What is parallel plate?

A pillbox gives a uniform field E = σ_0 = Q_0 A between the plates. The potential difference is V = Ed = Qd_0 A, so

A coaxial Gaussian cylinder gives E = Q2π_0 L r; integrating from a to b gives V = Q2π_0 Llnba, so C = 2π_0 Lln(b/a).

Two capacitors, 3\,μF and 6\,μF, are in series. Find the equivalent capacitance. [2 points]

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How is capacitance defined, and how do you find it and the energy stored for parallel-plate and other geometries?

Topic 10.3 Capacitors: define capacitance, derive it for parallel-plate, spherical and cylindrical geometries, and find the stored energy and series and parallel combinations.

A calculus-based answer to AP Physics C E&M Topic 10.3, covering capacitance, the parallel-plate, spherical and cylindrical capacitor (via Gauss's law), energy stored, energy density, and series and parallel combinations.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Capacitance is the charge stored per unit potential difference, $C = \dfrac{Q}{V}$ , in farads (C/V). For a parallel-plate capacitor, $C = \dfrac{\varepsilon_0 A}{d}$ (area $A$ , separation $d$ ). Other geometries follow from Gauss's law and integration: a spherical capacitor gives $C = \dfrac{4\pi\varepsilon_0 ab}{b-a}$ and a cylindrical one $C = \dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$ . The energy stored is $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}QV = \dfrac{Q^2}{2C}$ , held in the field with energy density $u = \tfrac{1}{2}\varepsilon_0 E^2$ . Capacitors in parallel add directly, $C_{eq} = \sum C_i$ (same voltage); in series, the reciprocals add, $\dfrac{1}{C_{eq}} = \sum \dfrac{1}{C_i}$ (same charge).

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What this topic is asking
Defining capacitance
Deriving capacitance from Gauss's law
Energy stored and energy density
Series and parallel combinations
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What this topic is asking

The College Board (Topic 10.3) wants you to define capacitance, derive it for parallel-plate, spherical and cylindrical geometries using Gauss's law and integration, compute the energy stored, and combine capacitors in series and parallel. A capacitor stores charge and energy in the field between two conductors, and is the link from electrostatics to circuits.

Defining capacitance

A larger $C$ stores more charge for the same voltage. Because $C$ depends only on geometry (and any dielectric), you find it once for a given shape and use it for any charge.

Deriving capacitance from Gauss's law

The general recipe: put $\pm Q$ on the conductors, find the field with Gauss's law, integrate to get $V$ , then divide.

Parallel plate. A pillbox gives a uniform field $E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0 A}$ between the plates. The potential difference is $V = Ed = \dfrac{Qd}{\varepsilon_0 A}$ , so

C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}

Cylindrical. A coaxial Gaussian cylinder gives $E = \dfrac{Q}{2\pi\varepsilon_0 L r}$ ; integrating from $a$ to $b$ gives $V = \dfrac{Q}{2\pi\varepsilon_0 L}\ln\dfrac{b}{a}$ , so $C = \dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$ .

Energy stored and energy density

Charging a capacitor requires work, stored as field energy. Building up the charge $dq$ against the growing voltage $q/C$ and integrating gives

U = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C} = \tfrac{1}{2}CV^2 = \tfrac{1}{2}QV

This energy resides in the electric field itself, with energy density

u = \frac{1}{2}\varepsilon_0 E^2 \qquad(\text{J per m cubed})

Integrating $u$ over the field volume reproduces $U$ , confirming the energy lives in the field, not on the plates.

Series and parallel combinations

In parallel, all capacitors share the same voltage, so their charges add, and the equivalent capacitance is the sum:

C_{eq} = C_1 + C_2 + \cdots

In series, all carry the same charge, the voltages add, and the reciprocals add:

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

(Note this is opposite to resistors, where parallel reciprocals add.)

Energy and charge of a parallel-plate capacitor

A parallel-plate capacitor has plates of area $0.020$ m squared separated by $1.0$ mm, connected to a $12$ V battery. Find its capacitance, charge, and stored energy. Use $\varepsilon_0 = 8.85\times10^{-12}$ .

step 1 Find the capacitance

$C = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85\times10^{-12})(0.020)}{1.0\times10^{-3}} = 1.77\times10^{-10}$ F (about $177$ pF).

step 2 Find the charge

$Q = CV = (1.77\times10^{-10})(12) = 2.12\times10^{-9}$ C ( $2.12$ nC).

step 3 Find the stored energy

$U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(1.77\times10^{-10})(12)^2 = 1.27\times10^{-8}$ J.

step 4 Check with another form

$U = \tfrac{1}{2}QV = \tfrac{1}{2}(2.12\times10^{-9})(12) = 1.27\times10^{-8}$ J, agreeing.

Try this

Q1. Two capacitors, $3\,\mu\text{F}$ and $6\,\mu\text{F}$ , are in series. Find the equivalent capacitance. [2 points]

Cue. $\dfrac{1}{C_{eq}} = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{1}{2}$ , so $C_{eq} = 2\,\mu\text{F}$ .

Q2. A $5\,\mu\text{F}$ capacitor is charged to $20$ V. Find the stored energy. [2 points]

Cue. $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(5\times10^{-6})(20)^2 = 1.0\times10^{-3}$ J.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice). A parallel-plate capacitor is connected to a battery of fixed voltage. The plate separation is then doubled. The energy stored becomes (A) doubled (B) halved (C) quartered (D) unchanged. Justify your reasoning.

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A 1-point MCQ on capacitor energy at fixed voltage. The answer is (B).

At fixed $V$ , $U = \tfrac{1}{2}CV^2$ . For a parallel plate $C = \dfrac{\varepsilon_0 A}{d}$ , so doubling $d$ halves $C$ , which halves $U$ . The trap is using $U = \dfrac{Q^2}{2C}$ with fixed $Q$ (that would apply to an isolated, disconnected capacitor, giving a doubling). With the battery attached, $V$ is held fixed.

AP 2024 (style)6 marksSection II (FRQ, derivation). A spherical capacitor has an inner conducting sphere of radius

a

and a concentric outer shell of radius

b

, carrying

+Q

and

-Q

. (a) Use Gauss's law to find the field between the spheres. (b) Find the potential difference between them by integration. (c) Derive the capacitance.

Show worked answer →

A 6-point FRQ deriving a capacitance from Gauss's law and integration.

(a) Field (2 points): a Gaussian sphere of radius $r$ ( $a < r < b$ ) encloses $+Q$ , so $E(4\pi r^2) = \dfrac{Q}{\varepsilon_0}$ , $E = \dfrac{kQ}{r^2}$ , radial.
(b) Potential difference (3 points): $V = -\displaystyle\int_b^a E\,dr = \displaystyle\int_a^b \dfrac{kQ}{r^2}\,dr = kQ\left[-\dfrac{1}{r}\right]_a^b = kQ\left(\dfrac{1}{a} - \dfrac{1}{b}\right) = kQ\dfrac{b-a}{ab}$ .
(c) Capacitance (1 point): $C = \dfrac{Q}{V} = \dfrac{ab}{k(b-a)} = \dfrac{4\pi\varepsilon_0 ab}{b-a}$ .

Markers reward the Gauss field, the integral for $V$ , and dividing $Q$ by $V$ .

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)