A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is conne

Question

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system

Accepted Answer

1. **Initial State:** Let the capacitance be $C$ and the potential be $V$. The initial charge is $Q = CV$. The initial energy stored is $U_i = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$.
2. **Disconnection and Reconnection:** When the battery is removed, the charge $Q$ is conserved. Connecting an identical uncharged capacitor ($C$) in parallel creates a system with equivalent capacitance $C_{eq} = C + C = 2C$.
3. **Final State:** The total charge $Q$ is now distributed across the equivalent capacitance. The final energy is $U_f = \frac{Q^2}{2C_{eq}}$.
4. **Calculation:** Substituting $C_{eq} = 2C$, we get $U_f = \frac{Q^2}{2(2C)} = \frac{1}{2} \left( \frac{Q^2}{2C} 
ight) = \frac{U_i}{2}$.
5. **Conclusion:** The total electrostatic energy decreases by a factor of 2. The missing energy is dissipated as heat and electromagnetic radiation during the redistribution of charge (as mentioned in NCERT Example 2.10).

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