A capacitor of capacitance C = 900pF is charged fully by 100V...

Initial charge on $900pF$ capacitor will be $q = CV$

When the two equal capacitors are joined together as shown in figure (b). The charge will be equally divided among them. Therefore,

$q^{'} = \dfrac{q}{2} = \dfrac{CV}{2}$

Now the energy stored in them will be

$E = \dfrac{\left(q^{'}\right)^{2}}{2C} + \dfrac{\left(q^{'}\right)^{2}}{2C}$

$⇒ E = \left(\dfrac{CV}{2}\right)^{2} \times \dfrac{1}{C} = \dfrac{CV^{2}}{4} = \dfrac{900 \times {10}^{- 12} \times \left(100\right)^{2}}{4}$

$⇒ E = 2. 25 \times {10}^{- 6}J$