 PM.png|300 px]]
[[File:Screen Shot 2018-04-08 at 4.27.27 PM.png|300 px]]
[[File:Screen Shot 2018-04-08 at 4.34.27 PM.png|300 px]]

===Example Problem===

A neutral metal sphere enters a capacitor, as shown. Show the charge distribution on the sphere. If the sphere was instead made of plastic, show the the charge distribution.

[[File:Screen Shot 2018-04-08 at 4.37.07 PM.png|300 px]]

===Solution===
For a metal sphere:

[[File:Screen Shot 2018-04-08 at 4.38.17 PM.png|300 px]]

Since metals are conductors, there are mobile electrons within the sphere that are free to move about the sphere when they feel the electrostatic force from the capacitor.

For a plastic sphere:

[[File:Screen Shot 2018-04-08 at 4.39.59 PM.png|300 px]]

Plastic is an insulator, so the individual molecules will become polarized due to the capacitor and become dipoles.

==Connectedness==

[[File:Lightning bedning.png|thumb|Mako bending lightning to save Republic City in Avatar Legend of Korra]]

Okay, so I’m going to get super fan boy nerdy for a second and connect what we’ve learned about charging and discharging to my favorite cartoon of all time: Avatar The Last Airbender. For those of you not familiar with this amazing series – these resources might be help create some context (I highly recommend watching the original series and the spinoff with Avatar Korra): http://www.nick.com/avatar-the-last-airbender/

Within the world of Avatar and bending, each bending art has one or more specialized bending techniques that are considered to be very rare, highly coveted skills. Lightening bending (or lightning generation) is a special technique within the art of firebending. Lightning bending is an extremely difficult technique to master, as it requires the bender to be at total peace and void of all emotion (which is why it was always impossible for Zuko to learn from Uncle Iroh because that boy had so many emotional issues – which is completely understandable considering his circumstances).

Lightning bending connects to charging and discharging in a few ways. Lightning (and lightning generation) in and of itself alone is a representation of electrical discharge. Lightning is created when two dense pools of opposite charges reach out and connect to one another, creating a channel for electrical transfer of charge that results in each pool becoming neutral once that transfer is complete.

[[File:Strike1.gif]] [[File:Strike2.gif]] [[File:Strike3.gif]]

Photos from free U.S. government resource: http://www.srh.noaa.gov/jetstream/lightning/lightning.html

Before we continue, please watch the following video to see lightning bending in action: https://www.youtube.com/watch?v=6htOzNpBJv8

So let’s break down what we see in the video and how it relates to the physics of charging and discharging. Before we even see Uncle Iroh generate visible lightning, he “charges himself” by, “separating the energies of yin and yang” (according to Avatar mythology). In other words, Uncle Iroh separates mobile negatively charged particles (yin) from mobile positively charged particles (yang) within his body and immediate surroundings to set the stage for electrical charge transfer. When the amount of charge Uncle Iroh has built up in each pool is great enough to overcome the air and his body’s insulation of electric flow, this is the moment we begin to actually see lightning. The generated pools of charge connect and create a channel where charges begin to flow between pools in order for each pool to discharge and become neutral.

So all in all, in order for a firebender to lighting bend, the bender must be adept at charging and discharging to perform the technique. Once the lighting is generated, the bender then guides the discharging electric flow of energy in a desired direction (more than likely at an opponent in order to zap them).

Note: all representations of anything related to Avatar The Last Airbender are property of Nickelodeon and they rights are reserved.

While insulators are not useful for transferring charge, they do serve a critical role in electrostatic experiments and demonstrations. Conductive objects are often mounted upon insulating objects. This arrangement of a conductor on top of an insulator prevents charge from being transferred from the conductive object to its surroundings.

==History==

==See also==

===Further Reading===
*[http://www.physicsbook.gatech.edu/Charge_Transfer Charge Transfer]
*[http://www.physicsbook.gatech.edu/Insulators Insulators]
*[http://www.physicsbook.gatech.edu/Conductivity Conductivity]
*[http://www.physicsbook.gatech.edu/Polarization_of_a_conductor Polarization of a Conductor]

===External links===
*[http://www.schoolphysics.co.uk/age16-19/Electricity%20and%20magnetism/Electrostatics/text/Electric_charge_distribution/index.html Explanation of why charge concentrates at a point on a conductor]

==References==
*[http://www.physicsclassroom.com/class/estatics/Lesson-1/Conductors-and-Insulators http://www.physicsclassroom.com/class/estatics/Lesson-1/Conductors-and-Insulators]
*http://www.schoolphysics.co.uk/age16-19/Electricity%20and%20magnetism/Electrostatics/text/Electric_charge_distribution/index.html
*http://p3server.pa.msu.edu/coursewiki/doku.php?id=184_notes:charging_discharging
*http://avatar.wikia.com/wiki/Specialized_bending_techniques
*http://www.srh.noaa.gov/jetstream/lightning/lightning.html