Q. A capacitor of capacity $C_{1}$ is charged to the potential of $V_{0}.$ On disconnecting with the battery, it is connected with a capacitor of capacity $C_{2}$ as shown in the adjoining figure. The ratio of energies before and after the connection of switch $S$ will be
A) $(C_{1} + C_{2})/C_{1}$
B) $C_{1}/(C_{1} + C_{2})$
C) $C_{1}C_{2}$
D) $C_{1}/C_{2}$
Explanation
(a) Common potential $= \frac{C_{1}V_{0} + C_{2} \times 0}{C_{1} + C_{2}} = \frac{C_{1}V_{0}}{C_{1} + C_{2}}$
$U_{before} = \frac{1}{2}C_{1}V_{0}^{2}$
$U_{after} = \frac{1}{2}C_{1}\left( \frac{C_{1}V_{0}}{C_{1} + C_{2}} \right)^{2} + \frac{1}{2}\left( \frac{C_{1}V_{0}}{C_{1} + C_{2}} \right)^{2}$
$= \frac{1}{2}\left( \frac{C_{1}V_{0}}{C_{1} + C_{2}} \right)^{2}\left( C_{1} + C_{2} \right) \Rightarrow \frac{U_{Before}}{U_{After}} = \frac{C_{1} + C_{2}}{C_{1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.