An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d.When this capacitor is connected to a battery that maintains a constant potential difference between the plates,the energy stored in the capacitor is U₀.If the separation between the plates is doubled,how much energy is stored in the capacitor? A) 4U₀ B) 2U₀ C) U₀ D) U₀/2 E) U₀/4

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Quizplus · Accepted Answer

The energy stored in a parallel-plate capacitor is given by: U = 1/2 * C * V^2 where C is the capacitance and V is the potential difference between the plates. The capacitance of a parallel-plate capacitor is given by: C = ε0 * A / d where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between them. If the separation between the plates is doubled, then the capacitance is halved: C' = ε0 * A / 2d The potential difference between the plates remains the same, so the energy stored in the capacitor becomes: U' = 1/2 * C' * V^2 = 1/2 * (ε0 * A / 2d) * V^2 = 1/4 * (ε0 * A / d) * V^2 = 1/4 * U Therefore, the energy stored in the capacitor is halved, or U' = U/2. The correct answer is D) U₀/2.

An Ideal Parallel-Plate Capacitor Consists of a Set of Two