In a parallel-plate capacitor, doubling the plate area A and halving the separation d changes the capacitance by what factor?

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Multiple Choice

In a parallel-plate capacitor, doubling the plate area A and halving the separation d changes the capacitance by what factor?

Explanation:
Capacitance of a parallel-plate capacitor scales with area and inversely with separation, C = ε0 A / d. Doubling the plate area increases C by a factor of 2. Halving the separation makes the 1/d term twice as large, so C also increases by a factor of 2 from that change. Combined, the two effects multiply: 2 × 2 = 4. Therefore the capacitance becomes four times as large.

Capacitance of a parallel-plate capacitor scales with area and inversely with separation, C = ε0 A / d. Doubling the plate area increases C by a factor of 2. Halving the separation makes the 1/d term twice as large, so C also increases by a factor of 2 from that change. Combined, the two effects multiply: 2 × 2 = 4. Therefore the capacitance becomes four times as large.

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