A parallel plate capacitor is constructed using three different dielectric materials as shown in figure. The parallel plates across which a potential difference is applied, are of area $A=1 c{m}^2$ and are separated by a distance $d=2 mm$. If $K_1=4, K_2=6$ and $K_3=2$, find capacitance across $P$ and $Q$. (Take ${\epsilon }_0=8.8\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2}$).

The capacitances $C_2$ and $C_3$ are in series, each having area $\frac{A}{2}$ and separation $\frac{d}{2}$.

$$\begin{gathered} C_2=K_2\left(\frac{{\epsilon }_0{\displaystyle \frac{A}{2}}}{{\displaystyle \frac{d}{2}}}\right)=K_2\left(\frac{{\epsilon }_{0}A}{d}\right) \\ \Rightarrow C_3=K_3\left(\frac{{\epsilon }_0{\displaystyle \left(\frac{A}{2}\right)}}{{\displaystyle \frac{d}{2}}}\right)=K_3\left(\frac{{\epsilon }_{0}A}{d}\right) \end{gathered}$$

Equivalent capacitance of $C_2$ and $C_3$ is

$C\text{'}=\frac{C_2{C}_3}{C_2+C_3}=\frac{K_2{K}_3{\left({\displaystyle \frac{{\epsilon }_{0}A}{d}}\right)}^2}{\left(K_2+K_3\right)\left({\displaystyle \frac{{\epsilon }_{0}A}{d}}\right)}=\left(\frac{K_2{K}_3}{K_2+K_3}\right)\frac{{\epsilon }_{0}A}{d}$

Also $C_1=K_1\left(\frac{{\epsilon }_0\left({\displaystyle \frac{A}{2}}\right)}{d}\right)=K_1\left(\frac{{\epsilon }_{0}A}{2d}\right)$

From figure it is clear that $C_1$ is in parallel with $C\text{'}$ (Combination of $C_2$ and $C_3$). Hence net capacitance between $P$ and $Q$.

$$\begin{gathered} C=C_1+C\text{'}=K_1\left(\frac{{\epsilon }_{0}A}{2d}\right)+\left(\frac{K_2{K}_3}{K_2+K_3}\right)\frac{{\epsilon }_{0}A}{d} \\ \Rightarrow C=\frac{{\epsilon }_{0}A}{d}\left(\frac{K_1}{2}+\frac{K_2{K}_3}{\left(K_2+K_3\right)}\right) \end{gathered}$$

Substituting given values

$$\begin{gathered} C=\frac{8.8\times {10}^{-12}\times \left(1\times {10}^{-4}\right)}{2\times {10}^{-3}}\left(\frac{4}{2}+\frac{6\times 2}{6+2}\right) \\ \Rightarrow C=0.44\times {10}^{-12}\left(\frac{7}{2}\right) \\ \Rightarrow C=1.54\times {10}^{-12}F=1.54 pF \end{gathered}$$