Five identical capacitor plates are arranged such that they make four capacitors each of $2\mu F$. The plates are connected to a source of emf $10V$. The charge on plate $C$ is :

Here four capacitors are in parallel and charge on each is $q=CV$.
As the plate C is the common plate between the two capacitors so its charge will be double ,i.e,
$q=2CV$ and it is conned to positive terminal of cell so its charge should be positive.
thus, $q=2CV=2\times 2\times 10=40\mu C$