A parallel plate capacitor of capacitance 200 μF is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ____________J.

Solution

Physics sir · 4 steps

Battery connected means V is fixed; the slab doubles C, so 1/2CV² doubles too.

1. What stays fixed. The battery is still connected, so the voltage across the plates stays at 200 V. The charge on the plates is free to change.
2. Energy before. U₁ = 1/2CV² = 1/2 × 200 × 10⁻⁶ × (200)² = 4 J.
3. Energy after. Filling the gap with a dielectric of K = 2 makes the capacitance KC = 400 μF. With V unchanged, U₂ = 1/2 × 400 × 10⁻⁶ × (200)² = 8 J.

   Δ U = U₂ - U₁ = 1/2(K-1)CV² = 1/2(2-1)(200 × 10⁻⁶)(200)²

4. So. The stored energy rises from 4 J to 8 J, a change of 4 J.

The common slip: 2. Treating the charge as fixed, as if the battery were disconnected, makes the energy fall from 4 J to 2 J and gives a change of 2, not 4.

The idea to keep. First ask whether the battery is connected. If it is, V is fixed and U = 1/2CV² grows K times. If it is disconnected, Q is fixed and U = Q²/2C falls K times.

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