Q. If the charge on a capacitor is increased by 2 coulomb, the energy stored in it increase by 21%. The original charge on the capacitor is
A) 10 C
B) 20 C
C) 30 C
D) 40 C
Explanation
(b) Since $U = \frac{1}{2}\frac{Q^{2}}{C},\ \therefore 1.21U = \frac{1}{2}\frac{(Q + 2)^{2}}{C}$
$\therefore\frac{1.21}{1} = \frac{(Q + 2)^{2}}{Q^{2}} \Rightarrow \sqrt{\frac{1.21}{1}} = \frac{Q + 2}{Q}$
$\Rightarrow 1.1Q = Q + 2 \Rightarrow Q = 20\ C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.