Two condensers of capacities 4mFand 6mF are connected to two cells as shown. Then

${\mathrm{E}}_{\mathrm{ef}}=8+2=10\mathrm{V};{\mathrm{C}}_{\mathrm{ef}}=\frac{4\times 6}{4+6}=2.4\mathrm{\mu F}$
Charge on each capacitor
Q = E_efC_ef =10 x 2.4  = 24µC ; P D across 6μC
$=\frac{24\times {10}^{-6}}{6\times {10}^{-6}}=4\mathrm{V}$