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?
Best Answer
You got the combined capacitance right.
To find the voltage on the combined capacitor, realize that the charge will be conserved. The charge on the two capacitors after S2 closes must be the same as the charge on the single capacitor when S2 is open.
Yes, it's really that easy.