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

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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 separation between the plates of a capacitor is increased, its capacitance $C$ decreases because $C = \frac{\varepsilon_0A}{d}$, where $A$ is the area of the plates, $d$ is the separation between the plates, and $\varepsilon_0$ is the permittivity of free space. Since the capacitor is disconnected from the battery, the charge $Q$ remains constant, and the voltage $V = \frac{Q}{C}$ increases as $C$ decreases. Therefore, because the energy $U$ depends on the square of the voltage $V$, the energy stored in the capacitor increases as the plates are separated.

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