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) $ CV^{2}/4 $

B) $ CV^{2}/2 $

C) $ 3CV^{2}/4 $

D) $ CV^{2} $

Show Answer

Answer:

Correct Answer: D

Solution:

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



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