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 resulting system

When the capacitor is charged by a battery of potential V, then energy stored in the capacitor,

$U_i=\frac{1}{2}C{V}^2 . . . . \left(i\right)$

When the battery is removed and another  identical  uncharged capacitor is connected in parallel Common potential,

$V^{\text{'}}=\frac{CV}{C+C}=\frac{V}{2}$

$\therefore \text{ Then the energy stored in the capacitor, }$

$U_f=\frac{1}{2}\left(2C\right){\left(\frac{V}{2}\right)}^2=\frac{1}{4}C{V}^2 . . . . \left(ii\right)$

$\therefore \text{ From eqns. (i) and (ii) }$

$U_f=\frac{U_i}{2}$

that means the total electrostatic energy of resulting system will decreases by a factor of 2.