Electrostatics Question 41

Question: 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 [MP PET 1989]

Options:

A) CV2/4

B) CV2/2

C) 3CV2/4

D) CV2

Show Answer

Answer:

Correct Answer: D

Solution:

Extra charge Q = (2CV , CV) = CV flows through potential V of the battery. Thus W = QV = CV2



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