Physics 9702 · Paper 4

Sharing Charge, Losing Energy

Connect a charged capacitor across an uncharged one and the charge redistributes — one scenario, recycled across three sessions, with a counter-intuitive energy ending.

The two-rule method

Once connected, the pair is a parallel combination. Charge flows until the p.d.s are equal — that sentence is a mark by itself (9702/42/M/J/23 Q5(b)(i)) — and throughout, the total charge is conserved.

Worked example

The recycled C-and-3C problem

A capacitor of capacitance $C$ , charged to p.d. $V$ (charge $Q$ ), is connected across an uncharged capacitor of capacitance $3C$ . Find the final p.d. and charges — the identical setup appeared in 2023 and again in 2024 (9702/42/M/J/23 Q5(b)) (9702/42/O/N/24 Q7(c)).

Total capacitance in parallel: $C_T = C + 3C = 4C$ ; total charge still $Q = CV$ .
Shared p.d.: $V_f = \dfrac{Q}{C_T} = \dfrac{CV}{4C} = \dfrac{V}{4}$ .
Each charge from $Q = CV_f$ : the original keeps $Q/4$ , the new one takes $3Q/4$ — split in proportion to capacitance.

Answer

V_f = V/4;\quad Q/4\ \text{and}\ 3Q/4

The energy books do not balance: initially $\tfrac{1}{2}CV^2$ , finally $\tfrac{1}{2}(4C)(V/4)^2 = CV^2/8$ — three quarters of the energy is gone (9702/42/M/J/23 Q5(b)(iii)). It is dissipated as heat in the connecting wires while the charge surges across. Numerical version: 19 mJ became 14 mJ when a 470 $\mu\text{F}$ capacitor was connected to an uncharged 180 $\mu\text{F}$ (9702/41/O/N/25 Q6(b)(iii)).

Common mistake

Conserve charge, never energy. Students who set initial energy = final energy get

V/2

instead of

V/4

— and asked whether the stored energy afterwards is more, the same or less, the answer is always less (9702/42/O/N/24 Q7(c)(ii)).

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