A parallel plate air capacitor is charged to a potential difference of $V\, volts$. After disconnecting the charging battery, the distance between the plates of the capacitor is increased using an insulating handle. As a result, the potential difference between the plates
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Capacitance of a parallel plate capacitor

$C=\frac{\varepsilon_{0} A}{d} \quad \ldots \text { (i) }$

Also capacitance = potential difference$/$charge   $......(ii)$

When battery is disconnected and the distance between the plates of the capacitor is increased then capacitance increases and charge remains constant.

since capacitance $=$ potential difference/charge

$\therefore$ Potential difference increases.

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