Question 106 of 180

Two identical parallel plate capacitors, of capacitance $C$ each, have plates of area $A,$ separated by a distance $d$ . The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants ${K}_{1},{K}_{2}$ and ${K}_{3}$ . The first capaciitor is filled as shown in figure $I,$ and the second one is filled as shown in figure $II.$
If these two modified capacitors are charged by the same potential $V$ , the ratio of the energy stored in the two, would be $({E}_{1}$ refers to capacitor $\left(I\right)$ and ${E}_{2}$ to capacitor $(II)$ ):

A$\dfrac{{E}_{1}}{{E}_{2}}=\dfrac{9K{}_{1}K{}_{2}K{}_{3}}{(K{}_{1}K{}_{2}K{}_{3})(K{}_{2}K{}_{3}+K{}_{3}K{}_{1}K{}_{1}+K{}_{1}K{}_{2})}$B$\dfrac{{E}_{1}}{{E}_{2}}=\dfrac{(K{}_{1}K{}_{2}K{}_{3})(K{}_{2}K{}_{3}+K{}_{3}K{}_{1}K{}_{1}+K{}_{1}K{}_{2})}{{K}_{1}{K}_{2}{K}_{3}}$C$\dfrac{{E}_{1}}{{E}_{2}}=\dfrac{(K{}_{1}K{}_{2}K{}_{3})(K{}_{2}K{}_{3}+K{}_{3}K{}_{1}K{}_{1}+K{}_{1}K{}_{2})}{9{K}_{1}{K}_{2}{K}_{3}}$D$\dfrac{{E}_{1}}{{E}_{2}}=\dfrac{{K}_{1}{K}_{2}{K}_{3}}{(K{}_{1}K{}_{2}K{}_{3})(K{}_{2}K{}_{3}+K{}_{3}K{}_{1}K{}_{1}+K{}_{1}K{}_{2})}$

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Answer: (A) $\dfrac{{E}_{1}}{{E}_{2}}=\dfrac{9K{}_{1}K{}_{2}K{}_{3}}{(K{}_{1}K{}_{2}K{}_{3})(K{}_{2}K{}_{3}+K{}_{3}K{}_{1}K{}_{1}+K{}_{1}K{}_{2})}$

Energy stored in a capacitor $E=\dfrac{1}{2}C{V}^{2}$
$\Rightarrow \dfrac{{E}_{1}}{{E}_{2}}=\dfrac{{C}_{1}}{{C}_{2}}$

${C}_{1}={\left(\dfrac{d}{3 {\epsilon }_{0}{k}_{1}A}+\dfrac{d}{3{\epsilon }_{0}A{k}_{2}}+\dfrac{d}{3{\epsilon }_{0}A{k}_{3}}\right)}^{-1}$
$={\left(\dfrac{d}{3A{\epsilon }_{0}}\right)}^{-1}{\left(\dfrac{1}{{k}_{1}}+\dfrac{1}{{k}_{2}}+\dfrac{1}{{k}_{3}}\right)}^{-1}$
$={\left(\dfrac{d}{3A{\epsilon }_{0}}\right)}^{-1}{\left(\dfrac{{k}_{1}{k}_{2}+{k}_{2}{k}_{3}+{k}_{3}{k}_{1}}{{k}_{1}{k}_{2}{k}_{3}}\right)}^{-1}$

${C}_{1}=\dfrac{3A{\epsilon }_{0}}{d}\left(\dfrac{{k}_{1}{k}_{2}{k}_{3}}{{k}_{1}{k}_{2}+{k}_{2}{k}_{3}+{k}_{3}{k}_{1}}\right)$
${C}_{2}=\dfrac{{\epsilon }_{0}\left(\dfrac{A}{3}\right){k}_{1}}{d}+\dfrac{{\epsilon }_{0}\left(\dfrac{A}{3}\right){k}_{2}}{d}+\dfrac{{\epsilon }_{0}\left(\dfrac{A}{3}\right){k}_{3}}{d}$
$∴\dfrac{{C}_{1}}{{C}_{2}}=\dfrac{{E}_{1}}{{E}_{2}}=\dfrac{9 {k}_{1}{k}_{2}{k}_{3}}{\left({k}_{1}+{k}_{2}+{k}_{3}\right)\left({k}_{1}{k}_{2}+{k}_{2}{k}_{3}+{k}_{3}{k}_{1}\right)}$