Figure shows a part of a circuit. If all the capacitors have a capacitance of 2 $\mathrm{\mu }$F, then the

$$\begin{gathered} 8-\frac{{\mathrm{q}}_1}{2}-2-\frac{{\mathrm{q}}_3}{2}=3\text{ or }6-3=\frac{{\mathrm{q}}_1}{2}+\frac{{\mathrm{q}}_3}{2} \\ \mathrm{or} {\mathrm{q}}_1+{\mathrm{q}}_3=6 ....\left(\mathrm{i}\right) \end{gathered}$$
$8-\frac{{\mathrm{q}}_1}{2}+\frac{{\mathrm{q}}_2}{2}=2\text{ or }\frac{{\mathrm{q}}_1}{2}-\frac{{\mathrm{q}}_2}{2}=6$
or        ${\mathrm{q}}_1-{\mathrm{q}}_2=12 ....\left(\mathrm{ii}\right)$
Also,    ${\mathrm{q}}_1+{\mathrm{q}}_2={\mathrm{q}}_3 ....\left(\mathrm{iii}\right)$
Solving (i), (ii), and (iii), we get q₁ = 6 $\mathrm{\mu }$C, q₂ = -6$\mathrm{\mu }$C , and $q_3=0$.