Does current flow through a capacitor?

Not through the dielectric. Electrons flow onto one plate and off the other in the external circuit, so a current is measured there while charge builds up.

Why is the stored energy only half of QV?

The p.d. rises from zero as charge is added, so on average the charge is moved through half the final voltage. The rest of the battery’s energy is dissipated as heat.

What does the time constant tell you?

How quickly the capacitor discharges. After one time constant 37% of the charge remains; after five, under 1%.

Why plot V against time?

The exponential becomes a straight line, whose gradient -1/RC is easy to measure accurately.

Back to course

116–120ElectricityStretch

Capacitance

A $\textbf{capacitor}$ stores charge on two conducting plates separated by an insulator. Its $\textbf{capacitance}$ measures how much charge it holds per volt, and when it discharges through a resistor the charge drains away exponentially — fast at first, then ever more slowly.

40 min Topic 7 · Electric and Magnetic Fields

What you'll be able to do

Define capacitance and use $C = Q/V$
Calculate energy stored using $W = \tfrac12 QV$ and its equivalents
Describe exponential discharge and use $Q = Q_{0}e^{-t/RC}$
Use the time constant and the logarithmic form of the discharge equation

Capacitance

When a capacitor is connected to a supply, electrons are pushed onto one plate and drawn off the other, leaving equal and opposite charges $+Q$ and $-Q$ . Charging stops when the p.d. across it equals the supply voltage.

Capacitance is the charge stored per unit potential difference: $C = Q/V$ , in $\textbf{farads}$ . One farad is large; real capacitors are usually microfarads or picofarads.

The insulator — the $\textbf{dielectric}$ — prevents charge crossing between the plates. Current flows in the external circuit while charging, but not through the capacitor itself.

C = \dfrac{Q}{V}

Capacitance, in farads.

Example 1A

470\ \mu\mathrm{F}

capacitor is charged to 9.0 V. Find the charge stored.

Q = CV = 470\times10^{-6} \times 9.0

Answer

4.2\times10^{-3}\ \mathrm{C}

Tip — Convert µF to F by multiplying by $10^{-6}$ before calculating.

Energy stored

Charging a capacitor is harder as it fills, because each extra bit of charge must be pushed against the p.d. already built up. A graph of $V$ against $Q$ is a straight line through the origin, and the work done is the $\textbf{area under it}$ .

That triangle gives $W = \tfrac12 QV$ . Substituting $Q = CV$ gives the alternatives $\tfrac12 CV^{2}$ and $\tfrac12 Q^{2}/C$ .

The factor of a half means a battery charging a capacitor supplies $QV$ but only $\tfrac12 QV$ is stored — the other half is dissipated as heat in the circuit resistance.

W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2} = \tfrac{1}{2}\dfrac{Q^{2}}{C}

Energy stored.

Example 1A camera flash uses a

220\ \mu\mathrm{F}

capacitor charged to 300 V. Find the energy stored.

W = \tfrac12 \times 220\times10^{-6} \times 300^{2}

Answer

9.9\ \mathrm{J}

Discharging through a resistor

When a charged capacitor discharges through a resistor $R$ , the current at any instant is $V/R$ . As charge leaves, $V$ falls, so the current falls too — the discharge slows as it proceeds.

The rate of loss of charge is proportional to the charge remaining, which is the defining property of $\textbf{exponential decay}$ : $Q = Q_{0}e^{-t/RC}$ . The same form describes $V$ and $I$ , since both are proportional to $Q$ .

The $\textbf{time constant}$ $\tau = RC$ is the time for the charge to fall to $1/e$ , about 37%, of its initial value. After $5RC$ the capacitor is effectively fully discharged.

Q = Q_{0}e^{-t/RC}

Discharge — also for

V

and

I

\tau = RC

Time constant: fall to 37%.

Example 1A

100\ \mu\mathrm{F}

capacitor charged to 12 V discharges through

47\ \mathrm{k}\Omega

. Find the time constant and the p.d. after 8.0 s.

RC = 47\times10^{3} \times 100\times10^{-6} = 4.7

V = 12e^{-8.0/4.7} = 12e^{-1.702}

Answer

\tau = 4.7\ \mathrm{s}

;

V = 2.2\ \mathrm{V}

Tip — Check units give seconds: ohms times farads. Leaving kΩ or µF unconverted gives a time constant a thousand or a million times wrong.

The logarithmic form

Taking natural logarithms of $V = V_{0}e^{-t/RC}$ gives $\ln V = \ln V_{0} - \dfrac{t}{RC}$ .

This is a straight line when $\ln V$ is plotted against $t$ : gradient $-1/RC$ and intercept $\ln V_{0}$ . That is how the time constant, and hence an unknown capacitance, is found experimentally — a straight line is far easier to fit than a curve.

The same form lets you solve for the time to reach a given voltage: $t = RC\ln(V_{0}/V)$ .

\ln V = \ln V_{0} - \dfrac{t}{RC}

Straight line: gradient

-1/RC

t = RC\ln\dfrac{V_{0}}{V}

Time to fall to

V

Example 1For the capacitor above, how long until the p.d. falls to 3.0 V?

t = 4.7 \ln\dfrac{12}{3.0} = 4.7\ln 4

Answer

6.5\ \mathrm{s}

Equation recap

C = Q/V

Capacitance.

W = \tfrac12 CV^{2}

Energy stored.

Q = Q_{0}e^{-t/RC}

Discharge.

\tau = RC

Time constant.

\ln V = \ln V_{0} - t/RC

Straight-line form.

Common mistakes to avoid

Using

W = QV

for stored energy.

Voltage builds up from zero, so

W = \tfrac12 QV

Saying the time constant is when the capacitor is empty.

It is when charge has fallen to about 37%.

Saying current flows through the capacitor’s dielectric.

Current flows in the external circuit; the dielectric insulates.

Assuming discharge is linear.

It is exponential: fast at first, slowing continuously.

Key takeaways

Capacitance is charge per unit p.d., $C = Q/V$ , in farads.
Stored energy is the area under a $V$ – $Q$ graph: $\tfrac12 QV = \tfrac12 CV^{2}$ .
Discharge through a resistor is exponential: $Q = Q_{0}e^{-t/RC}$ .
The time constant $RC$ is the time to fall to 37%.
$\ln V$ against $t$ is a straight line of gradient $-1/RC$ .

Test yourself

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