An ideal parallel-plate capacitor consists of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference across the
Plates. If the separation between the plates is now doubled, the amount of electrical energy stored
On the capacitor will
A) double.
B) quadruple.
C) be cut in fourth.
D) be cut in half.
E) not change.

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. 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. If the separation $d$ is doubled, the new capacitance $C'$ becomes $\frac{\varepsilon_0A}{2d}$, which is half of the original capacitance. Since the voltage $V$ remains constant (because the capacitor is connected to a battery that maintains a constant potential difference), the new energy $U'$ stored in the capacitor is $\frac{1}{2}C'V^2 = \frac{1}{2} \cdot \frac{1}{2}CV^2 = \frac{1}{4}CV^2$. This means the energy is halved, not quartered, because the $1/2$ factor in the energy formula does not change with the capacitance.

An Ideal Parallel-Plate Capacitor Consists of Two Parallel Plates of Area