Q. Capacitor of a capacitor is 48$\mu F$. When it is charged from 0.1 C to 0.5 C , change in the energy stored is

A) 2500 J
B) $2.5 \times 10^{- 3}\ J$
C) $2.5 \times 10^{6}\ J$
D) $2.42 \times 10^{- 2}\ J$
Explanation

(a) Change in energy $\mathrm{\Delta}U = \frac{1}{2}\left\lbrack \frac{q_{1}^{2} - q_{2}^{2}\rbrack}{C} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\left\lbrack \frac{(0.5)^{2} - (0.1)^{2}}{48 \times 10^{- 6}} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\left\lbrack \frac{0.25 - 0.01}{48 \times 10^{- 6}} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\left\lbrack \frac{0.24}{48 \times 10^{- 6}} \right\rbrack = \frac{1}{2}\left\lbrack \frac{10^{4}}{2} \right\rbrack = 2500\ J$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.