Derive an expression for the energy stored in a charged capacitor. A 50 μF capacitor is charged to 200 V: find the energy it stores.

How to lay out the answer. Follow the frame, then write it in your own words.

1 · State the idea. The work done in moving charge onto the plates of a capacitor, against the field, is stored as electrostatic potential energy.
2 · Average voltage. As C is charged slowly to a charge q, the potential difference rises from 0 to V, so its average is V/2 = q/(2C).
3 · Work done. W = charge × average PD = q × q/(2C) = q²/(2C).
4 · All three forms. U = q²/(2C) = ½qV = ½CV², stored in the field between the plates.
5 · Numerical. U = ½ × 50 × 10⁻⁶ × (200)² = 25 × 10⁻⁶ × 4 × 10⁴ = 1 J.
6 · Close in one line. So the energy stored grows as the square of the voltage.

Scoring tip: Explain where the ½ comes from: the average voltage during charging. An answer that only writes the formula loses the derivation marks.