VPQ in the figure.

12 V 8 V 4 V Zero

4 V

C = $\frac{2 \times 4}{2+4}=\frac{4}{3}$ µF in one loop. q = $12 \times \frac{4}{3}$ = 16 µC across one set of 2 µF & 4 µF capacitors. ∴ $V_P=12-\frac{16}{4} = 8 V$ $V_Q=12-\frac{16}{2}=4 V$ ∴ VP – VQ = 8 – 4 = 4 V ∴ (C)