Capacitance from Geometry

In Physics 2 you used $C = Q/V$ and the parallel-plate formula $C = \varepsilon_0 A/d$. Now we can derive capacitance for any shape that Gauss's law can handle, by finding the field and then integrating it into a voltage.

Key idea

Put $+Q$ and $-Q$ on the two conductors, find $\vec{E}$ between them, integrate to get $\Delta V$, then $C = Q/\Delta V$. $$\text{plates: } \frac{\varepsilon_0 A}{d}, \qquad \text{coaxial: } \frac{2\pi\varepsilon_0 L}{\ln(b/a)}, \qquad \text{spheres: } 4\pi\varepsilon_0\frac{ab}{b - a}, \qquad \text{lone sphere: } 4\pi\varepsilon_0 R.$$

Why it works

The field between the conductors is proportional to $Q$ (Gauss's law), so $\Delta V = \int E\,dr$ is proportional to $Q$ too. In the ratio $Q/\Delta V$ the charge cancels, leaving only sizes and $\varepsilon_0$. Capacitance is a property of the shape, like the volume of a bottle. A smaller gap or a bigger area means less voltage for the same charge, so more capacitance.

A parallel-plate capacitor on a battery. Change the gap: $C$, and with it the charge, change, but $C$ never depends on the voltage.

Worked examples

Worked example: Parallel plates

Plates of area $100\ \text{cm}^2$ sit $1.0\ \text{mm}$ apart in air. Find $C$.

Between the plates $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$, uniform, so $\Delta V = Ed = Qd/(\varepsilon_0 A)$ and $C = \varepsilon_0 A/d$.
$C = \dfrac{(8.85 \times 10^{-12})(0.010)}{0.0010} = 88.5\ \text{pF}$.

Answer: $C \approx 88.5\ \text{pF}$

Worked example: A coaxial cable

A coaxial cable $10\ \text{m}$ long has an inner conductor of radius $0.50\ \text{mm}$ and an outer conductor of radius $2.0\ \text{mm}$, with air between. Find $C$.

With charge per length $\lambda = Q/L$ on the inner conductor, Gauss gives $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ between the conductors.
$\Delta V = \int_a^b \dfrac{\lambda}{2\pi\varepsilon_0 r}\,dr = \dfrac{\lambda}{2\pi\varepsilon_0}\ln\dfrac{b}{a}$, so $C = \dfrac{\lambda L}{\Delta V} = \dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$.
$C = \dfrac{2\pi(8.85 \times 10^{-12})(10)}{\ln 4} = 401\ \text{pF}$, about $40.1\ \text{pF}$ per meter.

Answer: $C \approx 401\ \text{pF}$

Worked example: Concentric spheres

An inner sphere of radius $4.0\ \text{cm}$ sits inside a shell of radius $5.0\ \text{cm}$. Find $C$, and compare with the inner sphere alone.

Between them $E = kQ/r^2$, so $\Delta V = kQ\left(\dfrac{1}{a} - \dfrac{1}{b}\right)$ and $C = \dfrac{ab}{k(b - a)} = 4\pi\varepsilon_0\dfrac{ab}{b - a}$.
$C = 4\pi(8.85 \times 10^{-12})\dfrac{(0.040)(0.050)}{0.010} = 22.2\ \text{pF}$.
Move the shell far away ($b \to \infty$): $C \to 4\pi\varepsilon_0 a = 4.45\ \text{pF}$. Bringing the second conductor close raises $C$ fivefold.

Answer: $C \approx 22.2\ \text{pF}$

Notice the pattern in all four formulas: capacitance grows when the conductors are large and close together. Close conductors let a lot of charge sit at a small voltage, because the opposite charges nearly cancel each other's fields everywhere but in the thin gap. That is why real capacitors roll up long strips of foil separated by a very thin film. A lone conductor has surprisingly little capacitance: Earth itself, treated as an isolated sphere of radius $6.37 \times 10^{6}\ \text{m}$, has only $4\pi\varepsilon_0 R \approx 709\ \text{μF}$, less than some capacitors that fit in your hand.

Common mistake

"Capacitance depends on the voltage applied." $C = Q/V$ looks as if more voltage means less capacitance, but $Q$ grows in step with $V$, and the ratio stays fixed. Check with the formulas above: none of them contains $V$ or $Q$. Doubling the battery voltage doubles the charge and leaves $C$ alone.

Capacitance from Geometry

Worked examples

Practice

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Sources and further reading