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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.Since the battery is disconnected,the charge $Q$ on the

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The potential difference between the plates increases.

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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.Since the battery is disconnected,the charge $Q$ on the plates remains constant.When the distance $d$ between the plates is increased,the capacitance $C$ decreases because $C \propto \frac{1}{d}$.The potential difference $V$ between the plates is given by $V = \frac{Q}{C}$.Since $Q$ is constant and $C$ decreases,the potential difference $V$ must increase.Therefore,the correct statement is that the potential difference between the plates increases.

$A$ parallel plate capacitor is charged by a battery and then the battery is disconnected. If the distance between the plates is increased using an insulated handle,then:

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