A capacitor of capacity C is connected with a battery of potential V in parallel. distance between i

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2. Electric Potential and Capacitance

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A capacitor of capacity $C$ is connected with a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential $V$ again, the energy given by the battery will be

A

$C{V^2}/4$

B

$C{V^2}/2$

C

$3C{V^2}/4$

D

$C{V^2}$

Solution

(d) Extra charge $Q$ = $(2CV -CV)$ = $CV$ flows through potential $V$ of the battery. Thus $W$ = $QV$ = $C{V^2}$

Standard 12

Physics

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