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(B) Let the initial capacitance be $C$ and the potential difference provided by the battery be $V$. The initial charge on the capacitor is $q = CV$.Th

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decreases by a factor of $2$

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(B) Let the initial capacitance be $C$ and the potential difference provided by the battery be $V$. The initial charge on the capacitor is $q = CV$.The initial electrostatic energy stored in the capacitor is $U_i = \frac{1}{2} CV^2$.When the battery is removed and an identical uncharged capacitor of capacitance $C$ is connected in parallel,the total charge $q$ is shared between the two capacitors. Since the capacitors are identical,the charge on each capacitor becomes $q' = q/2 = CV/2$.The common potential difference across the parallel combination is $V_c = \frac{q}{C_{eq}} = \frac{CV}{C + C} = \frac{V}{2}$.The final electrostatic energy of the system is the sum of the energy stored in both capacitors:$U_f = \frac{1}{2} C V_c^2 + \frac{1}{2} C V_c^2 = C V_c^2$.Substituting $V_c = V/2$:$U_f = C \left(\frac{V}{2}\right)^2 = C \left(\frac{V^2}{4}\right) = \frac{1}{4} CV^2$.Comparing the final energy with the initial energy:$U_f = \frac{1}{2} \left(\frac{1}{2} CV^2\right) = \frac{1}{2} U_i$.Thus,the total electrostatic energy of the resulting system decreases by a factor of $2$.

$A$ capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:

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