Unit 10 · Topic 10.6

10.6 Capacitors

A capacitor stores separated charge, and with it energy. For a parallel-plate capacitor, C = Q/ΔV = κε₀A/d depends only on the plates' size, spacing and the material between them. The field between the plates is uniform, so a charge there moves like a projectile.

Key terms

capacitance
parallel-plate capacitor
farad
dielectric constant
uniform electric field
energy stored in a capacitor

What a capacitor is

A parallel-plate capacitor is two flat conducting plates facing each other with a small gap. When it's charged, one plate holds +Q and the other −Q, so the capacitor's net charge is zero. “The charge on a capacitor” means Q, the size of the charge on either plate.

Capacitance measures how much charge the capacitor holds for each volt across it:

$C = \frac{Q}{\Delta V}$

The unit is the farad (1 F = 1 C/V). A farad is huge; real capacitors are usually microfarads (μF) or picofarads (pF = 10⁻¹² F).

What capacitance depends on

For a parallel-plate capacitor:

$C = \frac{\kappa\varepsilon_0 A}{d}$

A is the area of one plate, d is the gap, and κ (kappa) is the dielectric constant of the material between the plates (κ = 1 for air or vacuum). Capacitance depends only on these physical features, not on how much charge is stored or what battery is used. Bigger plates or a smaller gap mean more capacitance.

The field between the plates

Away from the edges, the field between the plates is uniform: the same size and direction everywhere, pointing from the + plate to the − plate. With air between the plates, its size is:

$E = \frac{\Delta V}{d} = \frac{Q}{\varepsilon_0 A}$

A charged particle between the plates feels a constant force qE, so it has a constant acceleration a = qE/m. That's just like projectile motion in Earth's gravity: a particle that enters moving parallel to the plates keeps a constant speed in that direction and speeds up steadily toward one plate, tracing a parabola. For electrons and protons, gravity is negligible next to the electric force.

Stored energy and dielectrics

Charging a capacitor takes work, because each new bit of charge has to be pushed onto a plate that already repels it. That work is stored as electric potential energy:

$U_C = \frac{1}{2}Q\Delta V = \frac{1}{2}C(\Delta V)^2$

A dielectric is an insulator slipped between the plates. The plates' field polarizes it, and the polarized charges create a field inside the dielectric that points opposite to the plates' field. This reduces the net field for the same charge, so ΔV drops, and C = Q/ΔV goes up by the factor κ.

Change	Battery still connected (ΔV fixed)	Battery disconnected (Q fixed)
Gap d doubled	C halves, Q halves, E halves, U halves	C halves, ΔV doubles, E same, U doubles
Dielectric added	C × κ, Q × κ, U × κ	C × κ, ΔV ÷ κ, U ÷ κ

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
A simple capacitor
A capacitor has square plates of area 0.010 m² separated by 1.0 mm of air. It's connected to a 12 V battery. Find its capacitance, the charge on each plate, the field between the plates and the stored energy.
Show the solutionHide the solution
1. Step 1: $C = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85 \times 10^{-12})(0.010)}{1.0 \times 10^{-3}} = 8.85 \times 10^{-11}$ F, about 89 pF.
2. Step 2: Q = CΔV = (8.85 × 10⁻¹¹)(12) ≈ 1.06 × 10⁻⁹ C.
3. Step 3: $E = \dfrac{\Delta V}{d} = \dfrac{12}{1.0 \times 10^{-3}} = 1.2 \times 10^4$ V/m.
4. Step 4: $U_C = \frac{1}{2}Q\Delta V = \frac{1}{2}(1.06 \times 10^{-9})(12) \approx 6.4 \times 10^{-9}$ J.
Answer: C ≈ 89 pF, Q ≈ 1.1 × 10⁻⁹ C, E = 1.2 × 10⁴ V/m, U ≈ 6.4 × 10⁻⁹ J
Example 2Calculator allowed
Electron between plates
An electron enters the uniform field between two plates moving parallel to them at 2.0 × 10⁶ m/s. The field is 200 N/C and the plates are 5.0 cm long. How far sideways is the electron deflected by the time it leaves? (m = 9.11 × 10⁻³¹ kg)
Show the solutionHide the solution
1. Step 1: Time between the plates (constant speed along them): t = 0.050 m / 2.0 × 10⁶ m/s = 2.5 × 10⁻⁸ s.
2. Step 2: Sideways acceleration: $a = \dfrac{eE}{m} = \dfrac{(1.60 \times 10^{-19})(200)}{9.11 \times 10^{-31}} \approx 3.5 \times 10^{13}$ m/s², toward the positive plate (opposite the field).
3. Step 3: Deflection, starting from zero sideways speed: $y = \frac{1}{2}at^2 = \frac{1}{2}(3.5 \times 10^{13})(2.5 \times 10^{-8})^2 \approx 0.011$ m.
Answer: About 1.1 cm, toward the positive plate
Example 3
Pulling the plates apart (classic trap)
A capacitor is charged by a battery, then disconnected. The plates are pulled apart to twice their original separation. What happens to C, Q, ΔV and the stored energy?
Show the solutionHide the solution
1. Step 1: Disconnected means the charge has nowhere to go: Q stays the same.
2. Step 2: C = κε₀A/d, so doubling d halves C.
3. Step 3: ΔV = Q/C, so ΔV doubles.
4. Step 4: U = ½QΔV, so U doubles. The extra energy comes from the work you did pulling the attracting plates apart.
5. Step 5: The trap is assuming ΔV stays fixed. That's only true while the battery is still connected.
Answer: C halves, Q stays the same, ΔV doubles, U doubles

Common mistakes

Thinking capacitance depends on the charge or voltage. C is set by A, d and κ; Q and ΔV change together.
Mixing up connected and disconnected cases. Connected: ΔV is fixed. Disconnected: Q is fixed.
Forgetting to convert mm to m or cm² to m² in C = κε₀A/d.
Expecting a charge between the plates to move in a circle or at constant speed. The force is constant, so the path is a parabola.

On the exam

Factor-of-change questions on C, Q, ΔV, E and U when the plates move or a dielectric is added are very common; first decide what stays fixed.
Expect projectile-style questions about charges between plates, sometimes compared with a ball in Earth's gravitational field.

Connected topics

Videos

AP Physics 2 - Unit 10 - Lesson 8 - Capacitors
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
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Capacitors and capacitance | Circuits | Physics | Khan Academy
khanacademymedicineWatch on YouTube (opens in a new tab)
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AP Physics 2 - Unit 10 - Lesson 9 - Dielectrics
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The Organic Chemistry TutorWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 10.6 Capacitors. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

An air-filled parallel-plate capacitor has plates of area 0.020 m² separated by 1.0 mm. What is its capacitance?

Question 2 of 4Calculator allowed

A 2.0 μF capacitor is connected to a 9.0 V battery and fully charged. How much charge is on its positive plate?

Question 3 of 4Calculator allowed

How much energy is stored in a 2.0 μF capacitor charged to 9.0 V?

Question 4 of 4Calculator allowed

A parallel-plate capacitor is charged and then disconnected from the battery. The plates are then pulled apart to twice their original separation. What happens to the energy stored in the capacitor?

0 of 4 answered