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
Solution
<p>Given, $K = 2,\,C = 40\mu F,\,V = 100\,V$</p>
<p>$\Delta q = (KC)V - CV$ (As $q = CV$)</p>
<p>$= (K - 1)CV$</p>
<p>$= (2 - 1) \times 40 \times {10^{ - 6}} \times 100 = 4 \times {10^{ - 3}}C$</p>
<p>$\Delta q = 4\,mC$</p>
<p>$\Delta u = {1 \over 2}C'{V^2} - {1 \over 2}C{V^2}$</p>
<p>$= {1 \over 2}{V^2}(C' - C) = {1 \over 2}{V^2}(KC - C)$</p>
<p>$$ = {1 \over 2}{V^2}C(K - 1) = {1 \over 2} \times {10^4} \times 40 \times {10^{ - 6}}(2 - 1)$$</p>
<p>$= \Delta u = 2 \times {10^{ - 1}} = 0.2\,J$</p>
<p>$\Rightarrow \Delta u = 0.2\,J$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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