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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 capacitor is filled as shown in fig. $I$, and the second one is filled as shown in fig. $II$. 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 $(I)$ and $E_2$ to capacitor $(II)$)
$\frac{{{E_1}}}{{{E_2}}} = \frac{{{K_1}{K_2}{K_3}}}{{\left( {{K_1} + {K_2} + {K_3}} \right)\,\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}$
$\frac{{{E_1}}}{{{E_2}}} = \frac{{9{K_1}{K_2}{K_3}}}{{\left( {{K_1} + {K_2} + {K_3}} \right)\,\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}$
$\frac{{{E_1}}}{{{E_2}}} = \frac{{\left( {{K_1} + {K_2} + {K_3}} \right)\,\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}{{9{K_1}{K_2}{K_3}}}$
$\frac{{{E_1}}}{{{E_2}}} = \frac{{\left( {{K_1} + {K_2} + {K_3}} \right)\,\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}{{{K_1}{K_2}{K_3}}}$
Solution
$C_{1}=\frac{3 \varepsilon_{0} A K_{1}}{d}$
$C_{2}=\frac{3 \varepsilon_{0} A K_{2}}{d}$
$C_{3}=\frac{3 \varepsilon_{0} A K_{3}}{d}$
$\frac{1}{C_{e q}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$
$\Rightarrow \quad C_{e q}=\frac{3 \varepsilon_{0} A K_{1} K_{2} K_{3}}{d\left(K_{1} K_{2}+K_{2} K_{3}+K_{3} K_{1}\right)}………(i)$
$C_{1}=\frac{\varepsilon_{0} K_{1} A}{3 d}$
$C_{2}=\frac{\varepsilon_{0} K_{2} A}{3 d}$
$C_{3}=\frac{\varepsilon_{0} K_{3} A}{3 d}$
$C_{e q}^{\prime} =C_{1}+C_{2}+C_{3} $
$ = \frac{{{\varepsilon _0}A}}{{3d}}({K_1} + {K_2} + {K_3})………(ii)$
Now,
$\frac{{{E_1}}}{{{E_2}}} = \frac{{\frac{1}{2}{C_{eq}} \cdot {V^2}}}{{\frac{1}{2}C_{eq}^\prime {V^2}}} = \frac{{9{K_1}{K_2}{K_3}}}{{\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_1}{K_2} + {K_2}{K_3} + {K_3}{K_1}} \right)}}$
