In the figure shown below, the charge on the left plate of the $$10\mu F$$ capacitor is $$-30\mu C$$. The charge on the right plate of the $$6\mu F$$ capacitor is:

Given charge on the left plate of the $$10\ \mu\text{F}$$ capacitor is $$-30\ \mu\text{C}$$.

Magnitude of total charge flowing into the circuit combo: $$Q_{\text{total}} = 30\ \mu\text{C}$$

Since the $$10\ \mu\text{F}$$ capacitor is in series with the parallel combination of $$6\ \mu\text{F}$$ and $$4\ \mu\text{F}$$, the total charge splits between them:

$$Q_{\text{total}} = Q_6 + Q_4 = 30\ \mu\text{C}$$

Potential difference across the parallel branch is identical: $$V_p = \frac{Q_6}{6} = \frac{Q_4}{4} \implies Q_4 = \frac{4}{6}Q_6 = \frac{2}{3}Q_6$$

$$Q_6 + \frac{2}{3}Q_6 = 30 \implies \frac{5}{3}Q_6 = 30 \implies Q_6 = 18\ \mu\text{C}$$

The left plate of the $$10\ \mu\text{F}$$ capacitor is negative, meaning current/charge enters from the right side of the circuit, making the right plates of the parallel combination positive.