A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is ___________ nJ.

Solution

Physics sir · 4 steps

Charge is conserved when the capacitors are joined, but energy is not. Find the common voltage first.

1. Initial energy. Before joining, Uᵢ = 1/2CV² = 1/2 × 50 × 10⁻¹² × (100)² = 2.5 × 10⁻⁷ J = 250 nJ.
2. Share the charge. The charge Q = CV spreads over a total capacitance of 2C, because the charge has nowhere else to go. So the common voltage is V' = CV/2C = 50 V.
3. Final energy. Each capacitor now stores 1/2 × 50 × 10⁻¹² × (50)² = 62.5 nJ. Both together store U_f = 125 nJ, exactly half of the start.

   Δ U = Uᵢ - U_f = 250 nJ - 125 nJ = 125 nJ

4. So. The electrostatic energy lost as heat and radiation in the connecting wires is 125 nJ.

The common slip: 250. Using CV² without the 1/2 doubles every energy and gives a loss of 250, not 125.

The idea to keep. When a charged capacitor is joined to an identical uncharged one, the charge stays the same, the voltage halves, and exactly half the stored energy is lost.

Answer from the NTA final answer key. Solution by PrepWiser.