The potential difference between A and B $\left(V_A-V_B\right)$ is

Here, $2\mu F$ and $3\mu F$ capacitors are connected in series. Their equivalent capacitance is $\frac{1}{C_s}=\frac{1}{2}+\frac{1}{3}or{C}_S=\frac{6}{5}\mu F$

$Netvoltage,V=16V-6V=10V$

The equivalent circuit diagram as shown in figure below.

 Charge on each capacitor, $q=C_{s}V=\frac{6}{5}\times 10=12\mu C$

The potential difference between $AandB=\frac{12\mu C}{2\mu F}=-6V$