In the given circuit, the charge on 4 μF capacitor will be:

$$C_p = 1\ \mu\text{F} + 5\ \mu\text{F} = 6\ \mu\text{F}$$

This $$6\ \mu\text{F}$$ equivalent combination is in a series branch with the $$4\ \mu\text{F}$$ capacitor. Find the equivalent capacitance ($$C_s$$) of this top branch:

$$C_s = \frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4\ \mu\text{F}$$

Since this entire top branch is connected directly in parallel across the $$10\ \text{V}$$ battery, the total voltage across the branch is $$10\ \text{V}$$. The total charge ($$Q$$) flowing through this series branch, which is the same as the charge stored on the $$4\ \mu\text{F}$$ capacitor:

$$Q = C_s \times V = 2.4\ \mu\text{F} \times 10\ \text{V} = 24\ \mu\text{C}$$