homework and exercises - Capacitors in parallel final potential difference

I was provided with the following problem.

So I first calculated the total capacitance for (i), which was $$4.5 + 1.5 = 6.0 \mu F$$

Now part (ii) is the question i'm struggling at. I know that the charge on the $4.5 \mu F$ capacitor is $6.3 \times 4.5 = 28.35 \mu F$

So why is the p.d. across both capacitors equal to

$$V = \frac{Q}{C} = \frac{28.35\times 10^{-6}}{(4.5+1.5)\times 10^{-6}} = 4.7V$$

I understand that the charge has to be the same on both plates but why do you add the capacitance values together? Wouldn't the different values of capacitances (C) mean that $V = \frac{Q}{C}$ are different across each capacitor?

Answer

If the circuit is drawn differently, it can be observed that the capacitors are actually in parallel.

Hence the equation is correct.

Assuming the capactiors are set up as standard in the circuit above.