Physics 9702 · Paper 4

Energy Stored

Charging a capacitor is work: every extra electron is pushed onto a plate that already repels it. The total shows up as a triangle on a graph.

Where the ½ comes from

V against Q is a straight line through the origin (gradient 1/C). The energy stored is the area under it — a triangle, which is where the ½ in W = ½QV comes from.

Moving charge $Q$ through a fixed p.d. $V$ costs $W = QV$ . But a capacitor's p.d. grows from zero as it charges, so the work is the area under the p.d.–charge graph — a triangle, giving $W = \tfrac{1}{2}QV$ . Both the area statement and the result are separate marks (9702/42/O/N/24 Q7(b)).

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

Symbols

$W$ = energy stored by the capacitor (J)
$Q$ = charge stored (C)
$V$ = p.d. across the capacitor (V)
$C$ = capacitance (F)

Three spellings of one formula, linked by $Q = CV$ — pick whichever matches the given quantities. A 2025 part gave $W$ and $C$ : $V = \sqrt{2W/C} = \sqrt{2 \times 19\times10^{-3} / 470\times10^{-6}} = 9.0$ V, then $Q = CV = 4.2$ mC (9702/41/O/N/25 Q6(b)).

Common mistake

An energy change is

\Delta W = \tfrac{1}{2}C\left(V_2^2 - V_1^2\right)

— difference of squares, never

\tfrac{1}{2}C(V_2 - V_1)^2

. Raising a

45\ \mu\text{F}

capacitor from 8.0 V to 9.6 V stores an extra

\tfrac{1}{2} \times 45\times10^{-6} \times (9.6^2 - 8.0^2) = 6.3\times10^{-4}

J (9702/42/F/M/24 Q4(b)) — the wrong version gives an answer 11 times too small.

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