Physics - Capacitor Question with Solution | TestHub

As given in the figure, $6\mu \mathrm{F}$ and $4\mu \mathrm{F}$ are in parallel. Now using charge conservation Charge on $6\mu F$ capacitor $=\frac{6}{6+4}\times 30=18\mu C$ Since charge is asked on right plate therefore is $+18\mu C$