Two square plates $(\mathrm{l}\times \mathrm{l})$ and dielectric $\left(\frac{\mathrm{l}}{2}\times \frac{\mathrm{t}}{2}\times \mathrm{l}\right)$ arranged as shown in figure. Find the equivalent capacitance of the structure.

We can make equivalent circuit of given system in two ways as in Figs. (a) and (b). But (b) is wrong and (a) is correct.

Here ${\mathrm{C}}_1=\frac{{\mathrm{\epsilon }}_0\mathrm{A}/2}{\mathrm{t}/2}=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{t}}$

${\mathrm{C}}_2=\frac{{\mathrm{K\epsilon }}_0\mathrm{A}/2}{\mathrm{t}/2}=\frac{{\mathrm{K\epsilon }}_0\mathrm{A}}{\mathrm{t}} \text{ and }{\mathrm{C}}_3=\frac{{\mathrm{\epsilon }}_0\mathrm{A}/2}{\mathrm{t}}=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{2\mathrm{t}}$

Now, ${\mathrm{C}}_{\mathrm{eq}}=\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}+{\mathrm{C}}_3$

$=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{t}}\left[\frac{3\mathrm{K}+1}{2(\mathrm{K}+1)}\right]$