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 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 I and $E_2$ to capacitor II):
Answer: (A) $\dfrac{E_1}{E_2} = \dfrac{9K_1K_2K_3}{(K_1 + K_2 + K_3)(K_2K_3 + K_3K_1 + K_1K_2)}$
With $C = \dfrac{\varepsilon_0A}{d}$:
Figure I: three layers of thickness $d/3$ in series:
$$\frac1{C_1} = \frac{1}{3C}\left(\frac1{K_1} + \frac1{K_2} + \frac1{K_3}\right) \Rightarrow C_1 = \frac{3CK_1K_2K_3}{K_1K_2 + K_2K_3 + K_3K_1}$$
Figure II: three strips of area $A/3$ in parallel:
$$C_2 = \frac C3(K_1 + K_2 + K_3)$$
At the same $V$, $E = \frac12CV^2 \propto C$:
$$\frac{E_1}{E_2} = \frac{9K_1K_2K_3}{(K_1 + K_2 + K_3)(K_1K_2 + K_2K_3 + K_3K_1)}$$
Solution by Sreeraj P, M.Sc Physics