Verified Solution • 2025–26 Session moderate

QJEEM25Ph41aa: JEE Main 2025 Physics - Class 12 JEEM25Ph41aa Solution:   A parallel-plate capacitor of capacitance is connected…

 

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

Multiple Choice Options

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

Step-by-Step Verified Solution

amp;Δq=(KC−C)Vamp;=40×10−6×100amp;=4000×10−3=4mCamp;ΔU=12C′V2−12CV2=12( K−1)CV2amp;=12CV2(2−1)amp;=12CV2=12×40×10−6×10000amp;=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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