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A parallel plate capacitor is charged to V volt by a battery. The battery is disconnected and the separation between the plates is halved. The new potential difference across the capacitor will be
Option: 1 $\frac{V}{2}$
Option: 2 $V$
Option: 3 $2V$
Option: 4 $\frac{V}{4}$

Answers (1)

Charge before removing the battery

$\\Q=CV\\C=\frac{\epsilon_0A}{d}$

When the battery is removed the net charge remains the same

$C'=\frac{\epsilon_0 A}{d/2}=2C$

$\\Q'=2CV'\\Q=Q'\\CV=2CV'\\V'=V/2$

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Safeer PP

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