For parallel plates where the separation $d$ is much smaller than the dimensions of the plates (diameter for circular plates), and the potential difference between the plates is $V$, the electric field $E$ is given by
Where the electric field $E$ is directed from the + plate to the – plate.
Capacitance, $C$ is electrically defined as the amount of charge $q$ on the plates per volt across the plates, or
In terms of the physical characteristics of a capacitor, the capacitance is given by
Equating the last two equations gives us
Where $ε$ is the electrical per permittivity of the medium between the plates.
Substituting $V$ from the last equation for $V$ in the first equation gives us
The last equation shows that the electric field strength between the plates does not depend on the plate separation. Now returning to the first equation expressed in terms of potential difference we have
Since $E$ is constant, increasing the separation increases the electrical potential difference. This makes sense when you consider the following definition of potential difference, or voltage:
The potential difference, $V$, is defined as the work (joules) per unit charge $q$ (coulomb) required to move the charge between the points.
The force on a charge $q$ between the plates is $qE$. The work required to move the charge from one plate to another is
The work per unit charge is
Hope this helps.