A parallel-plate capacitor of capacitance 40 μ 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 -

Physics · Capacitors and R-C Circuits

JEE Main 2025 — 22 January, Morning Shift — Question 59

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 -

$\Delta \mathrm{q}=(\mathrm{KC}-\mathrm{C}) \mathrm{V}$

$=40 \times 10^{-6} \times 100$ $=4000 \times 10^{-3}=4 \mathrm{mC}$

$=\frac{1}{2} \mathrm{C}^{\prime} \mathrm{V}^{2}-\frac{1}{2} \mathrm{CV}^{2}$

$=\frac{1}{2} \mathrm{CV}^{2}(2-1)$

$=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 40 \times 10^{-6} \times 10000$ $=0.2 \mathrm{~J}$

Answer key and solution verified before publishing.

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