Figure shows a network of seven capacitors. If charge on 5 $\mathrm{\mu }$F capacitor is 10 $\mathrm{\mu }$C , find the potential difference between points A and C (in volts)

Charge across 5 $\mathrm{\mu }$F capacitor 10 $\mathrm{\mu }$C . 
Potential difference across 5 $\mathrm{\mu }$F capacitor is $\frac{\mathrm{q}}{\mathrm{C}}=\frac{10\times {10}^{-6}}{5\times {10}^{-6}}=2\text{ volt }$
As 3, 4 and 5$\mathrm{\mu }$F capacitors are in parallel, so potential difference across each of the them equals to 2V. 

Charge on 3$\mathrm{\mu }$F capacitor = CV = 3 x 10^-6C x 2V= 6$\mathrm{\mu }$C
Charge of 4$\mathrm{\mu }$F capacitor = 4 x 10^-6 x 2 = 8$\mathrm{\mu }$C 
Total charge flowing in upper branch of circuit is $10\mathrm{\mu C}+6\mathrm{\mu C}+8\mathrm{\mu C}=24\mathrm{\mu C}$
Potential difference across C₂ is $\frac{24\mathrm{\mu C}}{4\mathrm{\mu C}}=6\mathrm{V}$
Total potential difference across AB is =6+2=8V 

Equivalent capacitance of lower branch of circuit is $\frac{6\times 3}{6+3}=2\mathrm{\mu F}$
So, charge flowing is = 2 x 10^-6 x 8 = 16$\mathrm{\mu }$C
Therefore, potential difference between A and C is $\frac{16\mathrm{\mu C}}{3\mathrm{\mu F}}=\frac{16}{3}=5.33 \mathrm{V}$