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
A) $CV^{2}/4$
B) $CV^{2}/2$
C) $3CV^{2}/4$
D) $CV^{2}$
Explanation
(d) Extra charge $Q = (2CV - CV) = CV$ flows through potential $V$ of the battery. Thus $W = QV = CV^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.