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(D) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$.When the plates are moved farther apart,the distance $

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the voltage across the capacitor increases

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(D) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$.When the plates are moved farther apart,the distance $d$ increases,which causes the capacitance $C$ to decrease.Since the capacitor is disconnected from the battery,the charge $Q$ on the plates remains constant.Using the relation $Q = CV$,we can write $V = \frac{Q}{C}$.Since $Q$ is constant and $C$ decreases,the voltage $V$ across the capacitor must increase.Additionally,the energy stored $U = \frac{Q^2}{2C}$ increases because $C$ decreases.

$A$ parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles,then

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