A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to potential difference of 12.0 V. The battery is now disconnected and a dielectric slab (εᵣ=6) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________ ×10⁻¹² J.

Solution

Physics sir · 3 steps

Battery gone means charge is locked in, so use U = Q²/2C and watch C grow six times.

1. Energy before. Before the slab goes in, U₁ = 1/2CV² = 1/2 × 12.5 × 10⁻¹² × 12² = 900 × 10⁻¹² J. The charge on the plates is Q = CV = 150 pC.
2. What stays fixed. With the battery disconnected, no charge can flow on or off, so Q stays the same. The slab raises the capacitance to 6C, so write the energy in terms of Q.

   U₂ = Q²/2(6C) = U₁/6 = (900 × 10⁻¹²)/6 = 150 × 10⁻¹² J

3. So. The energy falls by U₁ - U₂ = 900 × 10⁻¹² - 150 × 10⁻¹² = 750 × 10⁻¹² J, so the answer is 750.

The common slip: 4500. Keeping V fixed, as if the battery were still connected, makes the energy rise to 5400 × 10⁻¹² J and gives a change of 4500, not 750.

The idea to keep. First ask what is held constant. Battery disconnected: Q is fixed and U = Q²/2C drops by a factor εᵣ. Battery connected: V is fixed and U = 1/2CV² rises by εᵣ.

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