Question

The equivalent capacitance of the arrangement shown in figure, if A is the area of each plate, is:

• A. $C=\frac{{\epsilon }_{0}A}{d}[\frac{K_1}{2}+\frac{K_2+K_3}{K_2{K}_3}]$
• B. $C=\frac{{\epsilon }_{0}A}{d}[\frac{K_2}{2}+\frac{K_2{K}_3}{K_2+K_3}]$
• C. $C=\frac{{\epsilon }_{0}A}{d}[\frac{K_3}{2}+\frac{K_2{K}_3}{K_2+K_3}]$
• D. $C=\frac{{\epsilon }_{0}A}{d}[\frac{K_1}{2}+\frac{K_2{K}_3}{K_2+K_3}]$

Answer 1

The correct option is A $C=\frac{{\epsilon }_{0}A}{d}[\frac{K_1}{2}+\frac{K_2+K_3}{K_2{K}_3}]$

$\begin{array}{c}{C}_1=\frac{K_1{\epsilon }_{0}A}{2d}\\ C_2=\frac{K_2{\epsilon }_{0}A}{d}\\ C_3=\frac{K_3{\epsilon }_{0}A}{d}\\ C_{eq}=\frac{C_2{C}_3}{C_2+C_3}+C_1\\ =\frac{K_2{K}_3{\epsilon }_{0}A}{\left(K_2+K_3\right)d}+\frac{K_1{\epsilon }_{0}A}{2d}\\ Hence,\\ optionAiscorrectanswer.\end{array}$