Physics - Capacitor Question with Solution | TestHub

The given capacitor can be viewed as a series combination of two capacitors $C_1$ and $C_2$ where $$C_1=\frac{{\epsilon }_0{K}_{1}A/2}{d};C_2=\frac{{\epsilon }_0{K}_{2}A/2}{d}$$ where $d$ is the separation between the plates, $\therefore$ The effective capacitance
$$\begin{array}{rl}& C=\frac{C_1{C}_2}{C_1+C_2}\\ & =\frac{\frac{{\epsilon }_0{K}_{1}A/2}{d}\times \frac{{\epsilon }_0{K}_{2}A/2}{d}}{\frac{{\epsilon }_0{K}_{1}A/2}{d}+\frac{{\epsilon }_0{K}_{2}A/2}{d}}=\frac{{\left(\frac{{\epsilon }_{0}A}{2d}\right)}^2{K}_1{K}_2}{\left(\frac{{\epsilon }_{0}A}{2d}\right)\left(K_1+K_2\right)}\\ & =\frac{{\epsilon }_{0}A}{2d}\left(\frac{K_1{K}_2}{K_1+K_2}\right)\end{array}$$