Find the effective capacitance of the following arrangement of four identical metallic plates between $P$ and $Q$. The area of each plate is $A$ and separation between successive plates is $d$.

The equivalent of Arrangement 1 is shown. Let $C$ be capacitance of each capacitor.

The net capacitance of $y$ and $z$ connected in series is

$C_{yz}=\frac{C}{2}$

Now $C_{yz}$ and capacitor $x$ are connected in parallel. Therefore net capacitance between $P$ and $Q$,

$C_{PQ}=C+\frac{C}{2}=\frac{3}{2}C=\frac{3}{2}\left(\frac{{\epsilon }_{0}A}{d}\right)$

The equivalent of Arrangement 2 is shown. 

The net capacitance of $x$ and  $y$ connected in parallel is

$C_{xy}=2C$

Now $C_{xy}$ and $z$ are connected in series. Therefore net capacitance between $P$ and $Q$,

$C_{PQ}=\frac{C{C}_{xy}}{C+C_{xy}}=\frac{C\left(2C\right)}{C+2C}=\frac{2}{3}C=\frac{2}{3}\left(\frac{{\epsilon }_{0}A}{d}\right)$