In the above circuit, the potential difference across AB is

2 V

Let us mark the capacitors as 1, 2, 3 and 4 for identifications. As is clear, 3 and 4 are in series, and they are in parallel with 2. Then 2, 3, 4 combine is in series with 1.

$C_{34}=\frac{C_3 \cdot C_4}{C_3+C_4}=4 \mu f, \quad C_{2,34}=8+4=12 \mu f$

$C_{eq}=\frac{8 \times 12}{8+12}=4.8 \mu f, \quad q=C_{eq} V=4.8 \times 10=48 \mu C$

The 'q' on 1 is 48 µC, thus $V_1=q / c=6 v \quad\left[v_1=\frac{48 \mu c}{8 \mu F}=6 v\right]$

⇒ V_PQ = 10 - 6 = 4 V

By symmetry of 3 and 4, we say, V_AB = 2V