Three capacitors are connected across a 45 V power supply as shown in fig. What is the charge on the 6 $\mathrm{\mu }$F capacitor ?

 

It is obvious from figure that 6 $\mathrm{\mu }$F and 12 $\mathrm{\mu }$F capacitors are in parallel to each other and the combination is in series with 9 $\mathrm{\mu }$F capacitor. So, we have 9 $\mathrm{\mu }$F and 18 $\mathrm{\mu }$F capacitors in series across 45 V power supply.
 Voltage drop across combination = 15 V
Now   $9\times {\mathrm{V}}_1={\mathrm{V}}_2\times 18$
So, charge on 6 $\mathrm{\mu }$F capacitor $=6 \mathrm{\mu F}\times 15=90 \mathrm{\mu C}$