Consider two uncharged capacitors of equal capacitance 200 pF. One...

Initial energy of the charged capacitor is given by $U_i = \dfrac{1}{2}CV^2$.

When it is connected to an identical uncharged capacitor, the common potential becomes $V' = \dfrac{CV}{C+C} = \dfrac{V}{2}$.

Final energy of the system is $U_f = \dfrac{1}{2}(2C)V'^2 = \dfrac{1}{2}(2C)\left(\dfrac{V}{2}\right)^2 = \dfrac{1}{4}CV^2$.

Energy lost during the process is $\Delta U = U_i - U_f = \dfrac{1}{2}CV^2 - \dfrac{1}{4}CV^2 = \dfrac{1}{4}CV^2$.

Substituting the given values $C = 200 \times 10^{-12}$ F and $V = 100$ V:

$\Delta U = \dfrac{1}{4} \times 200 \times 10^{-12} \times (100)^2$

$\Delta U = 50 \times 10^{-12} \times 10^4 = 0.5 \times 10^{-6}$ J.

Answer: $0.5 \times 10^{-6}$ J