A parallel-plate capacitor is connected to a battery and becomes fully charged. The capacitor is then disconnected, and the separation between the plates is increased in such a way that no charge
Leaks off. As the plates are being separated, the energy stored in this capacitor
A) decreases.
B) increases.
C) does not change.
D) become zero.

Question

Quizplus · Accepted Answer

The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$, where $U$ is the energy, $C$ is the capacitance, and $V$ is the voltage across the capacitor. When the capacitor is disconnected from the battery and the plate separation is increased, the charge $Q$ remains constant because it is isolated, and the voltage $V$ across the capacitor remains constant as well. However, the capacitance $C$ of a parallel-plate capacitor is given by $C = \frac{\varepsilon_0A}{d}$, where $\varepsilon_0$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the separation between the plates. As the separation $d$ increases, the capacitance $C$ decreases. Since the energy stored is inversely proportional to the capacitance (considering $V$ constant in this scenario), the energy stored in the capacitor increases as the plates are separated.

A Parallel-Plate Capacitor Is Connected to a Battery and Becomes