Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

increases by a factor of 4

decreases by a factor of 2

remains the same

increases by a factor of 2

Step-by-Step Solution

Initial State: Let the capacitance be $C$ and the potential be $V$ . The initial charge is $Q = CV$ . The initial energy stored is $U_i = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$ .
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 $C_{eq} = C + C = 2C$ .
Final State: The total charge $Q$ is now distributed across the equivalent capacitance. The final energy is $U_f = \frac{Q^2}{2C_{eq}}$ .
Calculation: Substituting $C_{eq} = 2C$ , we get $U_f = \frac{Q^2}{2(2C)} = \frac{1}{2} \left( \frac{Q^2}{2C} \right) = \frac{U_i}{2}$ .
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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