Question

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. $\frac{E_{1}}{E_{2}} = \frac{9K_{1}K_{2}K_{3}}{\left(K_{1} +K_{2}+K_{3}\right)\left(K_{2}K_{3} + K_{3}K_{1} +K_{1}K_{2}\right)}$
• B. $\frac{E_{1}}{E_{2}} = \frac{K_{1}K_{2}K_{3}}{\left(K_{1} +K_{2}+K_{3}\right)\left(K_{2}K_{3} + K_{3}K_{1} +K_{1}K_{2}\right)}$
• C. $\frac{E_{1}}{E_{2}} = \frac{\left(K_{1} +K_{2}+K_{3}\right)\left(K_{2}K_{3} + K_{3}K_{1} +K_{1}K_{2}\right)}{K_{1}K_{2}K_{3}}$
• D. $\frac{E_{1}}{E_{2}} = \frac{\left(K_{1} +K_{2}+K_{3}\right)\left(K_{2}K_{3} + K_{3}K_{1} +K_{1}K_{2}\right)}{9K_{1}K_{2}K_{3}}$

Answer 1

Answer: (D)

$\frac{E_{1}}{E_{2}} = \frac{\left(K_{1} +K_{2}+K_{3}\right)\left(K_{2}K_{3} + K_{3}K_{1} +K_{1}K_{2}\right)}{9K_{1}K_{2}K_{3}}$

Answer 2

$C_{1} = \frac{3\varepsilon_{0}AK_{1}}{d}$