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 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
$C{V^2}$
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