Question

Energy $E$ is stored in a parallel plate capacitor $C_1$. An identical uncharged capacitor $C_2$ is connected to it, kept in contact with it for a while and then disconnected. Then the energy stored in $C_2$ is:

• A. $\frac{E}{2}$
• B. $\frac{E}{3}$
• C. $\frac{E}{4}$
• D. Zero

Answer 1

Answer: (C)

$\frac{E}{4}$

Initial energy of capacitor $=\frac{Q^2}{2C}=E$
After the connection is made, since the capacitances are same and the potential across the capacitors also need to be equal, charge across both capacitors $=\frac{Q}{2}$
$\therefore$ final energy of a capacitor ($C_2$) $=\frac{Q^2}{4}\times \frac{1}{2C}=\frac{E}{4}$