JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 18)
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
8 mC and 2.0 J
4 mC and 0.2 J
2 mC and 0.2 J
2 mC and 0.4 J
Explanation
Given, $$K = 2,\,C = 40\mu F,\,V = 100\,V$$
$$\Delta q = (KC)V - CV$$ (As $$q = CV$$)
$$ = (K - 1)CV$$
$$ = (2 - 1) \times 40 \times {10^{ - 6}} \times 100 = 4 \times {10^{ - 3}}C$$
$$\Delta q = 4\,mC$$
$$\Delta u = {1 \over 2}C'{V^2} - {1 \over 2}C{V^2}$$
$$ = {1 \over 2}{V^2}(C' - C) = {1 \over 2}{V^2}(KC - C)$$
$$ = {1 \over 2}{V^2}C(K - 1) = {1 \over 2} \times {10^4} \times 40 \times {10^{ - 6}}(2 - 1)$$
$$ = \Delta u = 2 \times {10^{ - 1}} = 0.2\,J$$
$$ \Rightarrow \Delta u = 0.2\,J$$
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