Q69 / 75 · JEE Main 2025 (22 Jan Shift 1) · Physics

A parallel-plate capacitor of capacitance $40 \mu \mathrm{~F}$ is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant $\mathrm{K}=2$. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are

  • Correct answer
    A
    4 mC and 0.2 J
  • B
    8 mC and 2.0 J
  • C
    2 mC and 0.4 J
  • D
    2 mC and 0.2 J

Solution

$\begin{aligned} & \Delta \mathrm{q}=(\mathrm{KC}-\mathrm{C}) \mathrm{V} \\ & =40 \times 10^{-6} \times 100 \\ & =4000 \times 10^{-3}=4 \mathrm{mC} \\ & \Delta \mathrm{U}=\frac{1}{2} \mathrm{C}^{\prime} \mathrm{V}^2-\frac{1}{2} \mathrm{CV}^2=\frac{1}{2}(\mathrm{~K}-1) \mathrm{CV}^2 \\ & =\frac{1}{2} \mathrm{CV}^2(2-1) \\ & =\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \times 40 \times 10^{-6} \times 10000 \\ & =0.2 \mathrm{~J}\end{aligned}$

Hence, the correct answer is option: 'A'.

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