A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :

Solution

Physics sir · 3 steps

The battery is still connected, so V stays fixed and the energy grows in step with C.

1. What stays fixed. The capacitor stays connected to the battery, so the potential difference V across it does not change. Only the capacitance changes when the slab is swapped.
2. Capacitance scales with K. For a parallel plate capacitor filled with a dielectric, C = (Kε₀ A)/d, so C ∝ K. Going from K = 10 to K = 15 makes C grow by a factor of 15/10 = 1.5.

   U₂/U₁ = (1/2C₂V²)/(1/2C₁V²) = K₂/K₁ = 15/10 = 1.5

3. So. The new energy is 1.5 times the old one, an increase of 50%. That is option (a).

The idea to keep. First ask what stays constant. Battery connected means V is fixed, so use U = 1/2CV². Battery removed means Q is fixed, so use U = Q²/2C.

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