Energy in capacitors: 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?

Question

Quizplus · Accepted Answer

The energy stored in a capacitor is given by the formula $U=\frac{1}{2}CV^2$, where $C$ is the capacitance and $V$ is the potential difference between the plates. For an ideal parallel-plate capacitor, $C=\frac{\epsilon_0 A}{d}$, where $\epsilon_0$ is the permittivity of free space. When the separation between the plates is doubled, the capacitance is halved, since $C'=\frac{\epsilon_0 A}{2d}=\frac{1}{2}C$. The potential difference remains the same, since the battery maintains a constant potential difference between the plates. Therefore, the energy stored in the capacitor is: $U'=\frac{1}{2}C'V^2=\frac{1}{2}\cdot\frac{1}{2}CV^2=\frac{1}{2}U$ Therefore, the energy stored in the capacitor is halved when the separation between the plates is doubled, so the answer is (D) U₀/2.

Energy in Capacitors: an Ideal Parallel-Plate Capacitor Consists of a Set