Electrostatics Question 41
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 up to 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} $
BETA
