In the arrangement shown, the potential difference between points a and b is ______

$\begin{array}{l}{\mathrm{C}}_{\mathrm{e}}=\frac{3\times 6}{3+6}+\frac{3\times 6}{3+6}\mathrm{\mu F}\\ =4\mathrm{\mu F}\\ \mathrm{Q}=12\times 4\mathrm{\mu C}=48\mathrm{\mu C}\\ \therefore {\mathrm{Q}}_1={\mathrm{Q}}_2={\mathrm{Q}}_3={\mathrm{Q}}_4=24\mathrm{\mu C}\\ \therefore {\mathrm{V}}_{\mathrm{b}}+\frac{24}{6}-\frac{24}{3}={\mathrm{V}}_{\mathrm{a}}\\ \Rightarrow {\mathrm{V}}_{\mathrm{b}}-{\mathrm{V}}_{\mathrm{a}}=4\mathrm{volt}.\end{array}$