Physics Book - User contributions [en]

Faraday's Law

2020-11-23T19:27:21Z

Yalmonte3: /* Finding the direction of the induced conventional current */

Faraday's Law
focuses on how a time-varying magnetic field produces a "curly" non-Coulomb electric field, thereby inducing an emf.

==Faraday's Law==

Faraday's Law summarizes the ways voltage can be generated as a result of a time-varying magnetic flux. And it gives a way to connect the magnetic and electric fields in a quantifiable way (will elaborate later). Faraday's law is one of four laws in Maxwell's equations. It tells us that in the presence of a time-varying magnetic field or current (which induces a time-varying magnetic field), there is an emf with a magnitude equal to the change in magnetic flux. It serves as a succinct summary of the ways a voltage (or emf) may be generated by a changing magnetic environment. The induced emf in a coil is equal to the negative of the rate of change of magnetic flux times the number of turns in the coil. It involves the interaction of charge with the magnetic field.

==Curly Electric Field==

[[File:curly.jpg]]

===Mathematical Model===

'''Faraday's Law'''

emf = <math>{\frac{-d{{Φ}}_{mag}}{dt}}</math>

where emf = <math>\oint\vec{E}_{NC}\bullet d\vec{l}</math> and <math>{{Φ}}_{mag}\equiv\int\vec{B}\bullet\hat{n}dA</math>

In other words: The emf along a round-trip is equal to the rate of change of the magnetic flux on the area encircled by the path.

Direction: With the thumb of your right hand pointing in the direction of the ''-dB/dt'', your fingers curl around in the direction of Enc.

The meaning of the minus sign: If the thumb of your right hand points in the direction of ''-dB/dt'' (that is, the opposite of the direction in which the magnetic field is increasing), your fingers curl around in the direction along which the path integral of electric field is positive. Similarly, the direction of the induced current can be explained using Lenz's Law. Lenz's law states that the induced current from the non-Coulombic electric field is induced in such a way that it produces a magnetic field that opposes the first magnetic field to keep the magnetic flux constant.

'''Formal Version of Faraday's Law'''

<math>\oint\vec{E}_{NC}\bullet d\vec{l} = {\frac{-d}{dt}}\int\vec{B}\bullet\hat{n}dA</math> (sign given by right-hand rule)

===Fiding the direction of the induced conventional current===
To find the direction of the induced conventional current by the change in the magnetic flux one must find the direction of the Non-Coulomb electric filed generated by the change in flux as the conventional current is the direction of the Non-Coulomb electric field.
To find the direction of the the Non-Coulomb Electic field, one must find the direction of <math> \frac{-dB}{dt} </math>. Do this using the change in magnetic field as the basis of finding the <math> \frac{-dB}{dt} </math>.

As stated previously the negative sign in front of the change in magnetic flux in the Law is a representative of Lenz's law or in other words, it's there to remind us to apply Lenz's law. Lenz's law is basically there to make us abide by the law of conservation of energy. That said, thinking in terms of conservation of energy provides the simplest way to figure out the direction of the Non-Coulomb electric field.
The external magnetic field induces the Non-Coulomb electric field which drives the current which in turn creates a new magnetic field which we will call the induced magnetic field. This is the magnetic field whose direction we can deduce which in turn will help us find the direction of the current.
The easiest way to do this is to imagine a loop of wire with and an external magnetic field perpendicular to the surface of the plane of the loop. There is a change in magnetic flux generated by the change in the magnitude of the magnetic field. vector for the initial external magnetic field and a vector for the final magnetic field. Then, draw the change in magnetic field vector, <math> \Delta \mathbf{B} </math>, and then the negative vector of that change in magnetic field gives <math> \frac{-dB}{dt} </math>:

[[File:neg_change_B_dt.jpg]]

Pointing the thumb of your right hand in the direction of <math> \frac{-dB}{dt} </math> allows you to curl your fingers in the direction of <math> \mathbf{E_{NC}} </math>.

In this chapter we have seen that a changing magnetic flux induces an emf:

[[File:tips5.png]]

according to Faraday’s law of induction. For a conductor which forms a closed loop, the
emf sets up an induced current ''I =|ε|/R'' , where ''R'' is the resistance of the loop. To
compute the induced current and its direction, we follow the procedure below:

1. For the closed loop of area on a plane, define an area vector A and let it point in
the direction of your thumb, for the convenience of applying the right-hand rule later.
Compute the magnetic flux through the loop using

[[File:tips4.png]]

Determine the sign of the magnetic flux [[File:tips3.png]]

2. Evaluate the rate of change of magnetic flux [[File:tips2.png]] . Keep in mind that the change
could be caused by

[[File:tips.png]]

Determine the sign of [[File:tips2.png]]

3. The sign of the induced emf is the opposite of that of [[File:tips2.png]]. The direction of the
induced current can be found by using Lenz’s law or right-hand rule (discussed previously).

==Computational Model==
The following simulations demonstrate Faraday's Law in action.

==More on Faraday's Law==

Moving a magnet near a coil is not the only way to induce an emf in the coil. Another way to induce emf in a coil is to bring another coil with a steady current near the first coil, thereby changing the magnetic field (and flux) surrounding the first coil, inducing an emf and a current. Also, rotating a bar magnet (or coil) near a coil produces a time-varying magnetic field in the coil since rotating the magnet changes the magnetic field in the coil. The key to inducing the emf in the second coil is to change the magnetic field around it somehow, either by bringing an object that has its own magnetic field around that coil, or changing the current in that object, changing its magnetic field.

Faraday's law can be used to calculate motional emf as well. A bar on two current-carrying rails connected by a resistor moves along the rails, using a magnetic force to induce a current in the wire. There is a magnetic field going into the page. One way to calculate the motional emf is to use the [http://www.physicsbook.gatech.edu/Motional_Emf magnetic force], but an easier way is to use Faraday's law.

Faraday's law, using the change in magnetic flux, can be used to find the motional emf, where the changing factor in the magnetic flux is the area of the circuit as the bar moves, while the magnetic field is kept constant.

[[File:motionalemf.jpg]]

==Examples==

===Simple===

[[File:solenoid.ring.jpg|center|alt=Diagram for simple example]]

''Adapted from the'' Matter & Interactions ''textbook, variation of P12 (4th ed)''.

The solenoid radius is 4 cm and the ring radius is 20 cm. B = 0.8 T inside the solenoid and approximately 0 outside the solenoid. What is the magnetic flux through the outer ring?

''Solution:''

Because the magnetic field outside the solenoid is 0, there is no flux between the ring and solenoid. So the flux in the ring is due to the area of the solenoid, so we use the area of the solenoid to find the flux through the outer ring rather than the area of the ring itself:

<math> \phi = BAcos(\theta)</math>

<math>= (0.8 T)(\pi)(0.04 m)^2cos(0) </math>

<math>= 4.02 x 10^{-3} T*m^2 </math>

===Middle===

[[File:rectanglecoilsolenoid.jpg|center|alt=Diagram for simple example]]

''Adapted from the'' Matter & Interactions ''textbook, variation of P27 (4th ed)''.

A very long, tightly wound solenoid has a circular cross-section of radius 2 cm (only a portion of the very long solenoid is shown). The magnetic field outside the solenoid is negligible. Throughout the inside of the solenoid the magnetic field ''B'' is uniform, to the left as shown, but varying with time ''t: B'' = (.06+.02<math>t^2</math>)T. Surrounding the circular solenoid is a loop of 7 turns of wire in the shape of a rectangle 6 cm by 12 cm. The total resistance of the 7-turn loop is 0.2 ohms.

(a) At ''t'' = 2 s, what is the direction of the current in the 7-turn loop? Explain briefly.

(b) At ''t'' = 2 s, what is the magnitude of the current in the 7-turn loop? Explain briefly.

''Solution''

'''(a)''' The direction of the current in the loop is clockwise.

'''(b)'''

B(t) = (.06+.02<math>t^2</math>)

A = (π)(0.02 m)^2 = .00126 <math>m^2</math>

<math>|{ε}| = AN\frac{dB(t)}{dt}</math>

<math>|{ε}|</math> = (.00126 <math>m^2</math>)(7)<math>\frac{d(.06+.02t^2)}{dt}</math> = (.00882)(.02)(2t) = .0003528t

'''At ''t'' = 2 s:'''

<math>|{ε}|</math> = .0003528(2) = .0007056 V

<math>i = \frac{{ε}}{R}</math>

<math>i = \frac{{.0007056 V}}{0.2 ohms}</math> = '''.00353 A'''

===Difficult===

[[File:difficultfaraday.png]]

A square loop (dimensions L⇥L, total resistance R) is located halfway inside a region with uniform magnetic field B0. The magnitude of the magnetic field suddenly begins to increase linearly in time, eventually quadrupling in a time T.

'''(a) What current (magnitude and direction), if any, is induced in the loop at time T?
'''

<math> |emf| = \frac{-{Φ}_{B}}{Δt} = \frac{A(B_f - B_i)}{T} = \frac{L^2(4B_o - B_o)}{T} = \frac{3B_oL^2}{T}</math>

emf = IR = <math>\frac{3B_oL^2}{TR}</math>

'''(b) What net force (magnitude and direction), if any, is induced on the loop at time T?
'''

<math> F_{top} </math> and <math> F_{bottom} </math> cancel out.
<math> F_{left} </math> = 0 because the left side is out of <math> \vec{B} </math> region.

<math> \vec{F}</math> = <math> \vec{F}_{right} </math> = I <math> \vec{L} \times \vec{B} = (ILB)[(\hat{y} \times - \hat{z} )] = \frac{3B_oL^2}{TR}(4B_o L)(- \hat{x}) = \frac{3{B_o}^2 L^3}{TR}(- \hat{x})</math>

'''(c) What net torque (magnitude and direction), if any, is induced on the loop at time T?
'''

<math> \vec{τ} = \vec{μ} \times \vec{B} = 0 </math> because <math>\vec{μ}</math> and <math>\vec{B}</math> are anti-parallel.

==Connectedness==

Faraday's Law is one of Maxwell's equations which describe the essence of electric and magnetic fields. Maxwell's equations effectively summarize and connect all that we have learned throughout the course of Physics 2.

As an electrical engineer, Faraday's Law is relevant to my major.

== Faraday’s Law Applications ==

Physics 2 content has a lot of important concepts that we as engineers can use to make our jobs easier. Whether it be a direct application of a rule or some derivation of a rule. I know I personally struggle with a concept until I get a concrete real life application that I can see the material applied in. This section of the page will discuss how Faraday’s law is applied to concepts that you as students maybe more familiar with your day to day life.

== Hydroelectric Generators ==
Generators create energy by transforming mechanical motion into electrical energy, but hydroelectric generators use the power of falling water to turn a large turbine which is connected to a large magnet. Around this magnet is a large coil of tightly wound wire. The conceptual creation of electricity is the same as Faraday’s Law except alternating current is being produced, but the idea that a changing magnetic field in a coil of wire induces an electromotive force is still the same. The difference is the magnetic field changes sign and flips resulting in the same thing to occur in the induced EMF. Although the calculations here are slightly more difficult the concepts are the same.

== Transformers ==

Transformers use a similar concept for Faraday’s Law but it’s slightly different. The job of a transformer is to either step up or step down the voltage on the power line. Transformers have a constant magnetic field associated with it due to an iron core. The power supply voltage is adjusted by altering the number of turns of wire around the iron core which in turn alters the EMF of the electricity.

Cartoon of Hydroelectric Plant
https://etrical.files.wordpress.com/2009/12/hydrohow.jpg
Turbine Picture
http://theprepperpodcast.com/wp-content/uploads/2016/02/108-All-About-Hydro-Power-Generators-1054x500.jpg
Transformer Diagram https://en.wikipedia.org/wiki/Transformer#/media/File:Transformer3d_col3.svg

==History==

In 1831, eletromagnetic induction was discovered by Michael Faraday.

===Faraday's Law Experiment ===

[[File:experiment.png]]

Faraday showed that no current is registered in the galvanometer when bar magnet is
stationary with respect to the loop. However, a current is induced in the loop when a
relative motion exists between the bar magnet and the loop. In particular, the
galvanometer deflects in one direction as the magnet approaches the loop, and the
opposite direction as it moves away.

Faraday’s experiment demonstrates that an electric current is induced in the loop by
changing the magnetic field. The coil behaves as if it were connected to an emf source.
Experimentally it is found that the induced emf depends on the rate of change of
magnetic flux through the coil.

Test it out yourself [https://phet.colorado.edu/en/simulation/faradays-law here]

==See also==
===Further Readings===

''Matter and Interactions, Volume II: Electric and Magnetic Interactions, 4th Edition''

''The Electric Life of Michael Faraday'' (2009) by Alan Hirshfield

''Electromagnetic Induction Phenomena'' (2012) by D. Schieber

===External Links===

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

https://www.nde-ed.org/EducationResources/HighSchool/Electricity/electroinduction.htm

http://www.famousscientists.org/michael-faraday/

http://www.bbc.co.uk/history/historic_figures/faraday_michael.shtml

==References==

''Matter and Interactions, 4th Edition''

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

https://files.t-square.gatech.edu/access/content/group/gtc-970b-7c13-52a7-9627-cdc3154438c6/Test%20Preparation/Old%20Test/2212_Test4_Key-1.pdf

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

Magnetic Force

2020-11-23T15:13:42Z

Yalmonte3: /* Simple */

==The Main Idea==

An electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.

Magnetic forces are on moving particles, not stationary particles which means that the calculation of magnetic force '''MUST''' relate to the particle's velocity (we see this quantitatively with the Biot-Savart Law).

If the source charge is moving, it also generate a magnetic field; so not only is velocity involved in calculation of the magnetic force on a moving particle, or collection of moving particles (as we see in a rod or a wire), but this phenomenal relationship includes magnetic field as well.

Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. "If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest." These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. ''For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.''

'''The Main Idea - Aurora Borealis Edition'''

The Aurora Borealis or more commonly called, 'The Northern Lights' is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted.
==A Mathematical Model==

Suppose we have a moving particle. It has a charge given by ''q''. It has a velocity given by <math>{\vec{v}}</math>. It is also in the presence of a magnetic field given by <math>{\vec{B}}</math>. The force that this particle will experience is given by the following:

'''(1)''' <math>{\vec{F} = q\vec{v}\times\vec{B}}</math>

Therefore, for a particle at rest (<math>{\vec{v} = \vec{0}}</math>), the particle will experience a force given by <math>{\vec{F} = \vec{0}}</math>.

Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).

Note that the above equation '''(1)''' denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way:

'''(2)''' <math>{|\vec{F}| = q|\vec{v}||\vec{B}|sin(\theta)}</math>

In equation '''(2)''', the angle <math>{\theta}</math> represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at, and equation '''(2)''' gives the magnitude of the magnetic force. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. Similarly, if the velocity and magnetic field direction vectors are parallel to each other, and thus the angle spanning the two vectors is zero, then the value of theta is zero. Consequently, the magnitude of the magnetic force is zero. It's important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter.

What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire.

Recall... '''(1)''' <math>{\vec{F} = q\vec{v}\times\vec{B}}</math>

Thus, for some section of charge <math>{\Delta q}</math>... '''(3)''' <math>{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}</math>

... hence, for n charged particles, A cross sectional area, and sectional length <math>{\Delta L}</math>, we have... '''(4)''' <math>{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}</math>

... and now, by re-arranging the terms to collectively represent some current I, we have... '''(5)''' <math>{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}</math>

Equation '''(5)''' can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!

Recall that a moving charged particle generates a magnetic field <math>{\vec{B}}</math> given by the following:

'''(6)''' <math>{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}</math>

Equation '''(6)''' involves the vector <math>{\hat{r}}</math> which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It's important to remember that a charge won't enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.

For our purposes we're going to focus on two things: 1- The circular orbit of the electrons in the Earth's magnetic field 2- The helical orbit of the electrons in the Earth's magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights.

Let's imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation '''(1)'''. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth's magnetic field) then the charged particle will continue to move in a circular path in the x-z plane.

What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.

==A Computational Model==

The following Glowscript model displays a moving particle's path in the presence of a magnetic field. Initially, the particle moves in the negative x direction in the presence of a magneetic field that points in the positive y direction. Therefore, because there is a the particle is moving in some perpendicular component relative to the magnetic field, the particle, in this case an electron, experiences a magnetic force.

Initially when the particle moves in the negative x direction, the magnetic force is in the positive z direction since the cross product of particle's velocity and magnetic field yields a direction in the negative z direction. Because the particle is an electron, however, the particle experiences a force in the positive z direction. Now the question is, would the direction of the magnetic force always point in the positive z direction?

No, the direction of the magnetic force consistently changes since the direction of the particle's velocity continuously changes, and the direction of the magnetic force is dependent on the direction of the velocity of the electron. In fact, because the magnetic force is always perpendicular to the particle's velocity, the magnetic force also acts as a centripetal force that allows the electron to travel in a continuous circle as long as the magnetic field stays constant and no other outside forces suddenly begin to act on the particle.

[[https://trinket.io/glowscript/060ed7ba46?start=result&showInstructions=true Magnetic Force on a Moving Particle Perpendicular to the Magnetic Field]]

However, consider the case where the initial direction of the electron's velocity was not directly perpendicular to the direction of the magnetic field. Because the magnetic field is not completely perpendicular to the magnetic field, the velocity will have parallel and perpendicular components relative to the magnetic field. As a result, the parallel component of the velocity relative to the magnetic field causes the electron to move upwards as demonstrated in the glowscript simulation below rather than a simple circle on the x-z plane. The perpendicular component of the velocity, however, still contributes to the overall circular motion of the electron's path, and thus the overall path of the electron resembles that of a helix.

[[https://trinket.io/glowscript/894615d7dc?showInstructions=true Magnetic Force on a Moving Particle not Directly Perpendicular to the Magnetic Field]]

Also, take note of the iterative calculations made in the code. Within the code, we must initalize values for the initial velocity and momentum, position, mass, and charge of the particle, and magnetic field present in the location of the electron. In the iterative calculations, we must update the value of the magnetic force, as it is constantly changing directions since the electron's velocity is also changing in direction. Similarly, a net force causes a change in momentum, so we must update the momentum and velocity of the particle by utilizing the momentum principle where the derivative of momentum with respect to time is equivalent to the net force acting upon the particle. Furthermore, we update the particle's position and extend and append the trail with the particle's current location to display the path.

==Examples==

We can now consider several example problems related to this topic.

===Simple===

'''Question:'''

A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle?

'''Solution:'''

This situation involves a simple case of the velocity vector and the magnetic field vector appropriately combining to generate a force on our given particle. We have...

<math>{\vec{v} = <4 \times 10^5,0,0> m/s}</math>

<math>{\vec{B} = <0,0,0.2> T}</math>

<math>q = {1.6 \times 10^{-19} C}</math>

Therefore:

<math>{\vec{F} = q\vec{v}\times\vec{B}}</math>

<math>{\vec{F} = (1.6 \times 10^{-19}) <4 \times 10^5,0,0> \times <0,0,0.2>}</math>

The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector (middle finger) in the positive z-direction, then the resulting force vector (direction of your thumb) must be in the negative y-direction.

Thus...

<math>{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = <0, -1.28 \times 10^{-14}, 0> N}</math>

===Middling===

'''Question:'''

Suppose we have a situation where a positively charged particle (<math>{+ q}</math>) of mass ''m'' is in a region where a magnetic field (<math>{\vec{B}}</math>) is applied. It travels at a velocity (<math>{\vec{v}}</math>). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path in which this particle travels in terms of values given?

'''Solution:'''

This problem may appear complicated, but it's not as hard as it seems.

Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:

First... the magnetic force on the particle is given by the following:

<math>{\vec{F} = q\vec{v}\times\vec{B}}</math>

Because <math>{\vec{v}}</math> and <math>{\vec{B}}</math> are effectively perpendicular, the two vectors can be effectively combined in the following way:

<math>{|\vec{F}| = q|\vec{v}| |\vec{B}|}</math>

The force <math>{\vec{F}}</math> is constantly inward to generate a circular motion based path of the particle.

Recall that for circular motion with a constant inward force, the force is given by:

<math>{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}</math>

Thus, we can set the forces equal to each other:

<math>{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}</math>

Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem...

<math>{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}</math>

===Difficult===

'''Problem:'''

Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius <math>{R_1}</math> exists to the left of the origin a distance <math>{d_1}</math> which maintains a current <math>{I_1}</math>. Another current loop of radius <math>{R_2}</math> exists to the right of the origin a distance <math>{d_2}</math>, and it maintains a current <math>{I_2}</math>. The particle itself moves upward on the positive z-axis with a velocity <math>{\vec{v}}</math>.

Assume the following:

<math>{I_1 = I_2}</math>

<math>{R_1 = 0.5R_2}</math>

<math>{d_1 = 3d_2}</math>

<math>{d_1, d_2 >> R_1, R_2}</math>

The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right).

What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the variables staed above.

'''Solution:'''

We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging.

Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.

''For loop 1:''

<math>{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}</math> (This approximation can be used because of the fourth assumption made above, where <math>{d_1}</math> is considerably larger than <math>{R_1}</math>)

''For loop 2:''

<math>{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}</math> (This approximation can be used because of the fourth assumption made above, where <math>{d_2}</math> is considerably larger than <math>{d_2}</math>)

Now we combine the appropriate values for radius and distance in terms of <math>{R_2}</math> and <math>{d_2}</math>, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.

<math>{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}</math>

<math>{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}</math>

<math>{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}</math>

<math>{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}</math>

Now let's pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... ''but wait!'' The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:

<math>{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}</math>

We can now involve our determined magnetic field that was generated by the two current carrying rings.

<math>{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}</math>

The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y).

This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.

===Magnetic Forces in Wires===

Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation:

<math display = "block">|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus</math>

For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F.

====Simple====
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B, the magnetic field on the wire, is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the magnetic force on the wire?

<math>|\vec F_{mag}| = ILBsin(45)</math>

<math>|\vec F_{mag}| = (0.6)(0.005)(sin(45))</math>

<math>|\vec F_{mag}| = 0.002</math> N into the page

====Middling====
A horizontal bar is falling at a constant velocity v. B, the magnetic field, points into the page. What is the the magnitude and direction of current in the bar?

<math>|\vec F_{grav}| = mg</math>

<math>|\vec F_{grav}| = \vec F_{mag}</math> because there is no gravitational acceleration, the net force must equal zero.

<math>mg = I(L\times\vec B)</math>

<math>I = \frac{mg}{LB}</math>

To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.

I, the conventional current, flows to the left.

==Application (i.e. What Does This Have To Do With Anything?)==

This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle.

As a chemist (''The Astrochemist'', in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force.

These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics of the early universe. None of this would be possible without the dynamics of magnetic force!

The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits.

==History==

[[File:James Clerk Maxwell.png|thumb|James Clerk Maxwell]]The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861.

Additionally, the topic of magnetic force can't be explored without magnetic fields. Although magnetic fields had been known for a long time, the direct connection between electricity and magnetism wasn't discovered until the early 1800s by Hans Christian Oersted, who used compass needles. Experiments in the 1800s demonstrated that wires set adjacent together with currents in the same direction were attracted to each other, while those with opposing currents repelled each other.

Consequently, similar experiements were conducted with a static charge placed next to a current carrying wire, where no force was acted upon the static charge. Additionally, another experiment was conducted with a conductor placed in between two current carrying wires. Therefore, scientists could later come to a conclusion that magnetic fields are caused by moving charges, and later scientists determined that any charged particle with a velocity can produce a magnetic field, and magnetic forces can only act on moving charges.

Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820.

Hendrik Lorentz provided the actual "Lorentz Force Law" of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands.

In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions.

In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with help from a US Navy navigational satellite.

== See also ==

===Further reading===

[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]

[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]

[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration's explanation of the Northern Lights]

[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]

[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/ National Geographic - Heavenly Lights]
===External links===

[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]

[https://www.youtube.com/watch?v=X4dXXnUMHbQ&t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]

[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]

[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]

[http://www.youtube.com/watch?v=sENgdSF8ppA Magnetic force fields generated in copper (with more advanced and complex applications)]

==References==

1. Chabay, R.W; Sherwood, B.A.; ''Matter and Interactions''. '''2015'''. ''4''. 805-812.

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

Potential Difference of Point Charge in a Non-Uniform Field

2020-11-21T22:12:49Z

Yalmonte3: /* The Main Idea */

'''Claimed by dmengesha3 Dina''' Claimed by Qianyi Qu Spring 2017

==The Main Idea==
Non uniform electric fields are generated by point charges because the magnitude is different depending on the location at which the electric field is being observed (refer to the equation of the electric field made by a point charge). If one is trying to find the electric potential between two different locations due to a point charge the difference in electric fields at those two points must be considered. The potential difference in non-uniform electric fields can be calculated by adding up the contribution of potential difference of each individual electric field through certain distance we want to evaluate. Generally, in a non-uniform electric field one can write the change in potential difference between two locations to be:
<math> {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} </math> with being the initial location and f being the final location. And the sum of all the contribution of potential difference can be written as: <math> {\triangle V}=-\sum \vec{E} {\bullet} \triangle\vec{l} </math>. The electric field of the object can be that of a point charge, long rod, disk, capacitor, or sphere. Ideally, these equation can be used to calculate potential difference along curve path and situations where the electric field varies in magnitude and direction with respect to distance traveled. For the simplicity of calculation, we will only consider cases with varying electric field on a straight path in this class.

===A Mathematical Model===

'''Steps to calculate Potential Difference in a Non-Uniform Field:'''

1. Pick a path from initial to final location where the electric field can be evaluated at every location.

2. Divide the path into smaller pieces, with each pieces having its unique electric field and <math> \vec{l} </math>.

3. Identify the electric field of the piece.

4. Use the integral equation for potential difference shown above to calculate the potential difference of one piece.

5. Move on to calculate the next piece.

6. Add contribution of all pieces together to get the net potential difference in non-uniform field.

For example, we substitute E for the formula for the electric field of a point charge and dl for r the distance between the point charge and the observation location.

<math>\textstyle\int\limits_{i}^{f}(9x10^9)\frac{Q}{r^2}dr</math>

If we carry out the integration we find that Vf- Vi = <math>\vartriangle(9x10^9)\frac{Q}{r}</math> which can also be written <math>((9x10^9)\frac{Q}{rf})-((9x10^9)\frac{Q}{ri})</math>

This equation can be used to determine the potential difference of a point charge between any two different locations.

===A Computational Model===

'''Refer to this Website:'''

https://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html

To observe how the electric field of a point charge differs from one location to another because it is proportional to 1/r^2. Then play around with measuring the potential difference between two points and see how distance affects the calculation of potential difference. You should observe a 1/r relationship.

Exercises
1. Put down a positive charge. Put 3 E-Field Sensors down: one close to the charge, the next farther away, the next one even farther still

'''Notice that the magnitude of the arrows decreases as the distance from the point charge increases'''

2. Repeat the steps of #1 but replace the positive point charge with a negative point charge.

'''You should notice the same trend occur'''

3. Click "Show E-field" on the right menu bar and move the point charge around the plot

'''Notice that the Electric field gets larger in magnitude as the point charge moves closer to the observation location'''

4. Move the point charge closer to the observation location that measures potential difference

'''Notice that the closer the point charge is to this location the higher the magnitude of the potential difference'''

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
[[File:dipole pot diff.jpg|300px|thumb|right]]

1) If the path is along the line i to f what direction is the electric field of the dipole?

down

up

left

right

Cannot determine

'''Answer: '''

To the left. The electric field would point toward the negative end of the dipole

Is the potential difference Vf-Vi

positive

negative

zero

'''Answer: '''

It is negative because both deltal (displacement) and Electric field are pointing in the same direction which means that their product would have been positive and the negative sign in front of the expression -Edl would make the answer negative.

''Shortcut to remember: ''
E and dl same direction --> negative potential difference
E and dl opposite direction-->positive potential difference

[[File:Problem1 potential diff.jpg|500px|thumb|right|Problem 2 Diagram]]

2) Three charged metal disks are arranged as shown on the right. The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2<math> m^2</math> (this is the area of one side of the disk). Calculate the potential difference Vd-Va.

'''Answer: '''

Path: A <math>\rightarrow</math> D <math>\rightarrow</math> G <math>\rightarrow</math> H <math>\rightarrow</math> B <math>\rightarrow</math> C

Vd-Va and Vh-Vg are zero because E=0 in a metal. Vc-Vb is also zero because the path the perpendicular to <math>\vec{E_2}</math>.

Update Path: A <math>\rightarrow</math> B <math>\rightarrow</math> C

Electric Field of Capacitors: <math> E_{cap} = \frac{Q}{A\epsilon_0} </math>

Vc-Va = (Vb-Vh) + (Vg-Vd) = <math> -\vec{E_2} \bullet \triangle\vec{L_2} - \vec{E_1} \bullet \triangle\vec{L_1}</math>

= <math>-\frac{1.2e-7C*0.003m}{2m^2*8.85e-12 C^2/Nm^2}</math> + <math>-\frac{7e-8C*0.0015m}{2m^2*8.85e-12 C^2/Nm^2}</math>

= 14.4 V

===Middling===

3) Calculate the change in potential moving from location <math>(9x10^-9) m </math> to location <math> (9.7x10^-9) m </math> from a proton.

'''Answer: '''

the final location is: (9.7x10^-9) m

the intial location is: (9x10^-9) m

So if we plug these values into the above equation we found by integrating Edr.

We have:
<math>((9x10^9)\frac{1.6x10^-19}{9.7x10^-9})-((9x10^9)\frac{1.6x10^-19}{9x10^-9})</math>

Which yields a potential difference of -0.0115 V.

[[File:Problem2 pot diff.jpg|300px|thumb|right|Problem 4 Diagram]]

4) A very long, thin glass rod of length 2R carries a uniformly distributed charge +q. A very large plastic disk of radius R, carrying a uniformly distributed charge -Q, is located a distance d from the rod, where d<<R.

a. Predict the sign of the potential difference Vc-Va. Justify your answer.

'''Answer: '''

E and dl are both positive because they point to the right direction. According to the integral form of the potential difference formula for non-uniform field, <math> {\triangle V}=-\int\limits_{i}^{f} \vec{E} {\bullet} \vec{l} </math> , the change in V from position A to C has to be negative.

b. Calculate the potential difference Vc-Va. Location A is located a distance w fro the center of the rod. Location C is located at the disk's center a distance d from the rod.

'''Answer: '''

Assume d<<R. <math> E_{disk} = \frac{Q}{2A\epsilon_0} </math> & <math>E_{rod} = \frac{1}{4\pi\epsilon_0}</math> <math> \frac{2Q}{Lr}</math>

Vc-Va = <math> {\triangle V}=-\int\limits_{i}^{f} \vec{E_{net}} {\bullet} \vec{l} </math> = <math> \triangle {V_{rod}} + \triangle {V_{disk}}</math>= <math>-\int\limits_{i}^{f} \vec{E_{rod}} {\bullet} \vec{l}</math> <math>-\int\limits_{i}^{f} \vec{E_{disk}} {\bullet} \vec{l}</math>

<math> {\triangle V_{rod}}=-\int\limits_{w}^{d} \frac{1}{4\pi\epsilon_0} </math><math>\frac{2q}{xL} dx</math> = <math>\frac{-2q}{4\pi\epsilon_0L} [ln(x)\rbrack^d_w </math> = <math>\frac{-2q}{4\pi\epsilon_0L} ln(\frac{d}{w})</math>

<math> {\triangle V_{disk}}=-\int\limits_{w}^{d} \frac{Q}{2A\epsilon_0} dx</math>= <math>\frac{-Q}{2A\epsilon_0} [x\rbrack^d_w </math> = <math>\frac{-Q}{2A\epsilon_0L} (d-w)</math>

Vc-Va = -<math>[\frac{2q}{4\pi\epsilon_0L} ln(\frac{d}{w})</math> + <math>\frac{Q}{2A\epsilon_0L} (d-w)]</math>

===Difficult===

5) The potential calculated for a point charge in a particular region of space is written by V=14xy+6y-4z, what are the electric field components at location <x,y,z>?

'''Answer: '''

Remember that potential difference <math> \textstyle\int\limits_{i}^{f}-Edl </math> thus Electric field can be calculated as the <math>\frac {-dv}{dx}</math> So one would take the derivative of the potential difference with respect to each variable giving:

E= <-14y,-14x-6,4>

[[File:Problem7 pot diff.jpg|400px|thumb|right|Problem 6 Diagram]]

6) The diagram shows a spherical shell with a uniformed surface charge of -Q and a capacitor with uniformed distributed +Q2 and -Q2. R1<<R2, d<<R2. Calculate the potential difference V2-V1 and V3-V2.

'''Answer:'''

V2-V1:

<math>E_{sphere} =0</math> inside the sphere, so the only electric field acting along this distance is <math> E_{cap}</math>. <math> E_{cap} = \frac{Q_2}{A\epsilon_0}</math>

<math>\triangle V_{cap} = V_2-V_1 = -\int\limits_{0}^{-R_1} \frac{Q_2}{2A\epsilon_0} dy</math> = <math>\frac{-Q}{\pi\epsilon_o} [x\rbrack_0^{-R_1} = \frac{Q_2R_1}{\pi{R_1^2}\epsilon_0}</math>

V3-V2:

The net electric field is due to the sphere and the capacitor. <math> E_{sphere} = -\frac{1}{4\pi\epsilon_0} \frac{Q}{y^2}</math>

<math>\triangle V_{sphere} = -\int\limits_{d}^{R_1} \frac{1}{4\pi\epsilon_0} \frac{Q}{y^2} dy</math> = <math> -\frac{Q}{4\pi\epsilon_0} [\frac{1}{y}\rbrack_d^{R_1}</math> = <math> -\frac{Q}{4\pi\epsilon_0} (\frac{1}{R_1} - \frac{1}{d})</math>

<math>\triangle V_{cap} = -\int\limits_{R_1}^{d} \frac{Q}{A\epsilon_0} dy</math> = <math>-\frac{Q}{A\epsilon_0} (d-R_1) = -\frac{Q}{\pi{R_2^2}\epsilon_0} (d-R_1)</math>

<math>\triangle V_{sphere}+\triangle V_{cap} = V_3-V_2 = -\frac{Q}{4\pi\epsilon_0} (\frac{1}{R_1} - \frac{1}{d}) -\frac{Q}{\pi{R_2^2}\epsilon_0} (d-R_1)</math>

===Connectedness===
How is this connected to something you are interested in?

'''Connections in Chemistry'''

Electrostatic potential energy maps are made of molecules to portray the charge distribution of a molecule 3 dimensionally. These maps can be used to determine electronegativity, bond characteristics, and also help find the reactive sites of molecules. Reactive sites are defined as a charged region of a molecule that interact with other charged particles. This can become especially important in assessing what types of molecular interactions and reactions will take place between two elements or compounds.

How is it connected to your major?

I am a Biology major here at Tech. In biochemistry we use the principle of potential energies of half reactions to calculate the standard free energy <math>\vartriangle G </math> of a reaction in order to understand the reaction's energetics and determine whether the reaction is favorable or not. This predictive information allows biologists to know what chemical reactions take place in the body and which are able to take place in the body.

'''Industrial Application'''
Coulomb Barrier for Nuclear Fusion: For two particles (ex: 2 protons to fuse), they must be able to get close enough to one another for the nuclear strong force to overcome their electric repulsion. One must understand the potential difference of a point charge and the potential energy that creates a barrier between two point charges which can be calculated using U=ke^2/r where k =9e9 and e=1.6e-19 The r calculated using this formula determines the radius at which the nuclear attractive force becomes dominant.

==History==
[[File:volta.jpg]]

In 1800, Alessandro Volta (pictured above) found that metals such as zinc and copper could produce currents; thus, he constructed the first battery what is know called a voltic pile. It was constructed from pieces of zinc and copper in salt water which produced an electric current. The SI unit for potential difference is named after Alessandro Volta.

Refer to:

https://en.wikipedia.org/wiki/Alessandro_Volta

To read more about Volta's early works and life.

== See also ==

'''Calculating Electric Field of a Point Charge'''

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

'''Potential Difference in Uniform Electric Field'''

http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field

===Further reading===

Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions: Electric and Magnetic Interactions. 4th ed. Vol. II. Place of Publication Not Identified: John Wiley, 2015. Print. Chapter 16. Section 16.5

===External links===

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

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

==References==

[[Connectedness]]

http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/coubar.html#c1

http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/General_Principles_of_Chemical_Bonding/Electrostatic_Potential_maps

[[History]]

https://en.wikipedia.org/wiki/Volt

http://content.time.com/time/specials/2007/article/0,28804,1677329_1677708_1677755,00.html

Sign of a Potential Difference

2020-11-21T22:08:14Z

Yalmonte3: /* Harder Example */

Claimed by Junzhe Ruan(Fall 2018)

----
This article discusses the sign of potential difference, and how to determine the sign in different situations involving a particle and an electric field.

== Introduction ==

From previous sections, we know that the change in '''potential energy''' is equal to the charge multiplied by the change in '''potential difference'''(δU = q * δV). From Conservation of Energy(δV + δK = 0), the loss of kinetic energy must result in an increase in potential energy and vice versa in one same system. In another word, the change in the potential difference and kinetic energy can be positive ''or'' negative. For instance, when one object decreases its speed, its kinetic energy decreases. (δK < 0) Due to the conservation of energy, we can get δV > 0. However, we are mainly discussing the change of potential in an electrical system as shown below:[[File:Rsz_2212eq1.png|center]]

As seen from the equation above, two factors can affect the sign of potential energy. First, the charge of the particle being observed can be positive or negative (proton vs. electron). Second, and the topic of this page, the sign of the potential difference can also determine if a particle is gaining or losing potential energy.

[[File:Potential one.jpg|center]]

== Direction of Path vs. Direction of Electric Field ==

From the equation relating potential difference with electric field and motion, we can see that the sign of the potential difference is dependent on the direction of both the electric field and displacement vectors, as shown below.

[[File:Rsz 2212eq2.png|center]]

Consistent with the previous convention, the delta symbol (δ) indicates "final - initial." We will use this same notation in showing the direction of the path. For example δ V = Vb - Va ---> signifies the potential difference between location B (final) and location A (initial).

For this equation, note that the result of the cross product between the electric field and displacement is negated before finding the potential difference. Furthermore, it is important to note that the electric field and displacement vectors are multiplied by the dot product. '''A · B = A * B * cosθ''' Due to this formula, we will analyze 3 different scenarios: path in the direction of the electric field(θ = 0°), path in the ''opposite'' direction of the electric field(θ = 180°), and the path moving ''perpendicular'' to the direction of the electric field(θ = 90°).

== Quick Review of Delta symbol ==

In the equation for the potential difference, we see a Delta for the path, also written as dL sometimes. This delta means a change in the path as taken from two different reference points. These reference points are known as '''initial and final''' or sometimes '''source and observation location'''. To properly calculate a delta or change, you must take the final and subtract the initial from it. More simply, '''Delta is Final-Initial'''. In doing problems with electrical field or magnetic field involving potential difference, the first step we usually do is to calculate r = pos_final - pos_initial. Then use it in several advanced formula: [[File:Electric_field_formula.gif |center]]

Below are some practice problems to understand what dL would be.

[[File:Potential diff two.jpg|center]]

As seen from the problems, direction can differ in the x, y, and z dimensions, but as long as you remember '''Delta = Final-Initial''' or '''Delta = Observation-Source''', you will always get the location right! It is important to understand the dimensions of dL as we move forward with determining what the sign of potential difference is.

== Sign of Potential Difference ==

[[File:Rsz_2212eq3.png|center]]

For the last scenario, it will be critical to have a strong understanding of the dot product and how to calculate it. In a mathematical way, A·B = A*B*cos θ where θ is the angle between vector A and B. If A and B are in the same direction, '''θ = 0, cos θ = 1, then A·B = A*B'''. If A and B are in opposite direction, '''θ = 180, cos θ = -1, then A·B = -A*B'''. Last but not least, if A and B are perpendicular to each other, '''θ = 90, cos θ = 0, then A·B = 0'''. This means that when finding the dot product of two perpendicular vectors, the result will be '''zero'''.
To better understand this, imagine the electric field between two very long capacitor plates. The electric field points from one plate to the other, let's say in the +x direction. If you wanted to move a charge at any location between the plates in either the +y or -y direction, your displacement vector would point in one of these directions. Whether you moved the particle +y or -y direction, the dot product of the electric field and displacement, and therefore the potential difference, will be zero.

== Real Life Example to Explain Sign of Potential Difference ==

Imagine you are swimming in a stream. The energy the swimmer uses or does not use in this system will represent potential difference. If a swimmer is going along the stream then comes upon a current that starts to push him forward, he does not need to keep stroking in the water as the current can just carry him from one location to the other. This is a great representation of potential difference. Imagine the current in the swimming example is an electric field and that there are two set locations for the swimmer to go from and to. If the current and the change in location (final-initial) are in the same direction, then the swimmer had to use less energy swimming. This drop in energy can represent the negative potential difference. Both the current (electric field) and change in location (dL) were in the '''same direction''', so the swimmer let the current carry him rather than exert energy himself ('''negative potential difference''').

Now imagine if the same swimmer decided to test his endurance skills and swim upstream instead, or against the current. Now, the theoretical electrical field and change in location (dL) are opposing each other. This now means that the swimmer must exert a lot of his own energy to get from point A to B while opposing the current pushing against him. Looking at the energy in the scenario, the swimmer now has to provide a change in energy that he, himself, makes, so the potential difference would be positive. In reference to the swimmer, he is providing the energy as opposed to letting the current provide it like in the previous example. This shows that '''opposite directions''' mean a '''positive potential difference'''.

Finally, imagine if you were swimming perpendicular to the current. Now, the dL and electrical field equivalents are '''running perpendicular to each other'''. The swimmer does not have to exert any more or less energy to get from point A to B because the current is not in the same dimension as the initial and final locations. This shows why '''potential difference = 0''' when the dot product of Electric Field and dL is carried out.

Another real-life example is to imagine you are carrying a ball and prepare to throw it from the building. If you go downstairs (the same direction as the gravity.), the potential of the ball you carry is decreasing, because when you throw it, it takes a shorter time to reach the ground. Thus this movement has a negative potential difference. So on so forth, if you go upstairs (opposite direction as the gravity), the potential increases because it takes a longer time for the ball to reach the ground and when it reaches it, it has a larger energy. Due to δpotential = final potential - initial potential, the potential difference is positive. From this example, again, we can conclude that "opposite directions" means positive potential difference and same direction means negative potential difference. It should be the same on gravity potential field and electrical potential field.

== Easier Examples ==
Reminder:

Path going in direction of ''E Field'' ------> Potential is '''decreasing'''

Path going opposite to ''E Field'' -------> Potential is '''increasing'''

Path perpendicular to ''E Field'' --------> Potential '''does not change / =0'''

[[File:Old_examples.jpg|center]]

To better understand this, we could imagine in this way: E field makes positively charged particle to move along its path, as a positive particle, as it goes in direction of E Field, its potential of moving decrease. In another word, after the E field use its function (move the particle in a particular direction), the potential of it moving the particle decrease. And vice versa.

== Harder Example ==

[[File:Pot diff three.jpeg|center]]

Let's go through this problem step by step.

'''a)''' How would you get the dL of the path going from B to A? Well, recall that '''dL = final-initial''', so you must do: location of A - location of B:

<-2.5, 1, 0> - <4, -2.4, 0> = <-6.5, 3.4, 0> meters

Now you have the dL or the vector from B to A.

'''b)''' Now, how do you get the potential difference if a particle is moving from B to A given the electric field and the dL from part a)? You know it is just the dot product of Electric Field and dL:

∆V = -(''E'' · dL) = -((-6.5 x -450) + (3.4 x 300) + (0 x 0)) = -(2925 + 1020 + 0) = -3945 volts

'''c)''' Now that you know the potential difference, ∆V, how would you calculate the potential energy of the system if the particle was a proton? Well, now you just multiple the ∆V by the q, or charge of the proton.

∆U = ∆V x q = -3945 x 1.6 e-19 = -6.3 e-16 joules

'''d)''' Now, how would you calculate the potential energy of the system if the particle was an electron instead of a proton? Well, you know that the only difference between this and part c) is that the charge is now negative to account for the electron but the same number.

∆U = ∆V x q = -3945 x -1.6 e-19 = 6.3 e-16 joules

These questions are very simple as long as you think methodically about what you are doing and always keep in mind two very important things:

'''1)''' What is the initial and final for the particle's path? Again, '''dL is always final-initial'''!

'''2)''' Is the particle a proton or an electron? The number for charge will be the same but the sign will change dependent on which one it is. '''Sign is important!'''

== Summary ==

Determining potential difference and its sign is very simple if you remember these ideas:

'''1.''' Electric field and displacement vectors can be in all three dimensions, but the dot product makes the sum of all these dimensions the answer.

'''2.''' dL will '''ALWAYS''' be final-initial!

'''3.''' Never forget the negative in the ∆V = -(''E'' · dL) equation. This is probably the easiest way to mess up a problem because it is hardest to remember. Just recall the swimmer example from above if you get confused!

When determining the sign of the potential difference, there will be 3 different scenarios that will determine whether the sign of the potential difference is positive, negative, or zero. Pay close attention to notation when considering the change in a quantity (potential, displacement, etc..) in order to avoid confusing the wrong sign. Attention to detail when find the sign of potential difference will make solving the more difficult problems at the end of the chapter a little easier. The following summary of the 3 scenarios is extremely helpful:

[[File:Rsz_2212eq4.png|center]]

Potential Difference in a Uniform Field

2020-11-21T21:35:21Z

Yalmonte3: /* The Main Idea */

By: dachtani3

Claimed by SeongEun Park April, 2017

Electric potential is a scalar quantity that is used to describe the change in electric potential energy per unit charge. This page will elaborate on the change in electric potential in a uniform field.

==The Main Idea==

Potential difference is the change in electric potential between a final and initial location when work is done on a charge to affect its potential energy. The unit for electric potential is a volt (which is also joule/coulomb). Approximation of uniform field is useful for simple calculation.
It is important to remember that since energy is stored in a form of electric field, there will be no potential energy for a single point charge alone; potential energy, thus the potential difference, exists only when there are pairs of interacting particles.

===A Mathematical Model===

This equation represents potential difference, where V is electric potential, U is electric potential energy, and q is unit charge.
The unit for electric potential is 'V' (volt), where it takes 1 Joules of work to move 1 Coulomb of unit charge per 1 volt.
:<math>\Delta V</math> = <math>\frac{dU}{q} </math>

You can rearrange this equation to also show that electric potential energy is electric potential times unit charge.
:<math>\Delta U</math> = <math>q \Delta V</math>

In addition, you can express electric potential as the sum of the dot product of electric field and displacement in each dimension. This expression is as follows:
:<math>\Delta V</math> = -<math>\vec{E}</math>●<math>\Delta \vec{l}</math> = -(<math>E_{x}</math>●<math>\Delta x</math> + <math>E_{y}</math>●<math>\Delta y</math> + <math>E_{z}</math>●<math>\Delta z</math>)
You can also use dot product notation:
:<math>\Delta V</math> = -<math>\vec{E}</math>●<math>\Delta \vec{l}</math> = <math> (E_{x},E_{y},E_{z})●(dX, dY, dZ) </math>

Potential difference can be either positive or negative because each component of electric field and displacement can be positive or negative. Within this expression, you can also note that the units for the electric field are V/m. We originally learned that units for electric field are N/c, but V/m is also an appropriate unit.

===A Computational Model===

This video showcases 3D-models of the interaction of charges with electric potential and electric potential energy. This video is especially helpful if you want to learn these topics conceptually.

https://www.youtube.com/watch?v=-Rb9guSEeVE

In addition, this image showcases the relationship between electric fields and electric potential. If you take the negative gradient of electric potential, the result is the electric field. The gradient is the direction of the hills seen in the graph representing electric potential.

[[File:3-d_model.jpg]]

==Examples==

===Simple===

Questions 1 and 2 are based on the following situation:

A path consists of two locations. Location 1 is at <0.1,0,0>m and Location 2 is at <0.5,0,0>m. A uniform electric field of <300,0,0> N/C exists in this region pointing from Location 1 to Location 2.

Question 1: What is the difference in electric potential between the two points?

:<math>\Delta{L}</math> = final location - initial location = <0.5, 0,0 > - <0.1, 0, 0> = <0.4, 0, 0> m
:<math>\Delta{V}</math> = -<math>\vec{E}</math>●<math>\Delta \vec{l}</math> = -(<math>E_{x}</math>●<math>\Delta x</math> + <math>E_{y}</math>●<math>\Delta y</math> + <math>E_{z}</math>●<math>\Delta z</math>)
= -(300 N/C*.4 + 0*0 + 0*0)
= -120

Answer: -120 V

Question 2: What is the change in electric potential energy for a proton on this path?

:dU = dV*q = -120 V * 1.6e-19 C = -1.92e-17 J

Answer: - 1.92e-17 J

===Middling===
figure 1:
[[File:Slide1.jpg]]

Key Point to remember about signs:

1) If the path is going in the direction of the electric field, electric potential is decreasing.

:Ex: Refer to Simple Example
:Ex 2: Figure 1
- Consider the path going from point A to point B. The path is in the same direction as the electric field (red arrow).
Because ΔV = -E●Δl = -E●Δr(A->B) = -E●d. This then refers ΔV = Vb-Va < 0, meaning that the electric potential is decreasing.

2) If the path is going in the opposite direction of the electric field, electric potential is increasing.

:Question: Location 1 is at <.1,0,0> and Location 2 is at <.5,0,0>. The uniform electric field is <300,0,0> N/C. The charge travels from Location 2 to Location 1.

:new dl = <.1,0,0> - <.5,0,0> = <-.4,0,0>m
:dV = -Exdx = -(-.4*300) = +120V

:Answer: + 120 V

:Ex 2: Figure 1
- Refer back to the figure 1 above. If the path from point C to D was taken, it is going in the opposite direction of the electric field, which means that ΔV > 0;
therefore, the potential is increasing.

3) If the path is perpendicular to the electric field, electric potential does not change.

:Question: Location 1 is at <.1,0,0> and Location 2 is at <.5,0,0>. The uniform electric field is <0,300,0> N/C. The charge travels from Location 1 to Location 2.

:new dL = <.5,0,0> - <.1,0,0> = <.4,0,0>m
:dV = -(Exdx + Eydy + Exdz) = -(.4*0 + 0*300 +0*0) = 0 V

:Answer: 0V

:Ex 2: Figure 1
- Refer back to the figure 1 above. If the path is going from point B to C, or (point A to D), these points are perpendicular to the electric field. Because no work is required to move a
a charge between them, all points that are perpendicular to same electric field share same electric potential.

===Difficult===

The difficult example will refer to using a path that is not directly parallel to the electric field.

Question: Assume an electric field that has a magnitude of 300 N/C. The electric field is uniform. The path chosen is 5 m, 50 degrees away from the field. What is the difference in electric potential in this situation?

:dV = -E*dl = -E*l*cos(theta)
:dV = -(300 N/C)*(5 m)*cos(50) = -964 V

Answer: -964 V

==Connectedness==

'''How is this topic connected to something that you are interested in?'''

:Potential difference is extremely applicable and interesting to me because I am a computer engineer and as a computer engineer, I deal with many things regarding electricity. I am particularly interested in building electric guitar pedals. In addition to the digital signal processing and programming aspect of guitar pedals, there is a lot of physics involved with how they work. Potential difference and electric potential kick in because in addition to the signals that are transmitted throughout the TRS cables, current also runs through them. Obviously, energy and charge are also very prominent in this so they are directly applied to the construction of guitar pedals.

'''How is it connected to your major?'''

:As a computer engineer, in class and ideally in my job after college, I will be dealing with electronics and electricity in action. I am constantly dealing with breadboards, semiconductors, microprocessors, cables, etc. All of these things essentially require a basic understanding of physics topics especially related to electromagnetism.

'''Is there an interesting industrial application?'''

:An interested industrial application of electric potential is the ability to use electric potential sensors in human body electrophysiology. In a study conducted by the Center for Physical Electronics and Quantum Technology, the team utilized these sensors to detect electric signals in the human body. The sensors were mainly built with electrometer amplifiers. Specific to electric potential, the senors focus on displacement current rather than electric current at certain locations. The following signals were detected: electrocardiograms, electroencephalograms, and electro-oculograms, and the sensors were able to find the three signals without directly touching the human body. With these findings, the scientists are able to create "spatio-temporal array imaging" of different areas of the human body, including the heart and brain!

==History==

-Alessandro Volta, and Italian physicist, contributed many ideas and inventions to the field of electricity. He invented the first electric battery, the first electromotive series, and most notably, contributed to the idea of electric potential and its unit, the volt.

-In 1745, Volta was born. He spent most of his childhood experimenting with electricity in his friend's physics lab. When he was 18, he started communicating with physicists, Jean-Antonie Nollet and Giambatista Beccaria, who encouraged him to continue with his experiments.

-In 1775, Volta began teaching physics. He soon was able to isolate methane gas, which he discovered could produce electric sparks. In 1776, Volta put the two ideas together to conclude that he could send electric signals across Italy with the sparking machine.

-In 1778, Volta discovered electric potential, or voltage. He realized that the electric potential in a capacitor is directly proportional to the electric charge in that capacitor.

-In 1800, Volta combined all of his findings to create the voltaic pile, or the first electrochemical cell. This battery made of zinc and copper was able to produce a steady and constant electric current.

-Batteries today serve as the major practical application of electric potential. The unit for electric potential, the Volt, is named after Alessandro Volta and his contributions to the field of electricity.

== See also ==

===Further reading===

'''Books, Articles or other print media on this topic'''

-Electric Potential Difference across a cell membrane:
:https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=electric+potential+difference+journal+articles

-Electric Potential Difference between ion phases:
:http://pubs.acs.org/doi/abs/10.1021/j150300a003

===External links===

'''Internet resources on this topic'''

This video that walks through examples on potential difference in a uniform field:
:https://www.youtube.com/watch?v=mF3VAjcjvOA

This video further discusses electric potential and its concepts:
:https://www.youtube.com/watch?v=wT9AsY79f1k

This video on electric potential energy gives a good background on solving problems before learning specifically about electric potential:
:https://www.khanacademy.org/science/physics/electric-charge-electric-force-and-voltage/electric-potential-voltage/v/electric-potential-energy

==References==

-http://study.com/academy/lesson/what-is-electric-potential-definition-formula-quiz.html

-http://www.famousscientists.org/alessandro-volta/

-http://www.isipt.org/world-congress/3/269.html

-http://maxwell.ucdavis.edu/~electro/potential/overview.html

-Matter & Interactions Vol II: Electric and Magnetic Interactions textbook

[[Category:Fields]]

Charge Transfer

2020-08-26T13:30:12Z

Yalmonte3: /* Middling */

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 negative. 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?]]