A 600 pF capacitor is charged by 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. Electrostatic energy lost in the process is ____________ μ J

Solution

Physics sir · 4 steps

Charge is conserved when the capacitors share it, but energy is not. Find the common voltage first.

1. Charge on first. Q = CV = 600 × 10⁻¹² × 200 = 1.2 × 10⁻⁷ C. The initial energy is 1/2CV² = 1/2 × 600 × 10⁻¹² × (200)² = 12 μJ.
2. Common voltage. After connection, the same charge spreads over 600 + 600 = 1200 pF. So V' = (1.2 × 10⁻⁷)/(1200 × 10⁻¹²) = 100 V.
3. Final energy. U_f = 1/2 × 1200 × 10⁻¹² × (100)² = 6 μJ. The rest is lost as heat in the wires and as radiation.

   Δ U = (C₁ C₂)/2(C₁ + C₂)V² = 1/2 × 300 × 10⁻¹² × (200)² = 6 μJ

4. So. Energy lost = 12 - 6 = 6 μJ.

The common slip: 12. Reporting the initial stored energy as the loss gives 12, not 6. Only the difference between the initial and final energies is lost.

The idea to keep. When a charged capacitor shares charge with an identical uncharged one, exactly half the stored energy is lost. In general the loss is C₁C₂/2(C₁+C₂)(V₁-V₂)².

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