In the circuit shown below, the switch S is connected to position P...

When $S$ is connected to $Q$,

$i = \dfrac{2}{1 + 2}$

$i = \dfrac{2}{3}A$

Now, after distribution of potential,

${q}_{2} = CV$$= \left(1\mu F\right)\left(\dfrac{2}{3}V\right)$

${q}_{2} = \dfrac{2}{3}\mu C$

${q}_{2} = 0.67\mu C$