To solve this problem, we need to analyze the change in electrostatic energy when a charged capacitor is connected in parallel with an identical uncharged capacitor.$\\$Let's denote:$\\$• $C$ as the capacitance of each capacitor.$\\$• $V_0$ as the initial voltage across the charged capacitor.$\\$The initial energy stored in the charged capacitor is given by:$\\ U_{initial} = \frac{1}{2} C V_0^2 \\$When the charged capacitor is connected in parallel with the uncharged capacitor, the total capacitance becomes:$\\ C_{total} = C + C = 2C \\$The charge on the initially charged capacitor is $Q = C V_0. \\$This charge is now distributed over the total capacitance $C_{total}. \\$The new voltage across the capacitors is:$\\ V_{new} = \frac{Q}{C_{total}} = \frac{C V_0}{2C} = \frac{V_0}{2} \\$The new energy stored in the system is:$\\ U_{new} = \frac{1}{2} C_{total} V_{new}^2 = \frac{1}{2} \cdot 2C \cdot \left(\frac{V_0}{2}\right)^2 \\$Simplifying the expression:$\\ U_{new} = C \cdot \frac{V_0^2}{4} = \frac{1}{2} C \cdot \frac{V_0^2}{2} \\ U_{new} = \frac{1}{4} C V_0^2 \\$Comparing $U_{new}$ with $U_{initial}: \\ \frac{U_{new}}{U_{initial}} = \frac{\frac{1}{4} C V_0^2}{\frac{1}{2} C V_0^2} = \frac{1}{2} \\$Thus, the total electrostatic energy of the system decreases by a factor of 2.$\\$This corresponds to Option 1.