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I was provided with the following problem. enter image description here

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?

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3 Answers 3

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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.

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  • $\begingroup$ I'm confused though why do you treat it as a combined capacitor? Is the problem saying that the voltage across both capacitors is the same? $\endgroup$
    – vik1245
    Commented Jun 6, 2017 at 11:56
  • $\begingroup$ @Bob: You can see from the circuit that with S2 closed the voltage on both caps is forced to be the same. $\endgroup$
    – Olin Lathrop
    Commented Jun 6, 2017 at 11:58
  • $\begingroup$ why is the voltage on both capacitors the same? they should total 6.3V because of Kirchhoff's second law right? $\endgroup$
    – vik1245
    Commented Jun 6, 2017 at 12:02
  • $\begingroup$ @Bob: "Why is the voltage on both capacitors the same?" Because they are connected in parallel. $\endgroup$
    – Olin Lathrop
    Commented Jun 6, 2017 at 12:05
  • $\begingroup$ I can't see how they are connected in parallel. Once you close S2 and open s1, you have a loop from the 4.5 capacitor through the 1.5 capacitor and back to the 4.5 capacitor, so they are in series right? $\endgroup$
    – vik1245
    Commented Jun 6, 2017 at 12:15
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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.

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 $${Q_{4.5}}+{Q_{1.5}}=Q_0(=4.5V_0)$$

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  • $\begingroup$ why does charge flow until the voltages are equal, not the charges? $\endgroup$
    – vik1245
    Commented Jun 6, 2017 at 12:16
  • $\begingroup$ If the voltages are unequal then the potentials of either the top plates or the bottom plates are different. That means there will be an electromotive force on the charges and they will move. The charge will distribute until the net force on charges is zero. This is an electrostatic circuit. Remember that the capacitances are different, which implies either the plates are different sizes or the gaps are different. $\endgroup$
    – Bill N
    Commented Jun 6, 2017 at 12:35
  • $\begingroup$ I can't see how they are connected in parallel. Once you close S2 and open s1, you have a loop from the 4.5 capacitor through the 1.5 capacitor and back to the 4.5 capacitor, so they are in series right? $\endgroup$
    – vik1245
    Commented Jun 6, 2017 at 12:58
  • $\begingroup$ It doesn't matter. What's important is what parts are at equal potentials and what parts aren't. The tops are at the same potential as each other ; the bottoms are same potential. Therefore, the voltages are the same. If you want to call that "parallel" go ahead. If you want to say the sum of the voltages around the loop of two being zero means they are "in series", go ahead. Look at what the potentials have to be based on point to point connection. Actually, any two circuit elements which are connected in a loop will have equal voltages. To be unambiguous, you need 3 circuit elements. $\endgroup$
    – Bill N
    Commented Jun 6, 2017 at 13:12
  • $\begingroup$ I use the following guidance (these are non-official, but they seem to work): If two 2-port elements are connected to each other on both ends, they are in parallel with each other. If two 2-port elements are connected to each other only at one end, and no other element is connected at that connection point, they are in series. $\endgroup$
    – Bill N
    Commented Jun 6, 2017 at 13:13
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If the circuit is drawn differently, it can be observed that the capacitors are actually in parallel.

Hence the equation is correct.

enter image description here

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

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