Let a parallel plate capacitor be applied a voltage $V$ across its terminals due to which a charge  $Q$ develops across its ends. 

Separation between plates is $d$ and area of plates is $A$.

From gauss's theorem one can show electric field inside the capaciton plates $Q$,

$E=\frac{\sigma }{{ϵ}_o}$, where $\sigma =\frac{Q}{A}$ (charge / Area)

$\Rightarrow E=\frac{Q}{A{ϵ}_o}$

Now, voltage across the plates is related by :

$V=Ed$ [In general , $dV=-\vec{E}.d\vec{r}$ but we take as $\vec{E}$ is uniform]

$\Rightarrow V=\frac{Qd}{A{ϵ}_o}\Rightarrow \frac{Q}{V}=\frac{A{ϵ}_o}{d}$

Now, capacitance is defined by $C=\frac{Q}{V}$, thus we get :-

$C=\frac{A{ϵ}_o}{d}$