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. $\frac{E_1}{E_2}=\frac{\left(K_1+K_2+K_3\right)\left(K_2{K}_3+K_3{K}_1+K_1{K}_2\right)}{9K_1{K}_2{K}_3}$
• 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. $\frac{E_1}{E_2}=\frac{9K_1{K}_2{K}_3}{\left(K_1+K_2+K_3\right)\left(K_2{K}_3+K_3{K}_1+K_1{K}_2\right)}$

Answer 1

Answer: (D)

$<br/>\begin{array}{rl}<br/>\text{ I. }{C}_1& =\frac{3{ϵ}_{0}A{K}_1}{d}\text{ (see fig. I) }\\ <br/>C_2& =\frac{3{\in }_{0}A{K}_2}{d}\\ <br/>C_3& =\frac{3{\in }_{0}A{K}_3}{d}\\ <br/>\frac{1}{C_{eq}}& =\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\\ <br/>\Rightarrow C_{eq}& =\frac{3{\in }_{0}A{K}_1{K}_2{K}_3}{d\left(K_1{K}_2+K_2{K}_3+K_3{K}_1\right)}<br/>\end{array}<br/>$
II. $C_1=\frac{{€}_0{K}_{1}A}{3d}$ (see fig. II)
$<br/>\begin{array}{l}<br/>C_2=\frac{{\in }_0{K}_{2}A}{3d}\\ <br/>C_3=\frac{{\in }_0{K}_{3}A}{3d}\\ <br/>C_{eq}^{\prime }=C_1+C_2+C_3\\ <br/>=\frac{{\in }_{0}A}{3d}\left(K_1+K_2+K_3\right)<br/>\end{array}<br/>$
Now,
$<br/>\frac{E_1}{E_2}=\frac{\frac{1}{2}{C}_{eq}\cdot V^2}{\frac{1}{2}{C}_{eq}^{\prime }{V}^2}=\frac{9K_1{K}_2{K}_3}{\left(K_1+K_2+K_3\right)\left(K_1{K}_2+K_2{K}_3+K_3{K}_1\right)}<br/>$
so $\frac{E_1}{E_2}=\frac{\frac{1}{2}{C}_{eq}\cdot V^2}{\frac{1}{2}{C}_{eq}^{\prime }{V}^2}=\frac{9K_1{K}_2{K}_3}{\left(K_1+K_2+K_3\right)\left(K_1{K}_2+K_2{K}_3+K_3{K}_1\right)}$