A parallel-plate capacitor of capacitance 40μ F is connected to a 100… | JEE Main 2025 (22 Jan, Shift 1)

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Dielectrics

A parallel-plate capacitor of capacitance $40\mu\text{ F}$ 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

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Answer: (b)

Supply stays connected, so $V = 100$ V is fixed and $C' = KC = 80\ \mu\text{F}$ .
Extra charge $= (C' - C)V = 40 \times 10^{-6} \times 100 = 4 \times 10^{-3}$ C $= 4$ mC.
Change in energy $= \frac{1}{2}(C' - C)V^2 = \frac{1}{2} \times 40 \times 10^{-6} \times 10^4 = 0.2$ J.
So the answer is (b) 4 mC and 0.2 J.

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