A parallel plate capacitor is charged and then disconnected from the source of potential difference. If the…

Capacitance $C$ of a parallel plate air capacitor of plate area $A$ and plate separation $d$ is given by

$C=\frac{{\epsilon }_{0}A}{d} ....\left(1\right)$

When it is charged with the help of a source of potential difference $V$, then suppose the charge on it be $q$.

If the charging battery is disconnected and separation $d$ is increased, then charge $q$ on the capacitor remains constant.

Electric field between the plates is given by

$E=\frac{\sigma }{{\epsilon }_0}=\frac{q}{A{\epsilon }_0} ....\left(2\right)$

As charge remains constant, electric field also remains constant.

Also, from equation $\left(1\right)$, as $d$ increases, capacitance $C$ decreases.

Potential difference across the plates is given by

$V=Ed ....\left(3\right)$

$\therefore$ As $d$ increases, $V$ also increases.