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Physics Question on electrostatic potential and capacitance

Two plates of a parallel plate capacitor of capacity 50μF50\mu F are charged by a battery to a potential of 100 V. The battery remains connected and the plates are separated from each other so that the distance between them is doubled. The energy spent by the battery in doing so, will be

A

12.5×10−2J12.5\times {{10}^{-2}}J

B

−25×10−2J-25\times {{10}^{-2}}J

C

25×10−2J25\times {{10}^{-2}}J

D

−12.5×10−1J-12.5\times {{10}^{-1}}J

Answer

12.5×10−2J12.5\times {{10}^{-2}}J

Explanation

Solution

We know that when separation between the plates is doubled the capacitance becomes one half i.e., C=25 μFC=25\,\mu F The energy spend by the battery is given by =qV=(CV)V=CV2=qV=(CV)V=C{{V}^{2}} =25×10−6×(100)2=25\times 10-6\times {{(100)}^{2}} =25×10−2=25\times {{10}^{-2}}