JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2025
- Medium
- MCQ
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
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Answer & Solution
Correct Answer
(A) 4 mC and 0.2 J
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Delta \mathrm{q}=(\mathrm{KC}-\mathrm{C}) \mathrm{V} \\ & =40 \times 10^{-6} \times 100 \\ & =4000 \times 10^{-3}=4 \mathrm{mC} \\ & \Delta \mathrm{U}=\frac{1}{2} \mathrm{C}^{\prime} \mathrm{V}^2-\frac{1}{2} \mathrm{CV}^2=\frac{1}{2}(\mathrm{~K}-1) \mathrm{CV}^2 \\ & =\frac{1}{2} \mathrm{CV}^2(2-1) \\ & =\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \times 40 \times 10^{-6} \times 10000 \\ & =0.2 \mathrm{~J}\end{aligned}\)
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