JEE Main 2025 · 22 January · morning · Single correct · +4 / −1
A parallel-plate capacitor of capacitance 40 μ F is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K=2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are
- (a) 8 mC and 2.0 J
- (b) 4 mC and 0.2 J
- (c) 2 mC and 0.2 J
- (d) 2 mC and 0.4 J
Answer and solution
Answer (b) 4 mC and 0.2 J
Solution
Physics sir · 3 steps
The supply stays connected, so V is fixed: doubling C doubles both the charge and the energy.
- Before the dielectric. With C = 40 μF at V = 100 V, the charge is Q₁ = CV = 4 × 10⁻³ C = 4 mC. The energy is U₁ = 1/2CV² = 1/2 × 40 × 10⁻⁶ × 10⁴ = 0.2 J.
- After the dielectric. Filling the gap with K = 2 makes C' = KC = 80 μF. The supply holds V at 100 V, so Q₂ = 8 mC and U₂ = 0.4 J.
Δ Q = (K-1)CV = 4 mC, Δ U = 1/2(K-1)CV² = 0.2 J
- So. The supply pushes in an extra 4 mC and the stored energy rises by 0.2 J, which is option (b).
The idea to keep. First ask whether the battery is still connected. If it is, V stays fixed and Q and U both scale with K. If it is disconnected, Q stays fixed and U drops to U/K.
Answer from the NTA final answer key. Solution by PrepWiser.