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(A) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$,where $A$ is the area of the plates and $d$ is the dis

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increases

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(A) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.When the distance $d$ is increased,the capacitance $C$ decreases.Since the battery is disconnected,the charge $Q$ on the plates remains constant.The relationship between charge,capacitance,and potential difference is $V = \frac{Q}{C}$.Since $Q$ is constant and $C$ decreases,the potential difference $V$ must increase.

$A$ parallel plate air capacitor is charged to a potential difference of $V \text{ volts}$. After disconnecting the charging battery,the distance between the plates of the capacitor is increased using an insulating handle. As a result,the potential difference between the plates

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