JEE Main 2025 Jan 22 Shift 1, Physics Q19: Capacitance For Geometrical Figures With Dielectric

JEE Main2025Jan 22, Shift 1Physics

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
2 mC and 0.2 J
B
8 mC and 2.0 J
C
4 mC and 0.2 J
D
2 mC and 0.4 J

Solution

With the battery still connected, $V$ stays at 100 V. New capacitance $C^{'} = K C = 80 μ$ F.

Extra charge: $Δ q = (C^{'} - C) V = (K C - C) V = (2 - 1) \times 40 \times 1 0^{- 6} \times 100 = 4 \times 1 0^{- 3}$ C $= 4$ mC.

Change in energy: $Δ U = \frac{1}{2} (C^{'} - C) V^{2} = \frac{1}{2} (K - 1) C V^{2} = \frac{1}{2} \times 40 \times 1 0^{- 6} \times 1 0^{4} = 0.2$ J.

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JEE Main 2025 Jan 22 Shift 1, Physics Q19: Capacitance For Geometrical Figures With Dielectric

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