If charge on left plate of the  $5\mu F$ capacitor in the circuit segment shown in the figure is $-20\mu C$, the charge on the right plate of $3\mu F$ capacitor is 

$Q=20\mu C$ 

$Q_1+Q_2=Q=20\mu C\left(i\right)$

 And   $\frac{Q_1}{3}=\frac{Q_2}{4}\left(ii\right)$
Solving (i) and (ii) 
$Q_1=\frac{3}{7}\times 20=8.57\mu C$
Obviously, charge on the right plate of  $3\mu F$ capacitor is positive.