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:

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

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

$C = \frac{\epsilon_0 A}{d}$

$\text{Same potential} \Rightarrow E = \frac{1}{2}CV^2 \Rightarrow E \propto C$

$\text{Capacitor I (series dielectrics, each thickness } \frac{d}{3})$

$\frac{1}{C_1} = \frac{d/3}{\epsilon_0 K_1 A} + \frac{d/3}{\epsilon_0 K_2 A} + \frac{d/3}{\epsilon_0 K_3 A}$

$= \frac{d}{3\epsilon_0 A}\left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3}\right)$

$= \frac{d}{3\epsilon_0 A}\cdot \frac{11}{6} = \frac{11d}{18\epsilon_0 A}$

$C_1 = \frac{18}{11}\cdot \frac{\epsilon_0 A}{d} = \frac{18}{11}C$

$\text{Capacitor II (parallel dielectrics, each area } \frac{A}{3})$

$C_2 = \frac{\epsilon_0}{d}\left(K_1\frac{A}{3} + K_2\frac{A}{3} + K_3\frac{A}{3}\right)$

$= \frac{\epsilon_0 A}{3d}(1+2+3) = 2C$

$\frac{E_1}{E_2} = \frac{C_1}{C_2} = \frac{18/11}{2} = \frac{9}{11}$

The required ratio is $\frac{9}{11}$.