Unit 10 · Topic 10.3

10.3 Capacitors

A capacitor is a pair of conductors that hold equal and opposite charges. Its capacitance, $C = Q/\Delta V$ , depends only on its geometry and what fills the gap, and a charged capacitor stores energy in the electric field between its conductors.

Key terms

capacitance
farad
parallel-plate capacitor
spherical and cylindrical capacitors
energy stored in a capacitor

What capacitance means

Charge a capacitor and one conductor gets +Q while the other gets −Q. The potential difference between them, ΔV, is proportional to Q, and the ratio is the capacitance: $C = \frac{Q}{\Delta V}$

Capacitance is measured in farads (1 F = 1 C/V). A farad is huge, so real capacitors are measured in picofarads (10⁻¹² F) to microfarads (10⁻⁶ F). C doesn't depend on Q or ΔV. Double the charge and the voltage doubles too, leaving C the same. It's set by the conductors' sizes, shapes and spacing, plus whatever fills the gap.

Parallel plates

Two plates of area A, a small distance d apart, carry ±Q. Each plate on its own acts like a sheet of charge with field σ/(2ε₀) (8.6). Between the plates the two fields point the same way and add; outside they point opposite ways and cancel. So, away from the edges, the field between them is uniform: $E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0 A}$ . Outside the plates it is nearly zero. The potential difference is $\Delta V = Ed$ , so $C = \frac{\varepsilon_0 A}{d}$

Bigger plates hold more charge at the same voltage; a wider gap lowers C. With a dielectric filling the gap, multiply by κ (10.4).

Using Gauss's law to find C

For other shapes: assume charges ±Q, find E between the conductors with Gauss's law, integrate to get ΔV, then divide: C = Q/ΔV. The Q always cancels, which is a good check. The exam asks for this only for the three shapes below.

Capacitor	Capacitance
Parallel plates, area A, gap d	$\dfrac{\varepsilon_0 A}{d}$
Concentric spheres, radii a < b	$\dfrac{4\pi\varepsilon_0 ab}{b - a}$
Coaxial cylinders, radii a < b, length L	$\dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$

Energy stored

Charging a capacitor means moving charge against a growing voltage. The first bit of charge moves across almost no voltage and the last bit across the full ΔV, so the work done is the charge times the average voltage: $U_C = \tfrac{1}{2}Q\Delta V = \tfrac{1}{2}C(\Delta V)^2 = \frac{Q^2}{2C}$

Pick the form that uses what stays fixed. If the capacitor is isolated, Q is fixed, so use $Q^2/2C$ . If it stays connected to a battery, ΔV is fixed, so use $\tfrac{1}{2}C(\Delta V)^2$ . The energy is stored in the electric field between the plates.

Charged particles between the plates

The field between parallel plates is uniform, so a charge in the gap feels a constant force qE and moves like a projectile. For an electron, gravity is negligible. Example: in a field of 1.2 × 10⁴ V/m, an electron's acceleration is $eE/m \approx 2.1\times10^{15}$ m/s². Enter the gap sideways at 2.0 × 10⁷ m/s through plates 5.0 cm long, and it spends 2.5 ns between them and is deflected about 6.6 mm toward the positive plate.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
A parallel-plate capacitor
Two plates, each 0.020 m² in area, are 1.0 mm apart with air between them. The capacitor is connected to a 12 V battery. Find its capacitance, the charge on each plate, the field between the plates and the energy stored.
Show the solutionHide the solution
1. Step 1: $C = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85\times10^{-12})(0.020)}{1.0\times10^{-3}} \approx 1.8\times10^{-10}$ F (177 pF).
2. Step 2: $Q = C\Delta V = (1.77\times10^{-10})(12) \approx 2.1\times10^{-9}$ C.
3. Step 3: $E = \dfrac{\Delta V}{d} = \dfrac{12}{1.0\times10^{-3}} = 1.2\times10^{4}$ V/m.
4. Step 4: $U = \tfrac{1}{2}C(\Delta V)^2 = \tfrac{1}{2}(1.77\times10^{-10})(144) \approx 1.3\times10^{-8}$ J.
Answer: C ≈ 177 pF, Q ≈ 2.1 nC, E = 1.2 × 10⁴ V/m, U ≈ 1.3 × 10⁻⁸ J
Example 2Calculator allowed
Spherical capacitor from Gauss's law
A conducting sphere of radius a sits at the center of a thin conducting shell of radius b. Derive the capacitance, then evaluate it for a = 0.10 m and b = 0.12 m.
Show the solutionHide the solution
1. Step 1: Put +Q on the inner sphere and −Q on the shell. Between them (a < r < b), a Gaussian sphere encloses +Q, so $E = \dfrac{kQ}{r^2}$ , pointing outward.
2. Step 2: $\Delta V = V_a - V_b = \displaystyle\int_a^b\frac{kQ}{r^2}\,dr = kQ\left(\frac{1}{a} - \frac{1}{b}\right) = \frac{kQ(b-a)}{ab}$ .
3. Step 3: $C = \dfrac{Q}{\Delta V} = \dfrac{ab}{k(b-a)} = \dfrac{4\pi\varepsilon_0 ab}{b-a}$ . The Q canceled, as it must.
4. Step 4: Numbers: $C = \dfrac{4\pi(8.85\times10^{-12})(0.10)(0.12)}{0.02} \approx 6.7\times10^{-11}$ F.
Answer: $C = \dfrac{4\pi\varepsilon_0 ab}{b-a}$ , about 67 pF
Example 3Calculator allowed
Pulling the plates apart (classic trap)
A charged parallel-plate capacitor is disconnected from its battery. The plate separation is then doubled. What happens to C, Q, ΔV, E and the stored energy U? How would the answers change if the battery stayed connected?
Show the solutionHide the solution
1. Step 1: Disconnected: no path for charge, so Q stays the same. C = ε₀A/d halves.
2. Step 2: ΔV = Q/C doubles. E = Q/(ε₀A) depends only on Q and A, so E stays the same (and ΔV = Ed doubles, which agrees).
3. Step 3: U = Q²/(2C) doubles. The extra energy comes from the work you do pulling the attracting plates apart.
4. Step 4: Connected: ΔV is fixed by the battery. C halves, so Q = CΔV halves, E = ΔV/d halves, and U = ½C(ΔV)² halves.
5. Step 5: The trap is assuming ΔV stays fixed after the battery is removed. First ask which quantity can't change.
Answer: Isolated: C halves, Q same, ΔV doubles, E same, U doubles. Connected: C halves, ΔV same, Q, E and U all halve.

Common mistakes

Thinking C depends on Q or ΔV. It depends only on geometry and the material in the gap.
Forgetting the ½ in the stored energy, or using QΔV.
Using the battery's voltage after the capacitor has been disconnected. An isolated capacitor keeps its charge, not its voltage.
Converting units carelessly: mm to m and cm² to m² (1 cm² = 10⁻⁴ m²) are frequent slips.

On the exam

"What changes and what stays the same" questions are very common. Decide first whether Q or ΔV is fixed, then work through C, E and U.
Free-response questions often ask you to derive C for a spherical or cylindrical capacitor. Show the Gaussian surface, the field, the integral for ΔV and the final division.

Connected topics

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Check yourself

4 questions on 10.3 Capacitors. Pick an answer to see if you got it, and why.

A parallel-plate capacitor has square plates of area 0.020 m² separated by 1.0 mm of air. It is connected to a 12 V battery and fully charged. Treat air like a vacuum.

Described situation

Question 1 of 4Calculator allowed

What is the capacitance?

Question 2 of 4Calculator allowed

How much charge is on each plate?

Question 3 of 4Calculator allowed

What is the electric field between the plates?

Question 4 of 4Calculator allowed

How much energy does the capacitor store?

0 of 4 answered