A parallel plate capacitor of capacitance 2 F is charged to a potential V, The energy stored in the capacitor is E₁. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is E₂. The ratio E₂ / E₁ is :

Solution

Physics sir · 4 steps

Charge has nowhere to go, so keep Q fixed and let the capacitance double: energy Q²/2C halves.

1. Charge is conserved. The first capacitor holds Q = CV. When you join it to the uncharged one, no charge leaves the circuit. It only spreads over both plates, so the total stays Q.
2. New combination. Two identical capacitors in parallel give 2C. The common voltage is Q/2C = V/2.

   E₁ = Q²/2C, E₂ = Q²/2(2C) = E₁/2

3. Check with voltage. Using 1/2CV² gives the same result: E₂ = 1/2(2C)(V/2)² = 1/4CV², which is half of E₁ = 1/2CV².
4. So. E₂ / E₁ = 1 : 2. The missing half is lost as heat in the connecting wires while the charge redistributes. The value 2 F does not matter.

The idea to keep. When an isolated charged capacitor is connected to another, conserve charge and use Q²/2C. Energy always drops on sharing, and the loss goes into heat in the wires.

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