A parallel-plate capacitor of area A , plate separation d and...

Here the capacitance, (C=\frac{A k \epsilon_0}{d}) $C_{K_1}, C_{K_2}, C_{K_3}$ are in parallel combination and they are connected in series with (\mathrm{C}_{\mathrm{K}_4}) Here, (\mathrm{C}_{\mathrm{K}_1}=\frac{(\mathrm{A} / 3) \mathrm{K}_1 \epsilon_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{~K}_1}{3} \mathrm{C}), (\mathrm{C}_{\mathrm{k}_2}=\frac{(\mathrm{A} / 3) \mathrm{k}_2 \mathrm{\epsilon}_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_2}{3} \mathrm{C}) (C_{K_3}=\frac{(\mathrm{A} / 3) \mathrm{k}_3 \epsilon_0}{\mathrm{~d} / 2} \frac{2 \mathrm{k}_3 \mathrm{C}}{3} \mathrm{C}) (\mathrm{C}_{\mathrm{K}_4}=\frac{(\mathrm{A}) \mathrm{k}_4 \epsilon_0}{\mathrm{~d} / 2}=2 \mathrm{k}_4 \mathrm{C}) Now the equivalent capacitance for the combination of four capacitors is $$\begin{aligned} & \frac{1}{C_{c q}}=\frac{1}{\left(C_{k_1}+C_{k_2}+C_{k_3}\right)}+\frac{1}{C_{k_4}} \\ & C_{e q}=k C \end{aligned}$$) (\frac{1}{k C}=\frac{3}{2 C}$\left[\frac{1}{k_1+k_2+k_3}\right]+\frac{1}{2 k_4 C})$ or $(\frac{1}{k}=\frac{3}{2\left(k_1+k_2+k_3\right)}+\frac{1}{2\left(k_4\right)}) or (k=\frac{2\left(k_1+k_2+k_3\right)}{3}+2\left(k_4\right))