Identify the parallel combination.

The $3 \text{ µF}$ and $1 \text{ µF}$ capacitors are connected in parallel.
The equivalent capacitance for parallel capacitors ($C_p$) is the sum of their individual capacitances.

The $4 \text{ µF}$ capacitor is in series with the equivalent parallel capacitance ($C_p = 4 \text{ µF}$).
The effective capacitance for series capacitors ($C_{eff}$) is given by:

b) (ii) Calculate the charge on the 3 µF capacitor.

The total charge ($Q_{total}$) is stored by the effective capacitance ($C_{eff}$) at the total voltage ($V_{total}$).

In a series circuit, the charge on each component is the same as the total charge. So, the charge on the $4 \text{ µF}$ capacitor is $24 \text{ µC}$, and the charge on the parallel combination ($C_p = 4 \text{ µF}$) is also $24 \text{ µC}$.

The voltage across the parallel combination ($V_p$) is:

Since the $3 \text{ µF}$ capacitor is part of the parallel combination, the voltage across it is $V_p = 6 \text{ V}$.

c) Find the separation distance between two plates if the capacitance between them is $8.0 \times 10^{-12} \text{ F}$ and the area enclosed is $2.0 \text{ cm}^2$ (take $\epsilon_o = 8.85 \times 10^{-12}$).

Where $C$ is capacitance, $\epsilon_o$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the separation distance.

Rounding to three significant figures: