Capacitance and energy storedEdexcel A-Level Physics: Revision notes

A capacitor stores charge. The capacitance $C$ is the charge stored per unit potential difference across it:

$C = \frac{Q}{V}$

The unit is the farad (F), equivalent to C V⁻¹. A farad is a very large unit, so capacitors are usually rated in μF, nF or pF. The charge $Q$ is the magnitude of the charge on one plate; the total net charge on the capacitor is zero.

For a given capacitor $Q \propto V$, so a graph of $Q$ against $V$ is a straight line through the origin with gradient $C$.

As a capacitor charges, each extra small charge $\Delta Q$ has to be moved against the potential difference already across the plates, so the work done is $\Delta W = V\,\Delta Q$. The total work done is the area under the graph of potential difference against charge.

Because $V \propto Q$ the graph is a straight line through the origin, so the area is a triangle:

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

Using $Q = CV$ gives the other two forms:

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

Pick the form of the energy equation that fits the data you are given:

• Given $C$ and $V$: use $W = \tfrac{1}{2}CV^2$
• Given $Q$ and $V$: use $W = \tfrac{1}{2}QV$
• Given $Q$ and $C$: use $W = Q^2/2C$

For a fixed capacitor, doubling $V$ quadruples the energy stored. When a capacitor discharges from $V_1$ to $V_2$, the energy released is $\tfrac{1}{2}C(V_1^2 - V_2^2)$.

The half of the energy supplied by a supply that is not stored is transferred as thermal energy in the resistance of the circuit while the capacitor charges.

A 470 μF capacitor is charged by a 12 V battery.

Charge stored: $Q = CV = 470\times10^{-6}\times 12 = 5.64\times10^{-3}$ C.

Energy stored: $W = \tfrac{1}{2}CV^2 = \tfrac{1}{2}\times 470\times10^{-6}\times 12^2 = 3.4\times10^{-2}$ J.

Energy supplied by the battery: $QV = 5.64\times10^{-3}\times 12 = 6.8\times10^{-2}$ J. Half of this is dissipated as heat in the circuit.