JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Capacitance

The charged capacitor stores initial energy $U_i=\frac{1}{2}C{V}^2=\frac{1}{2}(600\times 10^{-12})(200{)}^2=12\mu \text{J}$. When it is connected across an identical uncharged capacitor the charge is conserved while the two equal capacitances share it, so the common potential falls to $V^{\prime }=\frac{Q}{2C}=\frac{V}{2}=100\text{V}$. The final energy stored in the parallel pair is $U_f=\frac{1}{2}(2C)V^{\prime 2}=\frac{1}{2}(1200\times 10^{-12})(100{)}^2=6\mu \text{J}$. The electrostatic energy lost as the charge redistributes is $U_i-U_f=6\mu \text{J}$.