Physics 9702 · AS & A Level · Energy stored in a capacitor

Energy stored in a capacitor — practice question

In Fig. 5.1, capacitors A and B are placed in the circuit shown. Capacitor A has capacitance $C$ and capacitor B has capacitance $3C$. The cell has electromotive force (e.m.f.) $V$. The two-way switch $S$ starts at position $X$, and capacitor B is initially uncharged.
(a(i))[1]

State, in terms of $V$ and $C$, the initial charge $Q_A$ on the plates of capacitor A.

(a(ii))[1]

State, in terms of $V$ and $C$, the initial energy $E_A$ stored in capacitor A.

(b(i))[2]

State and explain what happens to the charge that was initially on the plates of capacitor A.

(b(ii))[3]

Show that the final potential difference (p.d.) $V_B$ across capacitor B is given by $V_B = \frac{V}{4}$. Give your reasoning.

(b(iii))[2]

Determine an expression, in terms of $V$ and $C$, for the decrease $\Delta E$ in the total energy stored in the capacitors when the switch is changed over.

(ii)[3]

Show that the final potential difference (p.d.) $V_B$ across capacitor B is given by $V_B = \frac{V}{4}$. Give your reasoning.

(iii)[2]

Determine an expression, in terms of $V$ and $C$, for the decrease $\Delta E$ in the total energy stored in the capacitors as a result of the switch moving.

Worked solution & mark scheme

This 14-mark question has a full step-by-step worked solution and mark scheme. One marking point: So, $Q_A = CV$.

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