Find the capacitance of the system shown in figure.

The given capacitor is equivalent too 

${\mathrm{C}}_1=\frac{{\mathrm{\epsilon }}_0\frac{\mathrm{A}}{2}}{\frac{\mathrm{d}}{2}}=\frac{{\mathrm{A\epsilon }}_0}{\mathrm{d}}$

${\mathrm{C}}_2=\frac{2{\mathrm{\epsilon }}_0\frac{\mathrm{A}}{2}}{\frac{\mathrm{d}}{2}}=\frac{2{\mathrm{A\epsilon }}_0}{\mathrm{d}}$

${\mathrm{C}}_3=\frac{3{\mathrm{\epsilon }}_0\frac{\mathrm{A}}{2}}{\frac{\mathrm{d}}{2}}=\frac{3{\mathrm{A\epsilon }}_0}{\mathrm{d}}$

${\mathrm{C}}_4=\frac{4{\mathrm{\epsilon }}_0\frac{\mathrm{A}}{2}}{\frac{\mathrm{d}}{2}}=\frac{4{\mathrm{A\epsilon }}_0}{\mathrm{d}}$

${\mathrm{C}}_{\mathrm{eq}}=\frac{{\mathrm{C}}_1{\mathrm{C}}_3}{{\mathrm{C}}_1+{\mathrm{C}}_3}+\frac{{\mathrm{C}}_2{\mathrm{C}}_4}{{\mathrm{C}}_2+{\mathrm{C}}_4}$

${\mathrm{C}}_{\mathrm{eq}}=\left(\frac{1\times 3}{1+3}\right)\frac{{\mathrm{A\epsilon }}_0}{\mathrm{d}}+\left(\frac{2\times 4}{2+4}\right)\frac{{\mathrm{A\epsilon }}_0}{\mathrm{d}}$

${\mathrm{C}}_{\mathrm{eq}}=\frac{3}{4}\frac{{\mathrm{A\epsilon }}_0}{\mathrm{d}}+\frac{4}{3}\frac{{\mathrm{A\epsilon }}_0}{\mathrm{d}}=\frac{25}{24}\frac{{\mathrm{A\epsilon }}_0}{\mathrm{d}}$