Question

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_1,K_2,K_3,K_4$ arranged as shown in the figure. The effective dielectric constant K will be :

• A. $K=\frac{\left(K_1+K_2\right)\left(K_3+K_4\right)}{2\left(K_1+K_2+K_3+K_4\right)}$
• B. $K=\frac{\left(K_1+K_2\right)\left(K_3+K_4\right)}{\left(K_1+K_2+K_3+K_4\right)}$
• C. $K=\frac{\left(K_1+K_4\right)\left(K_2+K_3\right)}{2\left(K_1+K_2+K_3+K_4\right)}$
• D. $K=\frac{\left(K_1+K_3\right)\left(K_2+K_4\right)}{K_1+K_2+K_3+K_4}$

Answer 1

Answer: (D)

$C_{12}=\frac{C_1{C}_2}{C_1+C_2}=\frac{\frac{k_1{\in }_0\frac{L}{2}\times L}{d/2}.\frac{k_2\left[{\in }_0\frac{L}{2}\times L\right]}{d/2}}{\left(k_1+k_2\right)\left[\frac{{\in }_0.\frac{L}{2}\times L}{d/2}\right]}$
$C_{12}=\frac{k_1{k}_2}{k_1+k_2}\frac{{\in }_0{L}^2}{d}$
in the same way we get, $C_{34}=\frac{k_3{k}_4}{k_3+k_4}\frac{{\in }_0{L}^2}{d}$
$\therefore C_{eq}=C_{12}+C_{34}=\left[\frac{k_1{k}_2}{k_1+k_2}+\frac{k_3{k}_4}{k_3+k_4}\right]\frac{{\in }_0{L}^2}{d}$ ....(i)
Now if $k_{eq}=k,C_{eq}=\frac{k{\in }_0{L}^2}{d}$ .....(ii)
on comparing equation (i) to equation (ii), we get
$k_{eq}=\frac{k_1{k}_2\left(k_3+k_4\right)+k_3{k}_4\left(k_1+k_2\right)}{\left(k_1+k_2\right)\left(k_3+k_4\right)}$
This does not match with any of the options so probably they have assumed the wrong combination
$C_{13}=\frac{k_1{\in }_{0}L\frac{L}{2}}{d/2}+k_3{\in }_0\frac{L.\frac{L}{2}}{d/2}$
$=\left(k_1+k_3\right)\frac{{\in }_0{L}^2}{d}$
$C_{24}=\left(k_2+k_4\right)\frac{{\in }_0{L}^2}{d}$
$C_{eq}=\frac{C_{13}{C}_{24}}{C_{13}{C}_{24}}=\frac{\left(k_1+k_3\right)\left(k_2+k_4\right)}{\left(k_1+k_2+k_3+k_4\right)}\frac{{\in }_0{L}^2}{d}$
$=\frac{k{\in }_0{L}^2}{d}$
$k=\frac{\left(k_1+k_3\right)\left(k_2+k_4\right)}{\left(k_1+k_2+k_3+k_4\right)}$
However this is one of the four options. It must be a "Bonus" logically but of the given options probably they might go with (4)