We start with the most complicated part of the circuit. The 1 m F and 3 m F capacitors are in series. The equivalent capacitance is given by
1/C = 1/(1 m F) + 1/(3 m F) = 4/(3 m F)
and so Cseries = (3/4) m F.
Next we find the equivalent capacitance of the parallel combination. The capacitances add, so
Cparallel = 3 m F + (3/4) m F = (15/4) m F.
Finally, this capacitance is in series with another 3 m F capacitor, giving us an equivalent capacitance of:
1/C = 4/(15 m F) + 1/(3 m F) = 9/(15 m F)
or C = (5/3) m F.
The charge on the system is then:
Q = Ce = ((5/3) m F)(6 V) = 10 m C.
Now to find the charge on the 1 m F capacitor, we work backwards. The 10 m C charge is on the parallel combination, since capacitors in series have the same charge. But it is split between the upper and lower branches. The potential difference across each branch is the same, so:
Q1/C1 = Q2/C2
We also know that charge is conserved, so Q = Q1 + Q2 . Thus:
10 m C = Q1(1 + C2/C1) = Q1(1+ 4) = 5Q1
And so the charge on both capacitors in the lower branch is: 2 m C.
As a check we can compute the potential difference across each capacitor.
There is 8 m C on the 3 m
F capacitor in the upper branch of the parallel combination, giving a potential
difference of 8/3 V. There is 10 m C on the
single 3 m F capacitor, giving a potential difference
of 10/3 V. The total is 6 V, as required.