The potential difference between points A and B in the circuit shown in Fig is 

The batteries are in opposition as their positive terminals are connected together. Hence the effective voltage is 

V = V₁ - V₂ = 12 - 2 = 10 V 

As the capacitors C₁ and C₂ are in series, the effective capacitance of the circuit is given by

$$\begin{gathered} \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6} \\ or C =\frac{6}{5}=1.2 \mu F. \end{gathered}$$

Therefore, charge on capacitors is

Q = CV= 1.2 µF x 1OV = 12 µc

$\therefore$Potential difference across A and B = potential difference across capacitor C₂

$=\frac{Q}{C_2}=\frac{12\mu C}{2\mu F}=6 V$