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JEE Main 202522 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Morning Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2025 · 22 January · Morning Shift · Physics · Capacitance

A parallel-plate capacitor of capacitance 40 μ ~F is connected to a...

A parallel-plate capacitor of capacitance 40μ F40 \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 K=2\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

Options

  1. A4 mC and 0.2 J
  2. B8 mC and 2.0 J
  3. C2 mC and 0.4 J
  4. D2 mC and 0.2 J

Correct answer

A. 4 mC and 0.2 J

Step-by-step solution

Δq=(KC−C)V=40×10−6×100=4000×10−3=4mCΔU=12C′V2−12CV2=12( K−1)CV2=12CV2(2−1)=12CV2=12×40×10−6×10000=0.2 J\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}
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