Q. A parallel plate capacitor of capacity $C_{0}$ is charged to a potential $V_{0}$ $(i)$ The energy stored in the capacitor when the battery is disconnected and the separation is doubled $E_{1}$ (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_{2}.$ Then $E_{1}/E_{2}$ value is
A) 4
B) $3/2$
C) 2
D) $1/2$
Explanation
(a) Let $E = \frac{1}{2}C_{0}V_{0}^{2}$ then $E_{1} = 2E$ and $E_{2} = \frac{E}{2}$
So $\frac{E_{1}}{E_{2}} = \frac{4}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.