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Q. Assertion : If distance between the parallel plates of a capacitor is halved, then its capacitance is doubled
Reason: The capacitance depends on the introduced dielectric

Electrostatic Potential and Capacitance

Solution:

The capacitance of a parallel plates capacitor is given by $C=\frac{\varepsilon_{0} A}{d}$
where $ A$ is area of each plate, $d$ is the distance between plates
$\therefore \frac{C_{1}}{C_{2}}=\frac{\frac{\varepsilon_{0} A}{d}}{\frac{\varepsilon_{0} A}{d /2}}=\frac{1}{2} \therefore C_{2}=2C_{1}$
When dielectric of dielectric constant $k$ is introduced in between the plates then the capacitance $C=\frac{k\varepsilon_{0} A}{d}$ i.e. $C \propto K$
Capacitance $C$ depends on introduced dielectric