An air-filled parallel plate capacitor is attached to a voltage source and charged. The voltage source is removed, and then the plates are separated to double their previous distance. What happens to the energy stored in the capacitor when the plates are separated?

Question

Quizplus · Accepted Answer

When the plates of a charged capacitor are separated, the work done to separate them increases the potential energy stored in the capacitor. Since the capacitance $C = \frac{\varepsilon_0 A}{d}$ decreases as the distance $d$ between the plates increases (with $\varepsilon_0$ being the permittivity of free space and $A$ the area of the plates), and the energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$, where $V$ is the voltage across the capacitor, the decrease in $C$ leads to an increase in $U$ because $V$ remains constant after the voltage source is removed. The energy actually doubles because the work done to separate the plates (which is equal to the increase in stored energy) is proportional to the square of the separation.

An Air-Filled Parallel Plate Capacitor Is Attached to a Voltage