Sharing of Charge between Capacitors

Common potential $V_c = \frac{C_1V_1 + C_2V_2}{C_1 + C_2} = \frac{C \times 100 + C \times 0}{C + C} = 50$ V Energy stored in system (b) $= 2 \times \frac{1}{2}CV_c^2 = CV_c^2$ $= 900 \times 10^{-12} \times 50 \times 50 = 2.25 \times 10^{-6}$ J

When two capacitors are connected, charge flows until both are at the same (common) potential. The total charge is conserved.
Common potential: $V_{common} = \dfrac{\text{total charge}}{\text{total capacitance}} = \dfrac{C_1V_1 + C_2V_2}{C_1 + C_2}$
Here $V_1 = V$ and $V_2 = 0$ (uncharged):
$V_{common} = \dfrac{C_1V}{C_1 + C_2}$
So the final p.d. across each capacitor is $\dfrac{C_1V}{C_1 + C_2}$.

Initial energy: $U = \dfrac{q^2}{2C}$, where q is the charge on the capacitor.
After the battery is disconnected and an identical uncharged capacitor is connected in parallel, the charge q is shared. Common potential:
$V_{common} = \dfrac{q_1 + q_2}{C_1 + C_2} = \dfrac{q + 0}{C + C} = \dfrac{q}{2C}$
Energy on each capacitor $= \dfrac{1}{2}CV_{common}^2 = \dfrac{1}{2}C\left(\dfrac{q}{2C}\right)^2 = \dfrac{q^2}{8C} = \dfrac{1}{4} \cdot \dfrac{q^2}{2C}$
$= \dfrac{U}{4}$
(The total energy is $\dfrac{U}{2}$; the other half is lost as heat in the connecting wires.)