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 $6.3 imes 4.5 = 28.35 mu F$

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

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

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?

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edited Jun 6 “17 at 12:01

vik1245

asked Jun 6 “17 at 11:39

vik1245vik1245

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You got the combined capacitance right.

You are watching: What is the final potential difference across the capacitor?

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.

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answered Jun 6 “17 at 11:51

Olin LathropOlin Lathrop

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I understand that the charge has to be the same on both plates but why do you add the capacitance values together?

No, the charge is not the same on both plates. The total charge must be the same from before $S_2$ is closed to after it is closed. That means the **total** charge on the upper plates must equal $Q_0=28.35 mu F$.

But, the voltages across the capacitors must be the same because they are in parallel. The upper plates must be at a common potential (they are connected by a conductor), and the bottom plates at a different common potential, making the potential difference (voltage) across each capacitor the same. This is accomplished in a few microseconds after $S_2$ is closed by the charge being redistributed between the two capacitors.

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Wouldn”t the different values of capacitances (C) mean that V=QC are different across each capacitor?

It would *if* the charges become the same. Remember that only one capacitor initially has charge (and hence a non-zero voltage). When $S_2$ is closed, charge flows until the voltages are equal, not the charges.

Finally, $$V_{4.5}=V_{1.5}$$$$frac{Q_{4.5}}{4.5}=frac{Q_{1.5}}{1.5}$$and$${Q_{4.5}}+{Q_{1.5}}=Q_0(=4.5V_0)$$