# The capacitor is disconnected from the battery and connected to another $900\;pF$ capacitor [Fig]. What is the electrostatic energy stored by the system?

The charge on each capacitor is then $Q′ = CV'$. By charge conservation, $Q'=Q/2$.
This implies $V′ = V/2$.
The total energy of the system is $= U_1 + U_2$
$\qquad= 2 \times \large \frac{1}{2}$$Q' V \qquad= \large\frac{1}{2} . \frac{1}{2}$$QV$
$\qquad = \large\frac{1}{2}$$\times 4.5 \times 10^{-6}$
$\qquad=2.25 \times 10^{-6} J$