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

Question

Quizplus · Accepted Answer

The energy stored in a parallel-plate capacitor is given by the equation U = (1/2)CV^2, where C is the capacitance and V is the potential difference between the plates. Since the battery maintains a constant potential difference, V is constant. The capacitance of a parallel-plate capacitor is given by the equation C = εA/d, where ε is the permittivity of the material between the plates, A is the area of the plates, and d is the separation between the plates. When the separation between the plates is doubled, the capacitance is halved (since d is in the denominator of the equation for capacitance). Therefore, the energy stored in the capacitor is halved as well. So, the new energy stored in the capacitor is (1/2)U = (1/2)(1/2)CV^2 = (1/4)U, or U₀/2. The answer is D.

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