Capacitance

Capacitors store electric charge in a circuit.

A capacitor 电容器 stores charge. The simplest one is two parallel conductor 导体 plates with an insulator 绝缘体 (a dielectric 电介质, or just vacuum 真空 / air) between them. Connected to a battery, charge $+Q$ builds up on one plate and $-Q$ on the other, with a potential difference 电势差 $V$ across the gap.

A parallel-plate capacitor: equal and opposite charges on two plates with a p.d. $V$ across the gap

The capacitance 电容 $C$ of any capacitor (or any isolated conductor) is

This applies to:

• an isolated sphere holding charge $Q$ at potential $V$ (zero at infinity). For radius $r$, $V = Q/(4\pi\varepsilon_{0}r)$, so $C = 4\pi\varepsilon_{0} r$.
• a parallel-plate capacitor: charges $\pm Q$ on the plates, p.d. $V$ between them.

Unit: farad 法拉 (F) $= \text{C V}^{-1}$. A farad is huge, so real capacitors run from $\text{pF}$ to $\text{mF}$.

Capacitance is constant for a given capacitor (set by its size and dielectric). Doubling the charge doubles the voltage, so $C = Q/V$ stays the same.

Worked example. A $100\ \mu\text{F}$ capacitor is charged to $12\ \text{V}$. Find the charge stored.

Real capacitors range from large electrolytic cans (high capacitance) down to tiny film and ceramic types -- from pF up to mF

Combining capacitors

Capacitors in parallel share the same p.d. $V$. The total charge is the sum:

A parallel combination has larger capacitance than any one capacitor.

Capacitors in parallel share the same p.d.; the charges add

Capacitors in series carry the same charge $Q$. The total p.d. is the sum:

A series combination has smaller capacitance than any one capacitor.

Capacitors in series carry the same charge; the p.d.s add

Note: these rules are the opposite of those for resistors (resistors sum in series; capacitors sum in parallel), because $C = Q/V$ has $V$ on the bottom while $R = V/I$ has $I$ on the bottom.

Charging a capacitor from $0$ to $Q$ needs work, because each extra bit of charge is pushed against the p.d. already there. When the charge is $q$, the p.d. is $V(q) = q/C$, so adding a small charge $dq$ needs work $V\,dq$. The total work is

Using $V = Q/C$, this is the energy 能量 stored:

Worked example. Find the energy stored in a $100\ \mu\text{F}$ capacitor charged to $12\ \text{V}$.

Reading the $Q$–$V$ graph

A plot of $V$ against $Q$ is a straight line through the origin with gradient $1/C$. The energy stored is the area under the line up to a given charge $Q$, which is the triangle $\tfrac{1}{2} Q V$. The factor $\tfrac{1}{2}$ is there because the average p.d. during charging is $V/2$ (it grows from zero to $V$), not $V$.

The energy stored is the area under the potential–charge line (the triangle $\tfrac{1}{2}QV$)

Why charging is "half efficient"

Connect a capacitor $C$ to an ideal battery of e.m.f. $V$ through a wire. The capacitor stores $\tfrac{1}{2} C V^{2}$, but the battery supplies charge $Q = CV$ at e.m.f. $V$, giving out $QV = CV^{2}$. The other half is lost as heat in the wire — whatever the wire's resistance.

A capacitor $C$ charged to $V_{0}$ is connected through a switch to a resistor 电阻器 of resistance 电阻 $R$. When the switch closes at $t = 0$, the capacitor discharges.

A capacitor charges through switch A, then discharges through the resistor via switch B

Setting up the equation

By Kirchhoff's second law 基尔霍夫第二定律 around the loop, $V_{C} = V_{R}$. Using $V_{C} = Q/C$, $V_{R} = IR$ and $I = -dQ/dt$:

This is solved by an exponential decay 指数衰减 with time constant $RC$.

Discharge equations

Charge $Q$, p.d. $V$ and current $I$ all decay exponentially with the same time constant:

with $I_{0} = V_{0}/R$.

Charge decays exponentially during discharge, falling to $Q_0/e$ after one time constant $RC$

Time constant

is the time constant 时间常数 (in seconds: $\Omega \cdot \text{F} = \text{s}$). It is the time for a decaying quantity to fall to $1/e \approx 0.37$ (about 37%) of its starting value. After $2\tau$ it is at about 13.5%; after $5\tau$, below 1%.

Worked example. A $100\ \mu\text{F}$ capacitor charged to $12\ \text{V}$ is discharged through a $47\ \text{k}\Omega$ resistor. Find the time constant and the voltage after one time constant.

After one time constant the voltage falls to $1/e$ of its start: $V = 12 \times 0.37 \approx 4.4\ \text{V}$.

To find $\tau$ from a curve: read the time to fall to $1/e$ of the start. Or take logs: $\ln(V/V_{0}) = -t/(RC)$, so a plot of $\ln V$ against $t$ is a straight line with gradient $-1/(RC)$.

A graph of $\ln V$ against time is a straight line of gradient $-1/(RC)$

Reading graphs during discharge

• $Q$ against $V_{C}$: since $Q = C V$ always, this is a straight line through the origin with gradient $C$. Discharge moves the point from $(V_{0}, Q_{0})$ down to $(0,0)$.
• $I$ against $V_{C}$: since $I = V_{C}/R$, this is a straight line through the origin with gradient $1/R$, so you can find $R$.

Common exam questions

Given a discharge curve $V(t)$ or $Q(t)$:

• read the start value $V_{0}$ or $Q_{0}$ at $t = 0$.
• read the time to fall to $V_{0}/e$ → time constant $\tau = RC$.
• given $R$, find $C = \tau / R$ (or the other way round).
• predict a later value with the exponential formula.

• $C = Q/V$; energy stored $W = \frac{1}{2}QV = \frac{1}{2}CV^2$ (the $\frac{1}{2}$ is the area under the $Q$-$V$ graph).
• Combine capacitors the opposite way to resistors (parallel add, series reciprocal).
• Discharge is exponential: $Q = Q_0 e^{-t/RC}$; the time constant $\tau = RC$ is when the charge falls to $37\%$.