Physics - Capacitor Question with Solution | TestHub

$C_1$ and $C_2$ are in series.

The equivalent capacitance of $C_1$ and $C_2$ is

$C_{12}=\frac{{\in }_0\frac{A}{2}}{\frac{d}{2K_1}+\frac{d}{2K_2}}=\frac{{\in }_{0}A}{d}\frac{K_1{K}_2}{K_1+K_2}$

$C_{3 }\text{and} C_{4 }\text{are in series}$

$C_{34}=\frac{{\in }_0\frac{A}{2}}{\frac{d}{2K_3}+\frac{d}{2K_4}}=\frac{{\in }_{0}A}{d}\frac{K_3{K}_4}{K_3+K_4}$

$C_{eq}=C_{12}+C_{34}=\frac{{\in }_{0}A}{d}\left(\frac{K_1{K}_2}{K_1+K_2}+\frac{K_3{K}_4}{K_3+K_4}\right)$

$\therefore K_{eq}=\left(\frac{K_1{K}_2}{K_1+K_2}+\frac{K_3.K_4}{K_3+K_4}\right)$