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
AIPMT 2006, Easy
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Capacitance of a parallel plate capacitor
$C=\frac{\varepsilon_{0} A}{d}$ $...(i)$
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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