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

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Increases

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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.When the battery is disconnected,the charge $Q$ on the plates remains constant.The potential difference $V$ across the plates is given by $V = \frac{Q}{C}$.Substituting the expression for $C$,we get $V = \frac{Q d}{\varepsilon_0 A}$.Since $Q$,$\varepsilon_0$,and $A$ are constant,we have $V \propto d$.When the distance $d$ between the plates is increased,the potential difference $V$ increases.

$A$ parallel plate air capacitor is charged to a potential difference of $V$. After disconnecting the 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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