Energy stored by a capacitorAQA A-Level Physics: Revision notes

To charge a capacitor, work must be done to move charge onto the plates against the p.d. already present. A graph of charge Q against p.d. V for a capacitor is a straight line through the origin, because $C = Q/V$ is constant.

The area under the graph is the energy stored. The area is a triangle with base V and height Q:

$E = \tfrac{1}{2}QV$

The gradient of the graph is the capacitance.

Using $Q = CV$ in $E = \tfrac{1}{2}QV$ gives three equivalent forms:

$E = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C}$

• Use $\tfrac{1}{2}CV^2$ when you know C and V.
• Use $\tfrac{1}{2}QV$ when you know Q and V.
• Use $Q^2/2C$ when you know Q and C.

The unit is the joule.

A 470 µF capacitor is charged to 12 V.

• $Q = CV = 470 \times 10^{-6} \times 12 = 5.6 \times 10^{-3}$ C
• $E = \tfrac{1}{2}CV^2 = 0.5 \times 470 \times 10^{-6} \times 12^2 = 3.4 \times 10^{-2}$ J
• Check: $\tfrac{1}{2}QV = 0.5 \times 5.64 \times 10^{-3} \times 12 = 3.4 \times 10^{-2}$ J

If this energy is released in 0.50 s, the mean power is $3.4 \times 10^{-2} / 0.50 = 6.8 \times 10^{-2}$ W.

• At fixed C, $E \propto V^2$ and $E \propto Q^2$. Doubling the p.d. quadruples the energy.
• At fixed V, $E \propto C$. Doubling C doubles the energy.
• At fixed Q, $E \propto 1/C$.

Worked example. A 100 µF capacitor at 20 V stores $0.5 \times 100 \times 10^{-6} \times 400 = 2.0 \times 10^{-2}$ J. At 40 V it stores 8.0 × 10⁻² J, four times as much. To store 0.10 J at 20 V needs $C = 2E/V^2 = 0.20 / 400 = 5.0 \times 10^{-4}$ F.