Physics - Capacitor Question with Solution | TestHub

Here the capacitance, $C=\frac{Ak{ϵ}_0}{d}$ $C_{K_1},C_{K_2},C_{K_3}$ are in parallel combination and they are connected in series with ${\mathrm{C}}_{{\mathrm{K}}_4}$Here, ${\mathrm{C}}_{{\mathrm{K}}_1}=\frac{(\mathrm{A}/3){\mathrm{K}}_1{ϵ}_0}{\mathrm{d}/2}=\frac{2{\mathrm{K}}_1}{3}\mathrm{C}$, ${\mathrm{C}}_{{\mathrm{k}}_2}=\frac{(\mathrm{A}/3){\mathrm{k}}_2{ϵ}_0}{\mathrm{d}/2}=\frac{2{\mathrm{k}}_2}{3}\mathrm{C}$ $C_{K_3}=\frac{(\mathrm{A}/3){\mathrm{k}}_3{ϵ}_0}{\mathrm{d}/2}\frac{2{\mathrm{k}}_3\mathrm{C}}{3}\mathrm{C}$ ${\mathrm{C}}_{{\mathrm{K}}_4}=\frac{(\mathrm{A}){\mathrm{k}}_4{ϵ}_0}{\mathrm{d}/2}=2{\mathrm{k}}_4\mathrm{C}$ Now the equivalent capacitance for the combination of four capacitors is $\begin{array}{rl}& \frac{1}{C_{cq}}=\frac{1}{\left(C_{k_1}+C_{k_2}+C_{k_3}\right)}+\frac{1}{C_{k_4}}\\ & C_{eq}=kC\end{array}$ $\frac{1}{kC}=\frac{3}{2C}\left[\frac{1}{k_1+k_2+k_3}\right]+\frac{1}{2k_{4}C}$ or $\frac{1}{k}=\frac{3}{2\left(k_1+k_2+k_3\right)}+\frac{1}{2\left(k_4\right)}$ or $k=\frac{2\left(k_1+k_2+k_3\right)}{3}+2\left(k_4\right)$