To solve this problem, we need to analyze the effects of moving the plates of a parallel plate capacitor closer while connected to a battery.$\\$Let's consider the following:$\\$• The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon A}{d}, \\$ where $\varepsilon$ is the permittivity of the medium, $A$ is the area of the plates, and $d$ is the separation between the plates.$\\$• The charge $Q$ stored in the capacitor is $Q = CV, \\$ where $V$ is the voltage across the capacitor.$\\$• The energy $U$ stored in the capacitor is $U = \frac{1}{2} CV^2. \\$Now, let's analyze each statement:$\\$A. The charge stored in it, increases.$\\$• Since the capacitor is connected to a battery, the voltage $V$ remains constant.$\\$• If the plates are moved closer, $d$ decreases, which increases the capacitance $C. \\$• Therefore, the charge $Q = CV$ increases.$\\$• This statement is true.$\\$B. The energy stored in it, decreases.$\\$• With $C$ increasing and $V$ constant, the energy $U = \frac{1}{2} CV^2$ increases.$\\$• This statement is false.$\\$C. Its capacitance increases.$\\$• As $d$ decreases, $C = \frac{\varepsilon A}{d}$ increases.$\\$• This statement is true.$\\$D. The ratio of charge to its potential remains the same.$\\$• The ratio $\frac{Q}{V} = C$ changes because $C$ increases.$\\$• This statement is false.$\\$E. The product of charge and voltage increases.$\\$• The product $QV = CV^2$ increases as $C$ increases.$\\$• This statement is true.$\\$Based on the analysis, the correct option is:$\\$Option 4: A, C and E only