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(A) When a capacitor is charged and the battery is disconnected,the charge $q$ on the plates remains constant.The capacitance of a parallel plate capa

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The stored energy of the capacitor increases

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(A) When a capacitor is charged and the battery is disconnected,the charge $q$ on the plates remains constant.The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$. When the plates are moved apart,the distance $d$ increases,so the capacitance $C$ decreases.Since the charge $q$ is constant and $C$ decreases,the potential difference $V = \frac{q}{C}$ must increase.The energy stored in the capacitor is given by $W = \frac{q^2}{2C}$. Since $C$ decreases and $q$ is constant,the stored energy $W$ increases.

$A$ parallel plate capacitor is charged to a certain potential and the charging battery is then disconnected. Now,if the plates of the capacitor are moved apart,then:

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