Q. The energy of a charged capacitor is $U$. Another identical capacitor is connected parallel to the first capacitor, after disconnecting the battery. The total energy of the system of these capacitors will be
Explanation
(b) Common potential =$\frac{C_{1}V_{0} + C_{2} \times 0}{C_{1} + C_{2}} = \frac{C_{2}V_{0}}{C_{1} + C_{2}}$
$U_{before} = \ \frac{1}{2}C_{1}V_{0}^{2}$
$U_{after} = \frac{1}{2}C_{1}\left\lbrack \frac{C_{1}V_{0}}{C_{1} + C_{2}} \right\rbrack^{2} + \frac{1}{2}C_{2}\left\lbrack \frac{C_{1}V_{0}}{C_{1} + C_{2}} \right\rbrack^{2}$
$= \frac{1}{2}\left\lbrack \frac{C_{1}V_{0}}{C_{1} + C_{2}} \right\rbrack^{2}\ (C_{1} + C_{2})$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{U_{before}}{U_{after}} = \ \frac{C_{1} + C_{2}}{C_{1}}$
Here, $\ \ \ \ \ C_{1} = C_{2} = C$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{U_{before}}{U_{after}} = \ \frac{2C}{C}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U_{after} = \frac{U}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.