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
B
C
D
Text Solution
Verified by ExpertsThe correct answer is:
D
Extra charge Q = (2CV – CV) = CV flows through potential V of the battery. Thus W = QV = 
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