Energy stored by a capacitor (A-level only)

AQA·Physics·3.7.4.3

Energy stored in a capacitor

Charging a capacitor requires work to separate charge: charge is transferred from one plate to the other. The work done is stored as energy in the capacitor’s electric field.

As charge builds up on the plates, the potential difference between them increases. This means that more work is needed to transfer each further unit of charge. The energy stored therefore depends on how the potential difference rises during charging, not just on its final value.

For a capacitor with constant capacitance $$C$$, the charge and potential difference are related by $$Q = CV$$. Here, $$Q$$ is the magnitude of the charge on either plate and $$V$$ is the potential difference across the capacitor.

Energy and the charge–potential difference graph

A graph of charge $$Q$$ against potential difference $$V$$ is a straight line through the origin. Up to the final potential difference, the area under the line is a triangle with base $$V$$ and height $$Q$$.

The potential difference rises linearly from zero to its final value, so its average during charging is $$V/2$$. The total work done is therefore the charge transferred multiplied by this average potential difference. This equals the triangular area under the graph:

E=12QVE = \frac{1}{2}QV

Using $$Q = CV$$ gives two equivalent expressions for the stored energy:

E=12QV=12CV2=Q22CE = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}

Choose the expression that matches the quantities known. Energy $$E$$ is measured in joules, charge $$Q$$ in coulombs, potential difference $$V$$ in volts and capacitance $$C$$ in farads.