Capacitors and capacitance

A flash that needs a few seconds to charge

A camera flash takes a moment to be ready. A small battery slowly puts charge on a capacitor. Then the flash releases it in a blink. What can hold a charge like that, and how much can it hold?

The capacitor and its capacitance

A capacitor is a pair of conductors, with an insulator or a vacuum between them. The conductors are called plates. When we charge a capacitor, one plate gets $+Q$ and the other gets $-Q$. The total charge is zero. We call $Q$ the charge of the capacitor.

The potential difference $V$ between the plates is proportional to $Q$. Put twice the charge on the plates and the field doubles, and so does $V$. So the ratio does not change. It is the capacitance:

The capacitance depends only on the shape and the size of the plates, the gap and the material between them. It does not depend on $Q$ or $V$. It tells us how much charge the capacitor holds for each volt.

The unit is the farad (F), one coulomb for each volt. A farad is enormous, so we use smaller units: the microfarad $1\ \mu\text{F} = 10^{-6}$ F, the nanofarad $1\ \text{nF} = 10^{-9}$ F and the picofarad $1\ \text{pF} = 10^{-12}$ F.

A worked example. A capacitor of $470\ \mu\text{F}$ is connected across a $9.0$ V battery. How much charge is on its positive plate?

1. We want $Q$, so $Q = CV$.
2. $Q = 470 \times 10^{-6}\ \text{F} \times 9.0\ \text{V} = 4.23 \times 10^{-3}\ \text{C}$, that is $4.23$ mC.

A single conductor has a capacitance too

Take one isolated metal sphere of radius $R$ with a charge $Q$. We may think of the other plate as a large shell at infinity. The potential of the sphere is $V = \dfrac{Q}{4\pi\varepsilon_0 R}$, so

The capacitance grows in proportion to the radius. For the Earth, $R = 6.4 \times 10^{6}$ m gives $C = 7.1 \times 10^{-4}$ F. The whole Earth is less than one thousandth of a farad.

The parallel-plate capacitor

Two large flat plates, each of area $A$, face each other at a small distance $d$. One has $+Q$ and the other $-Q$. The picture shows the field.

Let us find $C$ in three steps.

1. The field. Each plate has the charge on each unit of area $\sigma = Q/A$. A large sheet makes a field $\sigma/2\varepsilon_0$ on each side. Between the plates the two fields point the same way and add. Outside they cancel. So between the plates $E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0 A}$.
2. The potential difference. The field is uniform, so $V = Ed = \dfrac{Qd}{\varepsilon_0 A}$.
3. The capacitance. $C = \dfrac{Q}{V}$, so $C = \dfrac{\varepsilon_0 A}{d}$.

We assumed that the field is uniform. This is true only when the gap is small beside the size of the plates. Near the edges the lines bend outward. We neglect this.

The formula tells us how to build a big capacitor. Make the area large and the gap small.

A worked example. Two square plates of side $10$ cm face each other with a gap of $0.50$ mm of air. They are joined to a $12$ V battery. Find the capacitance, the charge and the field between the plates.

1. The capacitance. $A = (0.10\ \text{m})^2 = 1.0 \times 10^{-2}\ \text{m}^2$. $C = \dfrac{\varepsilon_0 A}{d} = \dfrac{8.85 \times 10^{-12} \times 1.0 \times 10^{-2}}{0.50 \times 10^{-3}} = 1.77 \times 10^{-10}\ \text{F} = 177\ \text{pF}$.
2. The charge. $Q = CV = 1.77 \times 10^{-10}\ \text{F} \times 12\ \text{V} = 2.12 \times 10^{-9}\ \text{C}$, that is $2.12$ nC.
3. The field. $E = \dfrac{V}{d} = \dfrac{12\ \text{V}}{0.50 \times 10^{-3}\ \text{m}} = 2.4 \times 10^{4}\ \text{V/m}$.

Halve the gap and watch the readouts. Which of $C$, $Q$ and $E$ change, and by what factor? Then double the side of the plates.

A capacitor made of two spheres

A metal ball of radius $a$ sits inside a hollow metal shell of inner radius $b$, with the same centre. The inner ball has $+Q$ and the shell has $-Q$ on its inner wall. Between them the field is that of the inner ball only, and the potential difference is

So the capacitance is

When the shell is moved far away, $b \to \infty$ and $C \to 4\pi\varepsilon_0 a$. That is the sphere alone, as it should be.

A worked example. A ball of radius $5.0$ cm sits in a shell of inner radius $8.0$ cm.

$C = 4\pi\varepsilon_0 \dfrac{ab}{b - a} = 4\pi \times 8.85 \times 10^{-12} \times \dfrac{0.050 \times 0.080}{0.030} = 1.48 \times 10^{-11}\ \text{F}$, that is $14.8$ pF. The ball alone would have $5.6$ pF, so the shell nearby raises the capacitance.

The largest voltage

Air is an insulator only up to a limit. When the field in air passes about $3 \times 10^{6}$ V/m, the air breaks down and a spark jumps. This field is the dielectric strength of air. For two plates with a gap $d$, the voltage cannot pass

A gap of $2.0$ mm in air can take $3 \times 10^{6}\ \text{V/m} \times 2.0 \times 10^{-3}\ \text{m} = 6.0 \times 10^{3}\ \text{V}$. A thin gap makes a big capacitance, but it can hold only a small voltage. This is the reason why a real capacitor has a rating, such as $16$ V, on its side.

Practise with new numbers

Try it in Python

How big must the plates be for a given capacitance? The program finds the area of a parallel-plate capacitor for three cases.

Common mistakes

• Thinking $C$ depends on $Q$ and $V$. It is only the ratio. Change $Q$, and $V$ changes in the same proportion.
• Thinking the charge of a capacitor is $2Q$, or zero. One plate has $+Q$ and the other has $-Q$. We call $Q$ the charge of the capacitor.
• Mixing mm, cm and m. A gap of $0.5$ mm is $0.5 \times 10^{-3}$ m. An area of $10$ cm by $10$ cm is $1.0 \times 10^{-2}$ m².
• Using the length of a plate in place of its area. For a square of side $L$, the area is $L^2$.
• Putting the gap in the wrong place. The gap is at the bottom of $C = \varepsilon_0 A/d$. Halving the gap doubles $C$. It does not halve it.
• Writing $C$ in the wrong unit. $1$ pF is $10^{-12}$ F, not $10^{-9}$ F.

Remember these

• $C = Q/V$, in farad. $1\ \mu\text{F} = 10^{-6}$ F, $1\ \text{nF} = 10^{-9}$ F and $1\ \text{pF} = 10^{-12}$ F.
• $C$ depends on the shape, the size, the gap and the material. It does not depend on $Q$ or $V$.
• Isolated sphere: $C = 4\pi\varepsilon_0 R$.
• Parallel plates: $C = \dfrac{\varepsilon_0 A}{d}$, with $E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{V}{d}$ between the plates.
• Concentric spheres: $C = 4\pi\varepsilon_0 \dfrac{ab}{b - a}$.
• A gap in air breaks down at about $3 \times 10^{6}$ V/m, so $V_{\max} = E_{\max}d$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A capacitor holds $18\ \mu\text{C}$ when it is joined to a $6.0$ V battery. What is its capacitance?

2. The charge on a capacitor is doubled. What happens to its capacitance, and what happens to the voltage across it?

3. The area of the plates of a capacitor is doubled and the gap is also doubled. What happens to $C$?

4. Why can one single isolated sphere be said to have a capacitance?