In the arrangement of the capacitors shown, select the correct options

The equivalent capacitance of the arrangement here is 

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\Rightarrow C=\frac{20}{3}\mu F$

The potential difference across the combination is $90V$ and hence the total charge is

$Q=CV=\frac{20}{3}\times {10}^{-6}\times 90=6\times {10}^{-4}C=600\mu C$

Since all the capacitors are in series and hence all capacitors will have the same charge given by

$Q=CV$

The potential difference across each capacitor will be in the inverse ratio of their capacitances

So the potential difference across $C_1$ is $30V,$ the potential difference across $C_2$ is $20V$ and the potential difference across $C_3$ is $40V$