Two capacitors having capacitance C _ 1 and C _ 2 respectively are...

The initial charge on the capacitor ${C}_{1}$ will be $Q = {C}_{1}V$. If after closing the switch the charge flown through the circuit is $q$, the charge on ${C}_{2}$ will be $q$ and ${C}_{1}$ will be $Q - q$. The potential across the capacitors will be same. Hence,

$\dfrac{Q - q}{{C}_{1}} = \dfrac{q}{{C}_{2}} ⇒ q = \dfrac{{C}_{2}Q}{{C}_{1} + {C}_{2}} ⇒ q = \dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}}V$