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  4. 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

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Q. 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

Electrostatic Potential and Capacitance

Solution:

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