Q. The plates of a parallel plate capacitor of capacity$\ 50\mu C$ are charged to a potential of $100\ volts$ and then separated from each other so that the distance between them is doubled. How much is the energy spent in doing so
A) $25 \times 10^{- 2}\ J$
B) $- 12.5 \times 10^{- 2}\ J$
C) $- 25 \times 10^{- 2}\ J$
D) $12.5 \times 10^{- 2}\ J$
Explanation
(a) $W_{ext} = \frac{1}{2}C^{'}V^{'2} - \frac{1}{2}CV^{2}$
$= \left( \frac{1}{2} \right)\left( \frac{C}{2} \right)(2V)^{2} - \frac{1}{2}CV^{2} = \frac{1}{2}CV^{2}$
$W_{ext} = \frac{1}{2} \times 50 \times 10^{- 6} \times (100)^{2} = 25 \times 10^{- 2}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.