Q. A charge of $40\mu C$ is given to a capacitor having capacitance $C = 10\mu F.$ The stored energy in ergs is
A) $80 \times 10^{- 6}$
B) 800
C) 80
D) 8000
Explanation
(b) $U = \frac{Q^{2}}{2C} = \frac{\left( 40 \times 10^{- 6} \right)^{2}}{2 \times 10 \times 10^{- 6}} = 8 \times 10^{- 5}J$
$= 8 \times 10^{- 5} \times 10^{7} = 800\ erg$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.