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How does a capacitor store charge and energy, and what sets its capacitance?

Topic 10.6 Capacitors: relate charge, voltage and capacitance, find the capacitance of a parallel-plate capacitor, and calculate the energy stored.

A focused answer to AP Physics 2 Topic 10.6, covering capacitance as charge per volt, the parallel-plate capacitor and what sets its capacitance, the role of a dielectric, the uniform field between the plates, and the energy stored, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

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Quick answer

A capacitor stores charge and energy by holding equal and opposite charges on two conductors separated by an insulator. Its capacitance is the charge stored per volt, $C = \dfrac{Q}{V}$ , measured in farads (F $=$ C/V). For a parallel-plate capacitor, $C = \dfrac{\varepsilon_0 A}{d}$ : capacitance rises with plate area $A$ and falls with separation $d$ ( $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m). A dielectric (insulating material) inserted between the plates increases the capacitance by a factor $\kappa$ , letting more charge be stored per volt. The field between the plates is uniform, $E = \dfrac{V}{d}$ . The energy stored is $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}QV = \dfrac{Q^2}{2C}$ . Capacitance is a geometric property: it depends on the size, spacing and material of the plates, not on the charge or voltage applied. Capacitors are how circuits store electrical energy in an electric field.

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What this topic is asking
Capacitance: charge per volt
The parallel-plate capacitor and the dielectric
Energy stored in a capacitor
Try this

What this topic is asking

The College Board (Topic 10.6) wants you to relate charge, voltage and capacitance ( $Q = CV$ ), find the capacitance of a parallel-plate capacitor, describe the effect of a dielectric, and calculate the energy stored.

Capacitance: charge per volt

A capacitor is two conductors that hold equal and opposite charges; connecting them to a voltage source drives charge onto the plates until the voltage across them matches the supply. Capacitance measures how much charge the capacitor holds for each volt: a large capacitance stores a lot of charge at modest voltage. The defining relation $Q = CV$ is used constantly. Crucially, $C$ is fixed by the construction, so changing the voltage changes the charge in proportion, but not the capacitance.

The parallel-plate capacitor and the dielectric

The geometry rules are intuitive: bigger plates hold more charge, and a smaller gap means the plates' fields reach each other more strongly, both raising capacitance. A dielectric helps because its molecules polarize and partly oppose the field, so for the same voltage the plates can hold more charge. The field between the plates is the uniform field of Topic 10.3, which is why $E = V/d$ applies.

Energy stored in a capacitor

The energy stored is the work done charging the capacitor, building up its charge against the growing voltage:

U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}QV = \dfrac{Q^2}{2C}

The three forms are equivalent (use $Q = CV$ to convert), and the right one depends on what is held fixed. A subtle exam scenario is changing a capacitor's geometry after charging: if it stays connected to the supply the voltage is fixed (use $\tfrac{1}{2}CV^2$ ), but if it is disconnected the charge is fixed (use $\dfrac{Q^2}{2C}$ ). The factor of $\tfrac{1}{2}$ appears because the voltage rises from zero to $V$ as the capacitor charges, so the average voltage during charging is $V/2$ . The strategic point of this topic is that a capacitor stores energy in the electric field between its plates, the same field of Topic 10.3, with the voltage of Topic 10.5; this stored energy is what capacitors release in the circuits of Unit 11, where they smooth, time and store charge.

Charge and energy on a capacitor

A $20$ microfarad capacitor is connected to a $9.0$ V battery. Find the charge stored and the energy stored, then the new energy if the voltage is doubled.

step 1 Charge from Q = CV

$Q = CV = (20 \times 10^{-6})(9.0) = 1.8 \times 10^{-4}$ C.

step 2 Energy stored

$U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(20 \times 10^{-6})(9.0)^2 = \tfrac{1}{2}(20 \times 10^{-6})(81) = 8.1 \times 10^{-4}$ J.

step 3 Double the voltage

$U \propto V^2$ , so doubling $V$ multiplies the energy by $4$ : $U = 4 \times 8.1 \times 10^{-4} = 3.24 \times 10^{-3}$ J.

step 4 Check with the charge

At $18$ V, $Q = (20 \times 10^{-6})(18) = 3.6 \times 10^{-4}$ C (doubled), and $U = \tfrac{1}{2}QV = \tfrac{1}{2}(3.6 \times 10^{-4})(18) = 3.24 \times 10^{-3}$ J, consistent.

Try this

Q1. State what happens to the capacitance of a parallel-plate capacitor if the plate separation is halved. [1 point]

Cue. It doubles ( $C \propto 1/d$ ).

Q2. A $5.0$ microfarad capacitor holds $40$ microcoulombs. Calculate the voltage across it. [2 points]

Cue. $V = Q/C = (40 \times 10^{-6})/(5.0 \times 10^{-6}) = 8.0$ V.

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 2024 (style)7 marksSection II (long FRQ). A parallel-plate capacitor has plates of area

0.050

m squared separated by

2.0 \times 10^{-3}

m in vacuum. Take the permittivity of free space as

8.85 \times 10^{-12}

F/m. (a) Calculate the capacitance. (b) It is connected to a

12

V supply. Calculate the charge stored on each plate. (c) Calculate the energy stored. (d) State and justify what happens to the capacitance if a dielectric is inserted between the plates.

Show worked answer →

A 7-point FRQ on the parallel-plate capacitor.

(a) Capacitance (2 points): $C = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85 \times 10^{-12})(0.050)}{2.0 \times 10^{-3}} = 2.21 \times 10^{-10}$ F.
(b) Charge (2 points): $Q = CV = (2.21 \times 10^{-10})(12) = 2.66 \times 10^{-9}$ C.
(c) Energy (2 points): $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(2.21 \times 10^{-10})(12)^2 = 1.59 \times 10^{-8}$ J.
(d) Dielectric (1 point): the capacitance increases, because a dielectric reduces the field for the same charge, allowing more charge to be stored per volt.

Markers reward the parallel-plate formula, $Q = CV$ , $U = \tfrac{1}{2}CV^2$ , and the increase in capacitance from a dielectric.

AP 2023 (style)1 marksSection I (multiple choice). A charged parallel-plate capacitor is disconnected from its supply, then the plate separation is doubled. What happens to the energy stored (charge fixed)? (A) it halves (B) it doubles (C) it is unchanged (D) it quadruples. Justify your reasoning.

Show worked answer →

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

With charge fixed, $C = \varepsilon_0 A / d$ halves when $d$ doubles. The energy $U = \dfrac{Q^2}{2C}$ then doubles when $C$ halves. Physically, you do work pulling the attracting plates apart, adding energy. The trap is (A): the energy rises, because work is done separating the plates.

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Sources & how we know this

AP Physics 2: Algebra-Based Course and Exam Description — College Board (2024)