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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 $(II)$) 
  • A$\frac{{{E_1}}}{{{E_2}}} = \frac{{{K_1}{K_2}{K_3}}}{{\left( {{K_1} + {K_2} + {K_3}} \right)\,\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}$
  • B$\frac{{{E_1}}}{{{E_2}}} = \frac{{9{K_1}{K_2}{K_3}}}{{\left( {{K_1} + {K_2} + {K_3}} \right)\,\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}$
  • C$\frac{{{E_1}}}{{{E_2}}} = \frac{{\left( {{K_1} + {K_2} + {K_3}} \right)\,\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}{{9{K_1}{K_2}{K_3}}}$
  • D$\frac{{{E_1}}}{{{E_2}}} = \frac{{\left( {{K_1} + {K_2} + {K_3}} \right)\,\left( {{K_2}{K_3} + {K_3}{K_1} + {K_1}{K_2}} \right)}}{{{K_1}{K_2}{K_3}}}$
JEE MAIN 2019, Diffcult
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