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enter image description here

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I don’t understand that why the net charge of the two isolated plates must be zero. If it’s an isolated system then its net charge won’t change, which doesn’t imply its net charge must be zero.

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  • $\begingroup$ I do not see where it is mentioned in the problem that the two capacitors form an isolated system, do you? $\endgroup$
    – hyportnex
    Commented May 29, 2023 at 14:03
  • $\begingroup$ In the picture I post, not in the problem. The problem is just to show the charge on two capacitors in series is same. The answer says that. $\endgroup$
    – Xiang Li
    Commented May 29, 2023 at 16:13
  • $\begingroup$ Please edit your question to cite the work, or if there's a link to something that's not behind a paywall, provide that. $\endgroup$
    – TimWescott
    Commented May 30, 2023 at 4:12

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The claim follows from the fact that currents through capacitors in series are equal, and assumes the capacitors have no leakage. Specifically, $$\frac{dq_1}{dt} = i_1 = i_2 = \frac{dq_2}{dt}$$ $$q_1(t) = q_2(t) + \text{const.}$$ Note that $i=dq/dt$ is only strictly valid if there is no leakage current. If the capacitors are uncharged (or happen to have the same non-zero charge) at any time, the constant is zero and they are guaranteed to have the same charge at all other times.

Other answers have correctly explained that the charges don't have to be equal. This is either because of the constant term (i.e. if the capacitors aren't initially uncharged), or the fact that most capacitors tend to slowly lose charge through leakage, as John Doty mentions. For instance, if two capacitors with equal charge are in series but one has higher leakage, the charges won't be exactly equal at later times. In practice, if the capacitors are uncharged as you are putting them in your circuit (as they likely are), and the charges lost through leakage currents are insignificant at time scales of concern, the charges on the capacitors will be at least approximately equal.

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  • $\begingroup$ It’s clear. Thank you. $\endgroup$
    – Xiang Li
    Commented May 31, 2023 at 9:59
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This quote is simply wrong: imagine taking an uncharged capacitor and connecting it in series with a charged capacitor, but not closing the circuit. No current flows, so the charge doesn't change on either: one remains charged, the other uncharged.

Some circuit simulation programs fail if you have capacitors in series, because they cannot (ideally) tell what the charges on the individual capacitors should be. In real life, leakage currents will have their say...

But also note that "charge stored by a capacitor" is a different concept than "charge on an isolated object". We model capacitors as having zero net charge: what we call "charge" in this case is the charge difference between the plates.

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  • $\begingroup$ Ok. So under what conditions the two capacitors in series have same charge, both uncharged at first? The problem is from the book, electricity and magnetism by Purcell chapter 3, problem3.17. The picture I post is the answer he gave. I just don’t follow the reasoning. Can you explain it? Many thanks. $\endgroup$
    – Xiang Li
    Commented May 29, 2023 at 16:17
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    $\begingroup$ Although you're right that the statement should not be made as in the book, equal-charges is nevertheless a very good model to what would actually happen in lots of applications. Specifically, the case of pure AC / signals without DC bias and reasonably low leakage currents, the only way the capacitors can obtain charge is from the current going through both of them and always charging / discharging both of them by the same amount. Only with DC do leakage currents matter. And the case you're describing is arguably not just "two capacitors in series" but rather a cap, a switch and another cap. $\endgroup$ Commented May 30, 2023 at 12:45
  • $\begingroup$ I imagine the quote might be "wrong" simply because it's taken out of context, which likely discusses closed electric circuits. $\endgroup$ Commented May 30, 2023 at 12:55
  • $\begingroup$ @DmitryGrigoryev Even for a closed circuit, it's wrong, since the charge on the capacitors depends on the initial conditions. $\endgroup$
    – John Doty
    Commented May 30, 2023 at 14:21
  • $\begingroup$ @leftaroundabout In practice, capacitors in series are rarely used. When they are, there are usually resistors to balance the charge. JAXA likes to have spacecraft power filters using series capacitors to defend against short-circuit failures. They use resistors to balance the charge. $\endgroup$
    – John Doty
    Commented May 30, 2023 at 16:35
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The book is wrong, or it is right in a very narrow context which you missed, or the book authors missed.

It is right in that you can arrange things so that it is true. It is wrong in that it is not generally true.

The easiest way to do this in a thought experiment is to take two ideal, fully discharged capacitors and put them into a circuit. If that combination is treated as one capacitor and charged, the two plates that are connected together must have net charges that are negatives of one another.

There are other ways to arrange for the same effect (the reliable real-world way would involve deliberately putting resistances in parallel with the capacitors) -- but the point here is that the combinations of circuit elements and events that will make this happen are quite special, and will only be approximated in real life if you work at it.

In general, the charge (and, hence, voltage) on the common node between the capacitors will be unknown. In the real world it'll be uncontrollable without deliberately adding circuitry; in the world of theory and simulation it'll be undefined.

To illustrate, consider the case where one capacitor -- let it be C1 -- is discharged but the other -- let it be C2 -- is charged before going into the circuit. Let the charged capacitor have the charge $q_1$.

Now connect the two capacitors in series without connecting their free ends to anything. No current (or charge) could have flowed, so C1 is still discharged, and C2 is charged to $q_1$.

Now connect their free ends through a resistor. Current flows until the voltage across the combination of C1 and C2 is zero. If C1 and C2 have the same capacitance, this will happen when each has $\frac 1 2 q_1$ of charge.

If you started with the free end of C1 having $q_1$ of charge and the end connected to C2 having $-q_1$ of charge, then if you travel from the free end of C1 to the free end of C2, the charges will be $\frac 1 2 q_1$, $-\frac 1 2 q_1$, $-\frac 1 2 q_1$, and $\frac 1 2 q_1$.

You could arrive at a similar situation by wiring up a circuit where the free ends of C1 and C2 were connected to a voltage, and the common node between them were connected to a different voltage.

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  • $\begingroup$ I guess the author discussed the situation same as your thought experiment. If so, why the two plates that are connected together must have net charges that are negatives of one another. Why the two isolated plate from the rest circuit must have zero net charge as the author says? Many thanks. $\endgroup$
    – Xiang Li
    Commented May 30, 2023 at 0:53
  • $\begingroup$ I said that the two plates that are connected together may not have net charges that are negatives of one another. So there is no "why is this" because I have explained why it is not. I have also explained why two isolated plates may not have zero net charge, so there is no "must have zero net charge". If you can find where the author discusses this, then edit your question to include that discussion. Otherwise we're just arguing with an incorrect statement without the context that may make sense of it. $\endgroup$
    – TimWescott
    Commented May 30, 2023 at 4:11
  • $\begingroup$ Ok, I guess I understand now. Thank you. $\endgroup$
    – Xiang Li
    Commented May 31, 2023 at 9:59
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I agree with TimWescott's response-- the quote from the book is not necessarily true. However, as you have expounded on the quote, the net charge over each capacitor is in fact zero. Take for example a parallel plate capacitor: if you were to apply a voltage across the configuration, electrons would accumulate on one plate, while "holes" would accumulate on the other plate. In solid state physics and condensed matter physics, we refer to a hole basically as an absence of electron which results in a net positive charge of 1e.

In the simplest case, where potential difference (voltage) is applied across a parallel plate capacitor, a conducting metallic plate on one side will exhibit an absence of electrons (holes) from atomic (d) orbitals, such that the net charge of the plate, summing the protons and electrons, is net positive. On the other side of the parallel plate capacitor, another conducting metallic plate will collect the same number of electrons (and thus, the same magnitude of charge, but opposite sign) in vacant d orbitals. Since the magnitude of the charge collected on each side is the same (assuming no leakage), and the sign is opposite, the net charge over the entire capacitor sums to zero.

Now, if you were to align multiple capacitors in series, unless the capacitance for each capacitor were identical, upon applying a voltage, you would observe a different magnitude of charge collected on each side of different capacitors, but the same magnitude and opposite sign of charge on each end of each individual capacitor. However, upon adding the total charge of all capacitors, this would in fact net to zero, due to what I have said in the previous paragraph and also due to a phenomenon called "conservation of charge."

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  • $\begingroup$ Multiple capacitors in series will still store the same charge each even if their capacitances are not equal, barring leakage currents and assuming they are initially uncharged. $\endgroup$
    – Puk
    Commented May 30, 2023 at 18:22
  • $\begingroup$ Ok, many thanks for your explanations. $\endgroup$
    – Xiang Li
    Commented May 31, 2023 at 10:01
  • $\begingroup$ @Puk U=(1/2)*CV^2=V*q $\endgroup$
    – ndtech
    Commented May 31, 2023 at 18:12
  • $\begingroup$ @ndtech I agree (except $U = \frac12 q V$), but I'm not sure what your point is. I didn't say anything about energy storage and neither does your answer. $\endgroup$
    – Puk
    Commented May 31, 2023 at 18:44
  • $\begingroup$ Check again… Your equation is wrong… Not only that, but I proved your claim wrong in a one-liner, and you can’t even understand why… 😂 $\endgroup$
    – ndtech
    Commented Jun 1, 2023 at 22:00

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