Explanation

(a) The capacitance (C) of a capacity is defined as the ratio of the charge (q) given to the rise in the potential (V) of the conductor. When the plates are moved apart, the charge remains constant, hence
$q = CV = constant\ \ \ \ \ldots\ldots\ldots..(i)$
Capacitance (C) of a parallel plate capacitor is given by
$C = \frac{\varepsilon_{0}\ A}{d}\ \ \ldots\ldots\ldots..(ii)$
Where A is area of plates and d the separation between them.
From Eqs. (i) and (ii), we have
$q = \frac{\varepsilon_{0}\ A}{d}.V = constant\ $
When plates are moved apart, $d$ increases, hence value of C decreases and in order that charge (q) remains constant V increases.