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(B) Initially,the capacitor is charged to a potential $V_{old}$ by a battery. The charge on the plates is $Q = C_{old} V_{old}$.When the battery is di

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$0.5$

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(B) Initially,the capacitor is charged to a potential $V_{old}$ by a battery. The charge on the plates is $Q = C_{old} V_{old}$.When 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 separation $d$ is halved to $d' = \frac{d}{2}$,the new capacitance becomes $C_{new} = \frac{\epsilon_0 A}{d/2} = 2 \left( \frac{\epsilon_0 A}{d} \right) = 2 C_{old}$.Since the charge $Q$ is constant,the new potential difference $V_{new}$ is given by $V_{new} = \frac{Q}{C_{new}}$.Substituting $Q = C_{old} V_{old}$ and $C_{new} = 2 C_{old}$,we get $V_{new} = \frac{C_{old} V_{old}}{2 C_{old}} = \frac{1}{2} V_{old}$.Therefore,the ratio $\frac{V_{new}}{V_{old}} = 0.5$.

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