A

Step 1: Analyze the capacitance of a parallel plate capacitor. The formula for the capacitance (C) is given by:
$$ C = \frac{\epsilon_0 A}{d} $$
where $\epsilon_0$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the separation between the plates. Therefore, as the separation (d) increases, the capacitance decreases.
Step 2: Investigate the potential difference across the plates of a parallel plate capacitor. The potential difference is dependent on the charge (Q) on the plates and the capacitance (C) and is given by:
$$ V = \frac{Q}{C} $$
This shows that the potential difference is controlled by Q and C but is independent of the material of the plates.
Step 3: Consider the charge on the plates of a capacitor. Inserting a dielectric slab increases the capacitance which in turn increases the charge (Q) on the plates if connected to a power source. However, if it’s isolated, the charge remains constant.
Step 4: Assess the electrostatic potential energy of a capacitor given by:
$$ U = \frac{1}{2} C V^2 $$
In a scenario where the capacitor is connected to a power source, the potential energy will change as well, but if isolated once charged, the potential energy will remain constant as charge remains fixed.
Therefore, the correct matches are:
(i) - [A], (ii) - [B], (iii) - [C], (iv) - [D]. Hence, the answer is A.