Physics Book - User contributions [en]

Charged Capacitor

2026-04-26T01:13:25Z

Vtadakamalla3:

Claimed by Vivan Tadakamalla(Spring 2026)

==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==
There are many real world applications of charged '''capacitors''':

'''Medical Devices - Defibrillators'''

A life saving application of a charged '''capacitor''' is the cardiac defibrillator. In this device a '''capacitor''' is charged to a high voltage then discharged in milliseconds through the chest of a patient to restore and maintain normal sinus rhythm. This is done through the ability of the device to store and quickly release a precise amount of energy.

'''Energy Storage & Power Supplies'''

'''Capacitors''' are able to discharge electrical energy at instantaneous rates. This makes them important in power supply circuits where they smooth out the voltage fluctuations. Unlike batteries, '''capacitors''' are able to discharge their energy at microseconds giving them critical applications where rapid bursts of power are required.

'''Flash Photography'''

The camera flash seen across many devices is an application of a classic '''capacitor'''. Once the '''capacitor''' is charged by the device battery it is able to rapidly discharge to produce a burst of light. This is important as the '''capacitor''' is able to deliver far more instantaneous power than a battery alone.

'''Touchscreens'''

[[File:SurfaceCapacitance.jpg|400px|thumb|right|Diagram of a surface capacitive touchscreen showing electrode layers and electric current flow upon finger contact.]]

Most smartphones and tablets use a grid of '''capacitive''' touchscreens that are just behind the surface of the screen. When a finger (conductor) comes in contact with the screen, it is able to alter the specific electric field and therefore detect a precise touch at a given location.

'''Electric Vehicles (Supercapacitors)'''

Supercapacitors are high capacity versions of '''capacitors''' used in both hybrid and electric vehicles. Their main purpose is to capture the energy created in regenerative braking as well as rapidly release this energy during acceleration. In addition, '''capacitors''' are able to preserve the car battery health, which would otherwise degrade under rapid charge and discharge cycles.

==History==

[[File:LeydenJar.jpg|300px|thumb|right|A Leyden jar, the first capacitor, consisting of a glass jar coated in metal foil used to store static electric charge (1745).]]

The charged '''capacitor''' can be originated back to 1745 when '''Pieter van Musschenbroek''', Dutch physicist from the University of Leiden, first created the Leyden jar. This was a glass jar coated with metal foil and was designed to store static electric charge, technically making it the first '''capacitor'''. Later, '''Benjamin Franklin''' used these jars for his famous electrical experiments, which greatly advanced the field in understanding how charge accumulates and discharges. Later '''Michael Faraday''' in the 19th century was able to describe the science of capacitance in a more scientific way through introducing the concept of the dielectric constant. He was able to explain how insulating materials between the conductive plates were able to store charge. The modern definition of a '''capacitor''', C = Q/V, can be attributed to him with the Farad (F) being the SI unit of capacitance.

==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

How Equipment Works. (2025). How defibrillators work explained simply. https://www.howequipmentworks.com/defibrillator/

Usha Power. (2024). How capacitors affect power supply performance. https://ushapower.com/blog/how-capacitors-affect-power-supply-performance/

All About Circuits. (2022). Electric fields and capacitance. https://www.allaboutcircuits.com/textbook/direct-current/chpt-13/electric-fields-capacitance/

TechTarget. (2022). How does a capacitive touchscreen work? https://www.techtarget.com/whatis/definition/capacitive-touch-screen

Crow, M. (2025). Green energy management in electric vehicles with regenerative braking. Global NEST Journal. https://journal.gnest.org/publication/gnest_07439

Cinco Capacitor. (n.d.). Do you know the capacitor history? https://www.cincocapacitor.com/do-you-know-the-capacitor-history.html

[[Category: Electric Fields]]

File:LeydenJar.jpg

2026-04-26T01:10:47Z

Vtadakamalla3:

File:SurfaceCapacitance.jpg

2026-04-26T01:10:10Z

Vtadakamalla3:

Charged Capacitor

2026-04-26T01:01:38Z

Vtadakamalla3:

Claimed by Vivan Tadakamalla(Spring 2026)

==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==
There are many real world applications of charged '''capacitors''':

'''Medical Devices - Defibrillators'''

A life saving application of a charged '''capacitor''' is the cardiac defibrillator. In this device a '''capacitor''' is charged to a high voltage then discharged in milliseconds through the chest of a patient to restore and maintain normal sinus rhythm. This is done through the ability of the device to store and quickly release a precise amount of energy.

'''Energy Storage & Power Supplies'''

'''Capacitors''' are able to discharge electrical energy at instantaneous rates. This makes them important in power supply circuits where they smooth out the voltage fluctuations. Unlike batteries, '''capacitors''' are able to discharge their energy at microseconds giving them critical applications where rapid bursts of power are required.

'''Flash Photography'''

The camera flash seen across many devices is an application of a classic '''capacitor'''. Once the '''capacitor''' is charged by the device battery it is able to rapidly discharge to produce a burst of light. This is important as the '''capacitor''' is able to deliver far more instantaneous power than a battery alone.

'''Touchscreens'''

Most smartphones and tablets use a grid of '''capacitive''' touchscreens that are just behind the surface of the screen. When a finger (conductor) comes in contact with the screen, it is able to alter the specific electric field and therefore detect a precise touch at a given location.

'''Electric Vehicles (Supercapacitors)'''

Supercapacitors are high capacity versions of '''capacitors''' used in both hybrid and electric vehicles. Their main purpose is to capture the energy created in regenerative braking as well as rapidly release this energy during acceleration. In addition, '''capacitors''' are able to preserve the car battery health, which would otherwise degrade under rapid charge and discharge cycles.

==History==
The charged '''capacitor''' can be originated back to 1745 when '''Pieter van Musschenbroek''', Dutch physicist from the University of Leiden, first created the Leyden jar. This was a glass jar coated with metal foil and was designed to store static electric charge, technically making it the first '''capacitor'''. Later, '''Benjamin Franklin''' used these jars for his famous electrical experiments, which greatly advanced the field in understanding how charge accumulates and discharges. Later '''Michael Faraday''' in the 19th century was able to describe the science of capacitance in a more scientific way through introducing the concept of the dielectric constant. He was able to explain how insulating materials between the conductive plates were able to store charge. The modern definition of a '''capacitor''', C = Q/V, can be attributed to him with the Farad (F) being the SI unit of capacitance.

==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

How Equipment Works. (2025). How defibrillators work explained simply. https://www.howequipmentworks.com/defibrillator/

Usha Power. (2024). How capacitors affect power supply performance. https://ushapower.com/blog/how-capacitors-affect-power-supply-performance/

All About Circuits. (2022). Electric fields and capacitance. https://www.allaboutcircuits.com/textbook/direct-current/chpt-13/electric-fields-capacitance/

TechTarget. (2022). How does a capacitive touchscreen work? https://www.techtarget.com/whatis/definition/capacitive-touch-screen

Crow, M. (2025). Green energy management in electric vehicles with regenerative braking. Global NEST Journal. https://journal.gnest.org/publication/gnest_07439

Cinco Capacitor. (n.d.). Do you know the capacitor history? https://www.cincocapacitor.com/do-you-know-the-capacitor-history.html

[[Category: Electric Fields]]

Charged Capacitor

2026-04-26T00:59:28Z

Vtadakamalla3:

Claimed by Vivan Tadakamalla(Spring 2026)

==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==
There are many real world applications of charged '''capacitors''':

'''Medical Devices - Defibrillators'''

A life saving application of a charged '''capacitor''' is the cardiac defibrillator. In this device a '''capacitor''' is charged to a high voltage then discharged in milliseconds through the chest of a patient to restore and maintain normal sinus rhythm. This is done through the ability of the device to store and quickly release a precise amount of energy.

'''Energy Storage & Power Supplies'''

'''Capacitors''' are able to discharge electrical energy at instantaneous rates. This makes them important in power supply circuits where they smooth out the voltage fluctuations. Unlike batteries, '''capacitors''' are able to discharge their energy at microseconds giving them critical applications where rapid bursts of power are required.

'''Flash Photography'''

The camera flash seen across many devices is an application of a classic '''capacitor'''. Once the '''capacitor''' is charged by the device battery it is able to rapidly discharge to produce a burst of light. This is important as the '''capacitor''' is able to deliver far more instantaneous power than a battery alone.

'''Touchscreens'''

Most smartphones and tablets use a grid of '''capacitive''' touchscreens that are just behind the surface of the screen. When a finger (conductor) comes in contact with the screen, it is able to alter the specific electric field and therefore detect a precise touch at a given location.

'''Electric Vehicles (Supercapacitors)'''

Supercapacitors are high capacity versions of '''capacitors''' used in both hybrid and electric vehicles. Their main purpose is to capture the energy created in regenerative braking as well as rapidly release this energy during acceleration. In addition, '''capacitors''' are able to preserve the car battery health, which would otherwise degrade under rapid charge and discharge cycles.

==History==
The charged '''capacitor''' can be originated back to 1745 when '''Pieter van Musschenbroek''', Dutch physicist from the University of Leiden, first created the Leyden jar. This was a glass jar coated with metal foil and was designed to store static electric charge, technically making it the first '''capacitor'''. Later, '''Benjamin Franklin''' used these jars for his famous electrical experiments, which greatly advanced the field in understanding how charge accumulates and discharges. Later '''Michael Faraday''' in the 19th century was able to describe the science of capacitance in a more scientific way through introducing the concept of the dielectric constant. He was able to explain how insulating materials between the conductive plates were able to store charge. The modern definition of a '''capacitor''', C = Q/V, can be attributed to him with the Farad (F) being the SI unit of capacitance.

==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

2026-04-26T00:56:27Z

Vtadakamalla3:

Claimed by Vivan Tadakamalla(Spring 2026)

==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==
There are many real world applications of charged capacitors:

Medical Devices - Defibrillators

A life saving application of a charged capacitor is the cardiac defibrillator. In this device a capacitor is charged to a high voltage then discharged in milliseconds through the chest of a patient to restore and maintain normal sinus rhythm. This is done through the ability of the device to store and quickly release a precise amount of energy.

Energy Storage & Power Supplies

Capacitors are able to discharge electrical energy at instantaneous rates. This makes them important in power supply circuits where they smooth out the voltage fluctuations. Unlike batteries, capacitors are able to discharge their energy at microseconds giving them critical applications where rapid bursts of power are required.

Flash Photography

The camera flash seen across many devices is an application of a classic capacitor. Once the capacitor is charged by the device battery it is able to rapidly discharge to produce a burst of light. This is important as the capacitor is able to deliver far more instantaneous power than a battery alone.

Touchscreens

Most smartphones and tablets use a grid of capacitive touchscreens that are just behind the surface of the screen. When a finger (conductor) comes in contact with the screen, it is able to alter the specific electric field and therefore detect a precise touch at a given location.

Electric Vehicles (Supercapacitors)

Supercapacitors are high capacity versions of capacitors used in both hybrid and electric vehicles. Their main purpose is to capture the energy created in regenerative braking as well as rapidly release this energy during acceleration. In addition, capacitors are able to preserve the car battery health, which would otherwise degrade under rapid charge and discharge cycles.

==History==
The charged capacitor can be originated back to 1745 when Pieter van Musechenbroek, Dutch physicist from the University of Leiden, first created the Leyden jar. This was a glass jar coated with metal foil and was designed to store static electric charge, technically making it the first capacitor. Later, Benjamin Franklin used these jars for his famous electrical experiments, which greatly advanced the field in understanding how change accumulates and discharges. Later Michael Faraday in the 19th century was able to describe the science of capacitance in a more scientific way through introducing the concept of the dielectric constant. He was able to explain how insulating materials between the conductive plates were able to store charge. The modern definition of a capacitor, C = Q/V, can be attributed to him with the Farad (F) being the SI unit of capacitance.

==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

2026-04-25T05:21:41Z

Vtadakamalla3:

Claimed by Vivan Tadakamalla(Spring 2026)

==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]]

Steady State

2026-04-25T04:44:03Z

Vtadakamalla3: Undo revision 47986 by Vtadakamalla3 (talk)

'''Claimed Alex Dulisse (Spring 2018)'''

Imagine a circuit in which the only battery has been disconnected. In such a setup there would be no electric field in the circuit and therefore no moving charges so the circuit would be in static equilibrium. When the battery is reconnected, an electric field will quickly propagate through the circuit and the charges in the circuit will accelerate. After a short time, the charges will move at a constant speed through the circuit and the electric field at each part of the circuit will remain unchanged. This means that at any given moment, all properties of the circuit remain unchanged even though an electric field is acting on the charges. At this time, when the state of the circuit does not change with the passage of time, the circuit is said to be in '''steady state.'''

[[Image:current_thing.png|thumb|upright = 0.2|right|450px|Current Diagram]]

==The Main Idea==

If a circuit is in steady state the following will be true:
:* At each part of the circuit, the drift velocity of charges remains unchanged with time.
:* The maximum amount of energy has been stored in capacitors, inductors.
:* Excess charge only accumulates on the surface of the wire.
:* An unchanging voltage exists between any two points in the system.

Each of these bullets is a different way of saying the same thing, that the state of the circuitry acting on any charges remains unchanged with respect to time. It is important to remember that a circuit being in steady state does not mean that the drift speed is the same every where in the circuit, only that it is unchanging for each specific location in the circuit. Also, a circuit in which there is no current may still be in steady state (such as an RC circuit in which the capacitors are fully charged), as long as there is still a constant electric field in all conductive parts of the circuit.

'''A Conceptual Understanding'''

For charges to move through the circuit, there must be an applied electric field that causes the mobile charges to move. Since there cannot be excess charge in the interior of a conductor, the surface charges must be producing the electric field. This electric field must be parallel to the wire because of the properties of a conductor. In a conductor, mobile charges always move to cancel out an electric field, meaning that the only electric field in the circuit must be parallel to the direction of current flow.
It is also important to conceptually understand why this electric field eventually reaches steady state. When a battery is first connected to a circuit, the electric field increases in magnitude, the rate at which the charges move through the circuit increases in magnitude too. As charges move through the circuit, there is a drag force in the opposite direction of their motion (similar to friction) known as the resistance of the wire. Since <math>I=V/R</math>, eventually these opposing forces will be equal in magnitude, bringing the circuit to steady state. Once a circuit is described as being in the steady state, there must be a constant electric field in the wire, the electric field has uniform magnitude throughout wire sections with similar cross-sectional area, and the electric field is parallel to the wire at every location along the wire.

[[File:Wire_Charges2.png|thumb = Wire_Charges2.png|upright=0.1|400px|center|Surface Charge Distribution]]

'''Steady State vs. Other States'''
When circuits move from static equilibrium to steady state, they typically do so in a very small amount of time. However, the [[Non Steady State]], or Transient State, which takes place in between, is an important aspect of circuits. When a circuit is in the Non Steady State, over time its state will approach the steady state. After enough time has elapsed, the difference is so slim that the circuit can be considered to be in steady state, even though in theory it never quite reaches steady state. The equations to model Non Steady States are different depending on the type of circuit being analyzed.

The following chart compares common properties of circuits in Static Equilibrium, Steady State, and Non Steady States:

{| class="wikitable"
|-
!
! Static Equilibrium
! Steady State
! Non Steady State
|-
| <math>\bar{v}</math>
| <math>= 0</math>
| <math>\neq 0</math>
| <math>\neq 0</math>
|-
| <math>\frac{d\bar{v}}{dt}</math>
| <math>= 0</math>
| <math>= 0</math>
| <math>\neq 0</math>
|-
| <math>\sum E_{inside}</math>
| <math>= 0</math>
| <math>\neq 0</math>
| <math>\neq 0</math>
|-
|}

===A Mathematical Model===
'''Current'''

There are two ways to describe current in a circuit, the electron current (<math>i</math>), and the conventional current (<math>I</math>). Electron current describes the motion of charges as it actually happens in the circuit: electrons moving through the conductor. Conventional current has the same magnitude and opposite direction as electron current as it describes positive charges moving through a circuit. Although circuits will sometimes act like positive charges are moving through the circuit if there are “electron holes,” it is important to understand that there are not positively charged particles moving through circuits. Even though it is backwards from the physical reality, conventional current is more commonly used because the motion of electrons in circuits was not understood when the first discoveries of electricity were made.
The following formulas can be used to describe electron and conventional current, respectively:
<br><math>i= nA\bar{v}</math>

<math>I = \left | q \right |nA\bar{v}</math>

n = charge density, the number of charged particles per unit volume

A = cross sectional area of the wire

v = drift speed, the speed at which mobile charges move through a section of wire

q = charge of the mobile particles being described (only matters for conventional current)

Keep in mind that drift velocity <math>\bar{v}=\mu \left | E_{applied} \right |</math>

Where <math> \mu </math> = electron mobility.

If you have trouble thinking of this conceptually, imagine a pipe carrying water. For a certain amount of water to move through each part of the pipe per unit time, if the pipe is smaller, the water must move faster and for more water to be delivered, either the speed of the water must increase or the size of the pipe must increase.

'''Kirchoff's Node Rule'''

When a wire splits in to two, this junction is referred to as a node. Current flowing through the node must follow the equation <math>\Sigma I_{in} = \Sigma I_{out}</math>.
This is called the “[[Node Rule]].” All this is saying is that exactly what enters the node must exit. In other words, charge dose not disappear or come into existence just because one wire splits in two.

[[Image:NodeRule2.jpg|thumbnail = NodeRule2.jpg|upright=0.5|400 px|center|Node Rule]]

==Examples==

===Simple===
'''Question'''

What are the values of X and Y in the following circuit?
[[Image:Node_rule_simple_problem.png|thumbnail = Node_rule_simple_problem.png|upright=0.5|frame|center|Node Rule]]

'''Solution'''

Based on the node rule, 10=3+3+X. Therefore X=4.<br>
Y=3+X, therefore Y=3+4=7<br>

===Middling===

[http://cnx.org/contents/Ax2o07Ul@9.39:3ct4v3c5@4/Current| (Example taken from the OpenStax College Physics Textbook)]

'''Question'''

Calculate the drift velocity of electrons in a 12-gauge copper wire (which has a diameter of '''2.053 mm''') carrying a '''20.0-A''' current, given that there is one free electron per copper atom. The density of copper is <math>8.80\times 10^{3} kg/m^3</math>.

'''Solution'''

First, calculate the density of free electrons in copper. There is one free electron per copper atom. Therefore, is the same as the number of copper atoms per <math>m^3</math>. We can now find <math>n</math> as follows:

<math>n= \frac{1e-}{atom}\times \frac{6.02\times10^{23}atoms}{mol}\times \frac{1 mol}{63.54 g} \times \frac{1000 g}{kg} \times \frac{8.80 \times 10^3 kg}{1 m}</math>

<br><math>n= 8.342\times 10^{28}e−/m^3</math>
<br>The cross-sectional area of the wire is
<math>A= \pi r^2 = \pi(2.053\times0.5\times10^{−3})^2 = 3.3103 \times 10^{–6}m^2</math>

<br>Rearranging <math>I= \left | q \right |nA\bar{v}</math> to isolate drift velocity gives you:
<br><math>\bar{v}= \frac{I}{n\left |q \right |A}</math>
<br><math>\bar{v}= \frac{20.0A}{(8.342\times10^{28}/m^3) (1.60\times10^{-19}C)(3.310\times10^{-6}m^2)}</math>
<br><math>\bar{v}= 4.53\times 10^{–4}m/s </math>

===Difficult===

'''Question'''

A circuit made of two wires is in the steady state. The battery has an emf of '''1.5 V'''. Both wires are length '''25 cm''' and made of the same material. The thin wire's diameter is '''0.25 mm''', and the thick wire's diameter is '''0.31 mm'''. There are <math>4\times 10^{28}</math> mobile electrons per cubic meter of this material, and the electron mobility of this material is '''0.0006 (m/s)/(V/m)'''.

A) If the magnitude of the electric field in the thin wire is '''3.5 V/m''', what is the electric field in the thick wire?<br>
B) What is the drift velocity in the thin and thick wires?

[[File:sscircuit2.png]]

'''Solution'''
<br>Based on the Node Rule, we know that the electric current is equal in both wires:
<br><math>i_{thick}=i_{thin}</math>
<br><br>We also know that <math>i= nA\bar{v} = nA \mu E</math>. Therefore:
<br><math>nA_{thick} \mu E_{thick} = nA_{thin} \mu E_{thin}</math>
<br><br>When we rearrange this to solve for <math>E_{thick}</math>, the electron mobility (<math> \mu </math>) and electron density (<math>n</math>) cancel out and we are left with:
<br><math>E_{thick}= \frac{A_{thin}E_{thin}}{A_{thick}}</math>

<br>The cross-sectional area of the thin wire is <math>A_{thin}= \pi (d_{thin}/2)^2 = \pi(0.125\times10^{−3}m^2) = 4.91 \times 10^{–8}m^2.</math>
<br>The cross-sectional area of the thick wire is <math>A_{thick}= \pi (d_{thick}/2)^2 = \pi(0.155\times10^{−3}m^2) = 7.55 \times 10^{–8}m^2.</math>

<br><math>E_{thick}= \frac{(4.91 \times 10^{–8}m^2)(3.5 V/m)}{7.55 \times 10^{–8}m^2}</math>
<br><math>E_{thick}=2.28 V/m</math>

<br>We know that <math>\bar{v}=\mu \left | E_{applied} \right |</math>.
<br>The drift velocity of the thin wire is <math>\bar{v}=\mu E_{thin} = [0.0006 (m/s)/(V/m)][3.5 V/m]= 2.10 \times 10^{–3} m/s</math>.
<br>The drift velocity of the thick wire is <math>\bar{v}=\mu E_{thick} = [0.0006 (m/s)/(V/m)][2.28 V/m]= 1.37 \times 10^{–3} m/s</math>.

==Connectedness==
As an industrial engineer, I often do not use physics. However, when designing efficient systems, concepts similar to those of a circuit appear. For instance, if a number of people need to get from one floor of a building down to another floor and they have a large staircase and a small staircase connecting the two floors, it is most efficient for them to move like electrons would move in a parallel circuit with a strong and weak resistor. It makes sense that these similarities would exist because circuits are essentially charges moving from point A to point B while expending the least amount of energy possible. That being said, in specific circumstances understanding physics concepts can be essential to even the work of an industrial engineer, especially in laying out factories to use the least amount of energy.

== See also ==
===Further Reading===
* [[Current]]
* [[Fundamentals of Resistance]]
* [[Node Rule]]
* [[Loop Rule]]
* [[Ohm's Law]]
* [[RL Circuit]]
* [[Series Circuits]]
* [[Parallel Circuits]]
===External Links===
* https://www.youtube.com/watch?v=Byvz0MMH_fw
* https://www.amherst.edu/system/files/media/1182/Lecture15%252520slides.pdf

==References==

*[https://openstax.org/details/college-physics| OpenStax College Physics]
*[http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html| HyperPhysics]
*[https://www.codecogs.com/latex/eqneditor.php Formula code created with Online LaTeX Equation Editor]
*[https://en.wikipedia.org/wiki/Steady_state]

Steady State

2026-04-25T04:30:52Z

Vtadakamalla3:

'''Claimed Vivan Tadakamalla (Spring 2026)'''

Imagine a circuit in which the only battery has been disconnected. In such a setup there would be no electric field in the circuit and therefore no moving charges so the circuit would be in static equilibrium. When the battery is reconnected, an electric field will quickly propagate through the circuit and the charges in the circuit will accelerate. After a short time, the charges will move at a constant speed through the circuit and the electric field at each part of the circuit will remain unchanged. This means that at any given moment, all properties of the circuit remain unchanged even though an electric field is acting on the charges. At this time, when the state of the circuit does not change with the passage of time, the circuit is said to be in '''steady state.'''

[[Image:current_thing.png|thumb|upright = 0.2|right|450px|Current Diagram]]

==The Main Idea==

If a circuit is in steady state the following will be true:
:* At each part of the circuit, the drift velocity of charges remains unchanged with time.
:* The maximum amount of energy has been stored in capacitors, inductors.
:* Excess charge only accumulates on the surface of the wire.
:* An unchanging voltage exists between any two points in the system.

Each of these bullets is a different way of saying the same thing, that the state of the circuitry acting on any charges remains unchanged with respect to time. It is important to remember that a circuit being in steady state does not mean that the drift speed is the same every where in the circuit, only that it is unchanging for each specific location in the circuit. Also, a circuit in which there is no current may still be in steady state (such as an RC circuit in which the capacitors are fully charged), as long as there is still a constant electric field in all conductive parts of the circuit.

'''A Conceptual Understanding'''

For charges to move through the circuit, there must be an applied electric field that causes the mobile charges to move. Since there cannot be excess charge in the interior of a conductor, the surface charges must be producing the electric field. This electric field must be parallel to the wire because of the properties of a conductor. In a conductor, mobile charges always move to cancel out an electric field, meaning that the only electric field in the circuit must be parallel to the direction of current flow.
It is also important to conceptually understand why this electric field eventually reaches steady state. When a battery is first connected to a circuit, the electric field increases in magnitude, the rate at which the charges move through the circuit increases in magnitude too. As charges move through the circuit, there is a drag force in the opposite direction of their motion (similar to friction) known as the resistance of the wire. Since <math>I=V/R</math>, eventually these opposing forces will be equal in magnitude, bringing the circuit to steady state. Once a circuit is described as being in the steady state, there must be a constant electric field in the wire, the electric field has uniform magnitude throughout wire sections with similar cross-sectional area, and the electric field is parallel to the wire at every location along the wire.

[[File:Wire_Charges2.png|thumb = Wire_Charges2.png|upright=0.1|400px|center|Surface Charge Distribution]]

'''Steady State vs. Other States'''
When circuits move from static equilibrium to steady state, they typically do so in a very small amount of time. However, the [[Non Steady State]], or Transient State, which takes place in between, is an important aspect of circuits. When a circuit is in the Non Steady State, over time its state will approach the steady state. After enough time has elapsed, the difference is so slim that the circuit can be considered to be in steady state, even though in theory it never quite reaches steady state. The equations to model Non Steady States are different depending on the type of circuit being analyzed.

The following chart compares common properties of circuits in Static Equilibrium, Steady State, and Non Steady States:

{| class="wikitable"
|-
!
! Static Equilibrium
! Steady State
! Non Steady State
|-
| <math>\bar{v}</math>
| <math>= 0</math>
| <math>\neq 0</math>
| <math>\neq 0</math>
|-
| <math>\frac{d\bar{v}}{dt}</math>
| <math>= 0</math>
| <math>= 0</math>
| <math>\neq 0</math>
|-
| <math>\sum E_{inside}</math>
| <math>= 0</math>
| <math>\neq 0</math>
| <math>\neq 0</math>
|-
|}

===A Mathematical Model===
'''Current'''

There are two ways to describe current in a circuit, the electron current (<math>i</math>), and the conventional current (<math>I</math>). Electron current describes the motion of charges as it actually happens in the circuit: electrons moving through the conductor. Conventional current has the same magnitude and opposite direction as electron current as it describes positive charges moving through a circuit. Although circuits will sometimes act like positive charges are moving through the circuit if there are “electron holes,” it is important to understand that there are not positively charged particles moving through circuits. Even though it is backwards from the physical reality, conventional current is more commonly used because the motion of electrons in circuits was not understood when the first discoveries of electricity were made.
The following formulas can be used to describe electron and conventional current, respectively:
<br><math>i= nA\bar{v}</math>

<math>I = \left | q \right |nA\bar{v}</math>

n = charge density, the number of charged particles per unit volume

A = cross sectional area of the wire

v = drift speed, the speed at which mobile charges move through a section of wire

q = charge of the mobile particles being described (only matters for conventional current)

Keep in mind that drift velocity <math>\bar{v}=\mu \left | E_{applied} \right |</math>

Where <math> \mu </math> = electron mobility.

If you have trouble thinking of this conceptually, imagine a pipe carrying water. For a certain amount of water to move through each part of the pipe per unit time, if the pipe is smaller, the water must move faster and for more water to be delivered, either the speed of the water must increase or the size of the pipe must increase.

'''Kirchoff's Node Rule'''

When a wire splits in to two, this junction is referred to as a node. Current flowing through the node must follow the equation <math>\Sigma I_{in} = \Sigma I_{out}</math>.
This is called the “[[Node Rule]].” All this is saying is that exactly what enters the node must exit. In other words, charge dose not disappear or come into existence just because one wire splits in two.

[[Image:NodeRule2.jpg|thumbnail = NodeRule2.jpg|upright=0.5|400 px|center|Node Rule]]

==Examples==

===Simple===
'''Question'''

What are the values of X and Y in the following circuit?
[[Image:Node_rule_simple_problem.png|thumbnail = Node_rule_simple_problem.png|upright=0.5|frame|center|Node Rule]]

'''Solution'''

Based on the node rule, 10=3+3+X. Therefore X=4.<br>
Y=3+X, therefore Y=3+4=7<br>

===Middling===

[http://cnx.org/contents/Ax2o07Ul@9.39:3ct4v3c5@4/Current| (Example taken from the OpenStax College Physics Textbook)]

'''Question'''

Calculate the drift velocity of electrons in a 12-gauge copper wire (which has a diameter of '''2.053 mm''') carrying a '''20.0-A''' current, given that there is one free electron per copper atom. The density of copper is <math>8.80\times 10^{3} kg/m^3</math>.

'''Solution'''

First, calculate the density of free electrons in copper. There is one free electron per copper atom. Therefore, is the same as the number of copper atoms per <math>m^3</math>. We can now find <math>n</math> as follows:

<math>n= \frac{1e-}{atom}\times \frac{6.02\times10^{23}atoms}{mol}\times \frac{1 mol}{63.54 g} \times \frac{1000 g}{kg} \times \frac{8.80 \times 10^3 kg}{1 m}</math>

<br><math>n= 8.342\times 10^{28}e−/m^3</math>
<br>The cross-sectional area of the wire is
<math>A= \pi r^2 = \pi(2.053\times0.5\times10^{−3})^2 = 3.3103 \times 10^{–6}m^2</math>

<br>Rearranging <math>I= \left | q \right |nA\bar{v}</math> to isolate drift velocity gives you:
<br><math>\bar{v}= \frac{I}{n\left |q \right |A}</math>
<br><math>\bar{v}= \frac{20.0A}{(8.342\times10^{28}/m^3) (1.60\times10^{-19}C)(3.310\times10^{-6}m^2)}</math>
<br><math>\bar{v}= 4.53\times 10^{–4}m/s </math>

===Difficult===

'''Question'''

A circuit made of two wires is in the steady state. The battery has an emf of '''1.5 V'''. Both wires are length '''25 cm''' and made of the same material. The thin wire's diameter is '''0.25 mm''', and the thick wire's diameter is '''0.31 mm'''. There are <math>4\times 10^{28}</math> mobile electrons per cubic meter of this material, and the electron mobility of this material is '''0.0006 (m/s)/(V/m)'''.

A) If the magnitude of the electric field in the thin wire is '''3.5 V/m''', what is the electric field in the thick wire?<br>
B) What is the drift velocity in the thin and thick wires?

[[File:sscircuit2.png]]

'''Solution'''
<br>Based on the Node Rule, we know that the electric current is equal in both wires:
<br><math>i_{thick}=i_{thin}</math>
<br><br>We also know that <math>i= nA\bar{v} = nA \mu E</math>. Therefore:
<br><math>nA_{thick} \mu E_{thick} = nA_{thin} \mu E_{thin}</math>
<br><br>When we rearrange this to solve for <math>E_{thick}</math>, the electron mobility (<math> \mu </math>) and electron density (<math>n</math>) cancel out and we are left with:
<br><math>E_{thick}= \frac{A_{thin}E_{thin}}{A_{thick}}</math>

<br>The cross-sectional area of the thin wire is <math>A_{thin}= \pi (d_{thin}/2)^2 = \pi(0.125\times10^{−3}m^2) = 4.91 \times 10^{–8}m^2.</math>
<br>The cross-sectional area of the thick wire is <math>A_{thick}= \pi (d_{thick}/2)^2 = \pi(0.155\times10^{−3}m^2) = 7.55 \times 10^{–8}m^2.</math>

<br><math>E_{thick}= \frac{(4.91 \times 10^{–8}m^2)(3.5 V/m)}{7.55 \times 10^{–8}m^2}</math>
<br><math>E_{thick}=2.28 V/m</math>

<br>We know that <math>\bar{v}=\mu \left | E_{applied} \right |</math>.
<br>The drift velocity of the thin wire is <math>\bar{v}=\mu E_{thin} = [0.0006 (m/s)/(V/m)][3.5 V/m]= 2.10 \times 10^{–3} m/s</math>.
<br>The drift velocity of the thick wire is <math>\bar{v}=\mu E_{thick} = [0.0006 (m/s)/(V/m)][2.28 V/m]= 1.37 \times 10^{–3} m/s</math>.

==Connectedness==
As an industrial engineer, I often do not use physics. However, when designing efficient systems, concepts similar to those of a circuit appear. For instance, if a number of people need to get from one floor of a building down to another floor and they have a large staircase and a small staircase connecting the two floors, it is most efficient for them to move like electrons would move in a parallel circuit with a strong and weak resistor. It makes sense that these similarities would exist because circuits are essentially charges moving from point A to point B while expending the least amount of energy possible. That being said, in specific circumstances understanding physics concepts can be essential to even the work of an industrial engineer, especially in laying out factories to use the least amount of energy.

== See also ==
===Further Reading===
* [[Current]]
* [[Fundamentals of Resistance]]
* [[Node Rule]]
* [[Loop Rule]]
* [[Ohm's Law]]
* [[RL Circuit]]
* [[Series Circuits]]
* [[Parallel Circuits]]
===External Links===
* https://www.youtube.com/watch?v=Byvz0MMH_fw
* https://www.amherst.edu/system/files/media/1182/Lecture15%252520slides.pdf

==References==

*[https://openstax.org/details/college-physics| OpenStax College Physics]
*[http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html| HyperPhysics]
*[https://www.codecogs.com/latex/eqneditor.php Formula code created with Online LaTeX Equation Editor]
*[https://en.wikipedia.org/wiki/Steady_state]