Find the charge on 6 $\mu F$ capacitor in $\mu C$ .

If we see the charge on positive plate of  $6\mu F$ capacitor, then

$q_2=-q_1-(q_4-q_3)$.........(i)

Now, applying three loop equations, we have

$5-\frac{q_1}{3}+\frac{q_2}{6}=0$.....(ii)

$10-\frac{q_2}{6}-\frac{q_3}{2}=0$.........(iii)

$5-\frac{q_3}{2}-\frac{q_4}{4}=0$............(iv)

Solving these four equation, we can find $q_1,q_2,q_3 \text{and }{q}_4$