A2 Daily A Level Physics question

2026-03-08 OCR A Capacitance I (A2) 6.1.1 Capacitance: C = Q/V 6.1.2 Energy stored in a capacitor: 1/2 CV^2 (description) 6.1.3 Dielectrics (qualitative): effect on C; isolated vs connected capacitors

In a lab, a student charges an air-gap parallel-plate capacitor from a DC supply and then disconnects it so it is isolated (assume no leakage). Without changing plate area or separation, they fully slide a PTFE sheet (a high-permittivity dielectric) between the plates. Which statement must be true immediately after the sheet is in place?

A Voltage across the capacitor increases and the stored energy increases, because adding a dielectric increases the capacitance.
B Voltage across the capacitor stays the same but the stored energy decreases, because the electric field is reduced while charge cannot flow.
C Voltage across the capacitor decreases and the stored energy decreases, because the charge is fixed and the capacitance increases. (correct)
D Voltage across the capacitor decreases and the stored energy increases, because the dielectric does work on the capacitor.

Answer

The correct answer is C.

Correct: C — Voltage across the capacitor decreases and the stored energy decreases, because the charge is fixed and the capacitance increases. A is wrong because at fixed charge increasing capacitance makes the voltage drop, not rise, and the energy falls rather than increases. B is wrong because with fixed charge and increased capacitance the voltage cannot stay the same; it must decrease, even though the field and energy do reduce. C is correct because isolating the capacitor fixes charge; inserting a dielectric raises capacitance, so the voltage falls and 1/2 CV^2 decreases overall. D is wrong because although the voltage falls, the stored electrical energy does not increase; the system releases energy as the dielectric is pulled in when isolated.

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