JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Capacitance

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 -

Answer: (C)

Solution

The charge on the capacitor before the dielectric is inserted (constant-voltage supply) is

Q_{1} = C V = 40 \times 1 0^{- 6} \times 100 = 4 \times 1 0^{- 3} C = 4 mC .

Inserting a dielectric of constant

K = 2

increases the capacitance to

C^{'} = K C = 80 μ F

, and since the supply still holds the voltage at 100 V,

Q_{2} = C^{'} V = 80 \times 1 0^{- 6} \times 100 = 8 mC .

Extra charge drawn from the supply:

Δ Q = Q_{2} - Q_{1} = 8 - 4 = 4 mC .

The stored energy changes from

U_{1} = \frac{1}{2} C V^{2}

U_{2} = \frac{1}{2} C^{'} V^{2}

Δ U = \frac{1}{2} (K - 1) C V^{2} = \frac{1}{2} \times 40 \times 1 0^{- 6} \times 10 0^{2} = 0.2 J .

The extra charge is 4 mC and the change in electrostatic energy is 0.2 J.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option