A parallel plate capacitor is charged by conn | vedclass

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(D) $1$. Initially,the capacitor is charged to a potential $V$ by a battery. The charge on the capacitor is $Q = CV$,where $C = \frac{\epsilon_0 A}{d}

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$B$ and $C$ both

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(D) $1$. Initially,the capacitor is charged to a potential $V$ by a battery. The charge on the capacitor is $Q = CV$,where $C = \frac{\epsilon_0 A}{d}$.$2$. When the battery is disconnected,the charge $Q$ remains constant on the plates.$3$. When the plates are pulled apart to a new separation $d' = 2d$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2}$.$4$. Since the charge $Q$ is constant,the new potential difference $V'$ across the plates becomes $V' = \frac{Q}{C'} = \frac{Q}{C/2} = 2V$. Thus,the potential difference increases,making statement $C$ correct.$5$. When the capacitor is reconnected to the battery of potential $V$,the charge on the capacitor must return to $Q = CV$. Since the current charge is $Q = C'V' = (C/2)(2V) = CV$,the charge remains the same. However,the potential difference across the capacitor is $2V$,which is higher than the battery voltage $V$. Therefore,charge will flow from the capacitor into the battery until the potential across the capacitor drops to $V$. Thus,statement $B$ is correct.$6$. Since both $B$ and $C$ are correct,the correct option is $D$.

$A$ parallel plate capacitor is charged by connecting it to a battery. The battery is disconnected and the plates of the capacitor are pulled apart to make the separation between the plates twice. Again the capacitor is connected to the battery (with same polarity) then

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