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 two capacitors is filled with dielectrics of equal thickness and dielectric constants $K_1=1, K_2=2$ and $K_3=3$ as shown. If these two modified capacitors are charged by same potential, the ratio of energy stored $E_1$, (capacitor I) to $E_2$ (capacitor II) is:

$\frac{E_1}{E_2}=\frac{11}{9}$

The correct answer is Option (1) → $\frac{E_1}{E_2}=\frac{11}{9}$

Here, the ratio of electric field is given by -

$\frac{E_1}{E_2}=\frac{(K_1+K_2+K_3)(K_1K_2+K_3K_1+K_2K_3)}{9K_1K_2K_3}$

$=\frac{(1+2+3)(1×2+2×3+3×1)}{9×1×2×3}$

$=\frac{6×11}{9×6}=\frac{11}{9}$