A Discussion About Series CapacitorsFrom the PHYS-L Listserver - Sep. 1997

I have a question about the usual formula for capacitors in series.

Suppose I connect two capacitors in series across a battery. Label the four capacitor plates from left to right as A, B, C, and D. Okay, suppose A is connected to the positive terminal of the battery, so out goes charge +Q to it, and compensating charge off D, leaving -Q on it. My question is: why does the charge on B and C have to be -Q and +Q, respectively?

If the isolated circuit consisting of plates B and C and the wire between them is initially uncharged, then the sum of the charges presumably has to remain zero. But why couldn't I get -1.1Q and +1.1Q on these two plates, say? Why *exactly* -/+Q? Is it always exact: what if plates A and B have different shapes? Or what if I imagine distorting the wire between plates B and C, so that B and C are both portions of some larger object, say the two ends of a solid rectangular block, or even a sphere? Surely at some point the answer will no longer be -/+Q. At what point - in other words, what assumptions go into the usual derivation? I've looked in several textbooks and it's presented as though -/+Q is patently obvious.

Carl Mungan

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The key idea here is that the system of charges will redistribute itself in such a way as to minimize the resulting electrostatic potential energy of the system subject to the external constraint imposed by the battery potential. This is a consequence of the 2nd law. The system tries to dissipate as much of its energy as possible as (dare I say it) thermal energy which ultimately spreads throughout the universe raising its entropy as it goes. This means that only the irreducibly minimal energy remains in the electrostatic field configuration after the charges are through moving into their equilibrium positions. If the charges on plates B and C of the interior plates were not -Q and +Q respectively then there would be an electric field in the region between the capacitors and the mobile charges in the connecting conductor wire would feel an unbalanced restoring force which would tend to accelerate them toward the equilibrium configuration. If there was no dissipative damping the interior charges would slosh back and forth at a frequency given by the effective inductance and capacitance of the sloshing mode. The charges will move around dissipating energy until they are in the minimum potential energy configuration. This requires that both surfaces B and C be at the same potential with no electric field between them. The energy tied up in an electrostatic field is proportional to the volume integral of the square of the field strength. The greater the field strength and the greater the volume it occupies the greater the energy in it. If the interior charges are partitioned as -Q and + Q then all of the electric field is effectively confined to the interior of each capacitor and its fringe area. This minimizes the potential energy of the system and balances the forces on the mobile charges so they stay put there. If the geometry of the setup is distorted so the distance between B and C was comparable to or less than the capacitor gaps which would then have a huge fringe field volume, and if the geometry of the system significantly broke any left-right symmetry that the system might possess, then the values of the charges would not necessarily split up into exactly -Q and +Q on plates B and C but would also be some partial charge distributed along the surface of the conducting connector between them.

To solve for the final charge distribution in this case would entail, in general, solving Laplace's equation over all space with the requirement the the potential gradient at all metallic surfaces be locally perpendicular to the surface. This will zero out the electric field in the interior of the metal.

David Bowman

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If B & C are connected by a conductor they must have equal potentials. If their charges are not +/- Q there will be a potential difference and charges will flow until they are +/- Q. In your example, the +1.1 Q on C pull harder on B's electrons than the +Q on A, so some of them flow to C.

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This is an engineering-style response, so it may lack rigor in mathematical physics terms.

For two caps in series, the charge is carried by the conduction current in the conductors, and the displacement current through the caps. The series connection necessitates that the current is the same through the whole loop.

There is no reason why the caps cannot have a prior charge where the 'battery' ends ( A,D) are at zero potential, but the common capacitor point (B,C) has introduced charge to both caps at some previous time. (In this case, the charges in the two caps may be quite different, depending on the capacitance values; the charging path being in parallel for this case, i.e B or C to A and B or C to D.)

Even so, the extra series current provided by the change of voltage applied across the ends (A,D) is the same through both caps, and for the same time, and so it follows that the EXTRA charge is always equal at the two caps.

Regards, Brian Whatcott

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I would consider an application of Gauss' Law to the "closed surface" which bounds A&B (or C&D) together with the knowledge that the total charge is initially zero on the capacitor pair of plates A&B (or C&D). This information along with the same reasoning that gave us Qb=Qc requires that Qa=Qb (and also Qc=Qd). QED

John P. Ertel

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> Suppose I connect two capacitors in series across a battery. Label the four
> capacitor plates from left to right as A, B, C, and D. Okay, suppose A is
> connected to the positive terminal of the battery, so out goes charge +Q to
> it, and compensating charge off D, leaving -Q on it. My question is: why
> does the charge on B and C have to be -Q and +Q, respectively?

As you seem to have suspected, it doesn't.

Consider an extreme geometry such as one in which "plates" B and C are simply the ends of a short piece of wire in the region between plates A and D. The wire need not in any way be "centrally positioned" in the region and the plates A and D need not be planar or take on any particularly simple form. (It's not *even* necessarily the case that plates A and D will carry equal and opposite charges due to the effects of a net charge on the entire circuit or of other nearby charged objects, but let's assume for the moment that they do.)

In this extreme case, it isn't at all clear what we even mean by "plate B" or "plate C" (and it is important to recognize that this ambiguity is *always* present albeit to a very significantly reduced degree in practical cases involving real capacitors with large plates that face each other and are separated by a small distance.) There is certainly likely to be some separation of charge on the wire, but you will *not* in general be able to divide the wire in any way such that one piece carries net charge +Q and the other -Q.

The best way to see this is by invoking Gauss' Law: Some of the field lines starting on plate A will end on the nearer portion of the wire and some will end on plate D. Unless, *all* of the lines starting on plate A end on plate B (a situation that is *very, very* closely approximated in real capacitors), then plate B will not carry a charge of -Q.

The bottom line is that the textbook rules apply only as an *extremely good approximation* in the usual case in which the plates of *each* capacitor are large and close together, in which *different* capacitors are separated by distances that are large relative to the spacings of their plates, and in which we only consider the charge delivered to the plates via the action of the battery (and can neglect or ignore any preexisting charges that might result from violations of neutrality or external influences).

John Mallinckrodt

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Interestingly enough, I was just in private discussion with Ludwik about the same concerns Carl had; I've always felt uncomfortable lecturing about the standard textbook equal charge statements. Here are some of my thoughts on the topic and some responses to what else has just been written:

1) First a minor comment. I disagree with a statement David made. He wrote: "If the charges on plates B and C of the interior plates were not -Q and +Q respectively then there would be an electric field in the region between the capacitors . . ."

If we look at an ideal case where the two capacitors are well approximated by two infinite parallel plate capacitors far apart and connected by a very thin wire. We can model the static charge distribution as 4 sheets of charge and the field in between the two plates is zero, even if the interior charges are +/-1.1 Q.

2) I rather like John's response. Let's assume we are not dealing with an extreme geometry case; but rather a case where the textbook rule applies to very good approximation. I think the rule hinges on two points. One must argue that all field lines originating on plate A terminate on plate B. (This is what gets violated in extreme geometry cases). For situations like (1) above this isn't too hard to justify, because in a single capacitor one has already talked about this in the usual "what is a capacitor discussion". Secondly, you need to invoke Gauss' Law, put surfaces around each of plates A and B; the field line flux is the same (in magnitude, there is a sign difference of course), therefore the magnitudes of the net charge on plates A and B are the same.

Similar discussion applies at plates C and D.

3) Most textbooks I've looked at (Calculus level introductory texts) have abysmal explanations that amount to non-justification of the equal charge rule. A cursory look has shown me that most books have illogical and bogus statements that don't imply the rule, and in fact amount to just asserting that it is the case, i.e. they make it seem that it is "patently obvious" without true justification. Often they talk about the inner plates being one conductor that is overall neutral and invoke conservation of charge; this doesn't do IT as Carl's original thought of the inner plates having a plus/minus 1.1Q charge doesn't violate conservation of charge. One or two books I've looked at, at least refer to all the field lines on plate A terminating on B; but don't make the further crucial step of invoking Gauss' Law. They are missing a golden opportunity to use Gauss' Law in a profound way (and necessary way) to justify the equal charge law.

4) Thanks to Carl for bringing this up; he's braver than I.

Joel Rauber

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>1) First a minor comment. I disagree with a statement David made. He
>wrote: "If the charges on plates B and C of the interior plates were not -Q
>and +Q respectively then there would be an electric field in the region
>between the capacitors . . ."
>If we look at an ideal case where the two capacitors are well approximated
>by two infinite parallel plate capacitors far apart and connected by a very
>thin wire. We can model the static charge distribution as 4 sheets of
>charge and the field in between the two plates is zero, even if the
>interior charges are +/-1.1 Q.

No. If plate A has a +Q charge uniformly distributed on it, plate B has a -1.1*Q charge uniformly distributed on it, plate C has a + 1.1*C uniformly distributed on it, and plate D has a -Q charge uniformly distributed over it, then capacitor AB has a net -0.1*Q charge on it while capacitor CD has a net +0.1*Q charge on it This means that there would be field lines going from the net positively charged CD capacitor to the net negatively charged AB capacitor. Using your (nearly) infinite parallel plate approximation where A is the (actually finite) area of the plates and L is the separation between plates B and C where L^2 << A and L>> max(AB separation, CD separation), then these (effectively parallel and uniform) E field lines in the region between the B and C plates will approximately represent a field of strength 0.1*Q/([epsilon_0]*A) in the region between plates B and C. Because of this field there is a potential difference of about 0.1*Q*L/([epsilon_0]*A) between plates B and C. Since these plates are connected by a conducting wire the mobile charge in the metal is free to flow along this wire in response to the field/potential difference. In this case plate B attracts positive charge and plate attracts negative charge. There is thus a tendency for the excess charges to try to equalize via the wire. If there were no dissipation mechanism then once the charges equalize they would tend to continue to flow (er, or move) due to inertial (inductance) effects and this would cause a reversed imbalance to develop. The excess charges would then continue to oscillate across the wire and alternately overcharging/undercharging plates B and C. Dissipation effects guarantee that the charge eventually settles down into an equilibrium configuration of minimal potential energy. This minimum corresponds to a situation where there is charge neutrality on both capacitors AB and CD. (Of course, in describing the above senario I implicitly and incorrectly assumed that the charges on plates A and D remained constant during the equilibration process. Although including charge fluctuations on plates A and D through the battery during the equilibration process would be correct, the neglect of them doesn't change the result that the final minimal potential energy configuration is the one where each separate capacitor has no net charge, and this zeroes out the field (except for stray fringe effects) in the interior region between plates B and C.)

>2) I rather like John's response. Let's assume we are not dealing with an
>extreme geometry case; but rather a case where the textbook rule applies to
>very good approximation. I think the rule hinges on two points. One must
>argue that all field lines originating on plate A terminate on plate B.
> (This is what gets violated in extreme geometry cases). For situations
>like (1) above this isn't too hard to justify, because in a single capacitor
>discussion". Secondly, you need to invoke Gauss' Law, put surfaces around
>each of plates A and B; the field line flux is the same (in magnitude, there
>is a sign difference of course), therefore the magnitudes of the net charge
>on plates A and B are the same.

Notice that this requirement of no net excess electric flux external to each of the capacitors is equivalent to a requirement of a vanishing field in the region between plates B and C. The reason why even a single capacitor would want to be charged in such a way as to keep it overall charge-neutral is that if this condition did not hold then the net electric flux emanating from the capacitor would result in an electric field external to the capacitor and this would raise the potential energy of the system above the minimal value that occurs when nearly all the electric flux (and field) is confined to the relatively small region between the capacitor plates.

David Bowman

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David is absolutely correct, and boy do I feel embarrassed about my comment #1. I hope it doesn't detract from the other comments. And it is also a good lesson, in making hasty comments without a little thought. I realized my mistake before I even finished the first line of his response, I confused thinking about the interior region with the region of space exterior to the capacitors. OUCH!

Joel Rauber

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I contributed a small piece to this thread and followed with interest the contributions of the other folks. I am left profoundly disturbed by how plainly unpractical and inapplicable the models selected for explaining the sharing of charge between capacitors have been. People who deal with electronic design very soon develop excellent intuition into the voltages, currents, and to a minor extent charges to be expected by varying arrangements of serial and parallel capacitors. It has been surprising how willing the people here have been to divide the system inappropriately - by splitting a capacitor, plate from plate for instance, or visualizing a pathological cap design with one plate and one wire as electrodes. Even worse, the dominant models seem to have overwhelming difficulty in explaining the reasonable case of two caps of different C value, precharged to the same voltage ( i.e. with a different Q in each cap) and then connected in series. The engineer is immediately able to assign voltages to each point of the series arrangement; he can compute the charge in each; and when the series arrangement is connected to a battery, what the extra charge is and how it impacts the voltage at each point.

I am less than confident that physicists can execute this task after reading their commentaries. Care to try?

Procedure: Connect cap 1 and 2 to battery 3. Leads B and C to pos terminal Leads A and D to neg terminal.

Q's ???

Q's ???

Voltage at leads B and C with respect to D ???

(The answers comprise five numbers...) Regards

Brian Whatcott

------------------------------

> I am less than confident that physicists can execute this task after
> Care to try?

Sure; I love a good capacitor problem! I'll assume that these are ideal capacitors behaving according to all the textbook rules. You specify a "procedure" so I assume that the steps are carried out in sequence. There is, unfortunately, some ambiguity in your specification of the procedure so I will make my assumptions explicit:

I'll assume that the capacitors are initially uncharged.

> Procedure:
> Connect cap 1 and 2 to battery 3.
> Leads B and C to pos terminal
> Leads A and D to neg terminal.
> Q's ???

This is a simple parallel combination with both capacitors charged to 10 V so Cap 1 attains a charge of 30 microcoulombs and Cap 2 attains a charge of 60 microcoulombs.

> Then

By which I assume you mean "with the capacitors carrying the charges attained in the first part."

This will simply short out battery 4 unless I first disconnect leads A and D. So I'm *guessing* that you intended for me to completely disassemble the first circuit before beginning this part. Thus, I have formed a series circuit using Caps 1 and 2 and battery 4 with Cap 1 precharged to 10 V and its (initially) negative lead connected to the positive terminal of the battery and with Cap 2 also precharged to 10 V and its negative lead connected to the negative terminal of the battery.

> Q's ???

After the charges redistribute we will find that Cap 1 still carries 30 microcoulombs (although the polarity has changed) and Cap 2 now carries 120 microcoulombs (with the same polarity.)

> Voltage at leads B and C with respect to D ???

Leads B and C will be at a potential 20 V above that of lead D.

Unless I've really missed something, this problem is essentially the same as one I usually give--albeit, I hope, less ambiguously phrased--to my intro physics students. Many *do* find the second part difficult, but it is essentially meat (or a soy substitute) and potatoes.

John Mallinckrodt

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Brian Whatcott wrote:
>I am left profoundly disturbed by how plainly unpractical and
>inapplicable the models selected for explaining the sharing of charge
>between capacitors have been.

The original questioner was not interested in matters of practicality but in matters of principle and of underlying understanding. What do you mean by 'inapplicable'?

>People who deal with electronic design very soon develop excellent
>intuition into the voltages, currents, and to a minor extent charges
>to be expected by varying arrangements of serial and parallel capacitors.

I should hope so.

>It has been surprising how willing the people here have been to divide the
>system inappropriately - by splitting a capacitor, plate from plate for
>instance, or visualizing a pathological cap design with one plate and one
>wire as electrodes.

Inappropriate? Extreme cases can help aid understanding. The original questioner intentionally asked about unconventional geometric arrangements to aid in understanding matters of principle.

>Even worse, the dominant models seem to have overwhelming difficulty
>in explaining the reasonable case of two caps of different C value,
>precharged to the same voltage ( i.e. with a different Q in each
>cap) and then connected in series.

What difficulty?

>The engineer is immediately able to assign voltages to each point of the
>series arrangement; he can compute the charge in each; and when
>the series arrangement is connected to a battery, what the extra charge is
>and how it impacts the voltage at each point.

Again, that's to be expected. But do most engineers know when and *why* the formulae, algorithms and rules of thumb that they as so adept at using actually properly describe the situation at hand, and when and why they may break down? Maybe the engineer lumps such breakdowns into the category of inappropriate pathological cases that should not be considered.

>I am less than confident that physicists can execute this task after

Huh? I certainly did not get this impression. Of course, I did not read every response with the care and attention to detail that I should have.

>Care to try?

OK

>Procedure:
>Connect cap 1 and 2 to battery 3.
>Leads B and C to pos terminal
>Leads A and D to neg terminal.

>Q's ???

Let u = [mu] = 10^(-6) Q_1 = 30 uC (Q_A = -30 uC, Q_B = +30 uC), Q_2 = 60 uC (Q_C = +60 uC, Q_D = -60 uC)

>Then

>Q's ???

Q_1 = 30 uC (Q_A = +30 uC, Q_B = -30 uC), Q_2 = 120 uC (Q_C = +120 uC, Q_D = -120 uC)

>Voltage at leads B and C with respect to D ???

+20 V

(Assuming that the cap charge polarities are unambiguous)

(BTW, do you also think that physicists don't know how a voltage doubler power supply works?)

David Bowman

------------------------------

>>I am left profoundly disturbed by how plainly unpractical and
>>inapplicable the models selected for explaining the sharing of charge
>>between capacitors have been.
>The original questioner was not interested in matters of practicality but in
>matters of principle and of underlying understanding. What do you mean by
>'inapplicable'?

The original questioner was interested to know why the charge at point B and C have to be numerically equal, as quoted above.

The only person who explicitly contradicted this equality ( as far as I recall ) was John Mallinckrodt. You for example merely modelled the oscillation which could precede the new steady-state, did you not?

>>Even worse, the dominant models seem to have overwhelming difficulty
>>in explaining the reasonable case of two caps of different C value,
>>precharged to the same voltage ( i.e. with a different Q in each
>>cap) and then connected in series.
>What difficulty?

The difficulty of erroneously assigning equal charge on connected plates of series caps.

>do most engineers know when and *why* the
>formulae, algorithms and rules of thumb that they as so adept at using
>actually properly describe the situation at hand, and when and why they may
>break down?

I may have mentioned recently that I do NOT subscribe to the orthodox ( I deliberately do not say "old-fashioned") view that there are "natural laws" that scientists discover; I find it much more productive to suppose that there are man-made models - any of which may be more or less suitable to any particular case. This has the particular virtue that when physicists comfortably think they are dealing on a 'more fundamental' level, I can as easily see the flaws in their models as in any other man made elaboration.

>Let u = [mu] = 10^(-6)
>Q_1 = 30 uC (Q_A = -30 uC, Q_B = +30 uC), Q_2 = 60 uC (Q_C = +60 uC,
>Q_D = -60 uC)

So you can now easily admit a case where Qb is not numerically equal to Qc. This was the crux of Carl's question, was it not?

>(BTW, do you also think that physicists don't know how a voltage doubler
>power supply works?)

I am afraid that this was at the heart of my sadness. It was Cockroft and Walton who defined this arrangement for a voltage multiplier in the earliest accelerator experiments. They were physicists as I recall. But I don't suppose their names are known to teachers hereabouts.

(On careful deliberation, I have deleted the putative reason I offered for this. I hope you will accept this as a tribute to your private coaching in how to prevent public shows of sniping.)

Regards, Brian Whatcott

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> Suppose I connect two capacitors in series across a battery. Label the four
> capacitor plates from left to right as A, B, C, and D. Okay, suppose A is
> connected to the positive terminal of the battery, so out goes charge +Q to
> it, and compensating charge off D, leaving -Q on it. My question is: why
> does the charge on B and C have to be -Q and +Q, respectively?

Yes, but Carl went on to clarify his question saying:

> If the isolated circuit consisting of plates B and C and the wire between
> them is initially uncharged, then the sum of the charges presumably has to
> remain zero. But why couldn't I get -1.1Q and +1.1Q on these two plates,
> say? Why *exactly* -/+Q? Is it always exact: what if plates A and B have
> different shapes? Or what if I imagine distorting the wire between plates B
> and C, so that B and C are both portions of some larger object, say the two
> ends of a solid rectangular block, or even a sphere? Surely at some point
> the answer will no longer be -/+Q. At what point - in other words, what
> assumptions go into the usual derivation? I've looked in several textbooks
> and it's presented as though -/+Q is patently obvious. Carl

Making it plain to anyone who read his message to the end that he was interested in the *reason* for the result; in what its limits of applicability, if any, are; and *not at all* in what happens if the capacitors are initially charged. Frankly, I wouldn't be at all surprised to find that Carl, having landed a position as a professor of physics in a university physics department, knows how to solve simple introductory physics problems like the one you posed.

Brian goes on to write:

> The original questioner was interested to know why the charge at point
> B and C have to be numerically equal, as quoted above.

Again, no. Please take the time and have the courtesy to read his post in its entirety.

Quoting David Bowman, Brian goes on:

> What difficulty?
> The difficulty of erroneously assigning equal charge on connected plates of
> series caps.

Since the net charge on the connected plates was irrelevant to Carl's central question, he explicitly made the simplifying assumption that it was zero. Read his post.

Brian goes on:

> I may have mentioned recently that I do NOT subscribe to the
> orthodox ( I deliberately do not say "old-fashioned") view
> that there are "natural laws" that scientists discover;
> I find it much more productive to suppose that there are man-made models -
> any of which may be more or less suitable to any particular case.
> This has the particular virtue that when physicists comfortably think they
> are dealing on a 'more-fundamental' level, I can as easily see the flaws in
> their models as in any other man made elaboration.

Very interesting.

Continuing to quote David Bowman, Brian writes:

> Let u = [mu] = 10^(-6)
> Q_1 = 30 uC (Q_A = -30 uC, Q_B = +30 uC), Q_2 = 60 uC (Q_C = +60 uC,
> Q_D = -60 uC)
> So you can now easily admit a case where Qb is not numerically equal to Qc
> This was the crux of Carl's question, was it not?

No, quite clearly it was not.

> It was Cockroft and Walton who defined this arrangement for a
> voltage multiplier in the earliest accelerator experiments.
> They were physicists as I recall.
> But I don't suppose their names are known to teachers hereabouts.

Why on earth would you suppose such an absurdity?

> (On careful deliberation, I have deleted the putative reason I offered
> for this. I hope you will accept this as a tribute to your private coaching
> in how to prevent public shows of sniping)

And I must apologize for *this* public show of sniping. I guess I get a little exercised by remorseless public shows of nonsense.

John Mallinckrodt

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Brian Whatcott (among other things) wrote:

> The original questioner was interested to know why the charge at point
> B and C have to be numerically equal, as quoted above.
> The only person who explicitly contradicted this equality
> (as far as I recall ) was John Mallinckrodt.
> You for example merely modelled the oscillation which could precede
> the new steady-state, did you not?

At the risk of being accused of a public show of sniping I must mention that this is not correct. Although my post did mention some of the details that must occur for the charge re-equilibration process to proceed, the point of it was to show how and why the second law makes the system minimize its macroscopic electrostatic potential energy in equilibrium. I *did* discuss a case for which the charges at B and C were not (oppositely) equal to either each other or to the charges on the other cap plates at A and D. This was an explicit consideration of an extreme geometry case similar to that which John Mallinckrodt considered and to which Carl Mungan asked about. I wrote:

>If the geometry of
>the setup is distorted so the distance between B and C was comparable to or
>less than the capacitor gaps which would then have a huge fringe field
>volume, and if the geometry of the system significantly broke any left-right
>symmetry that the system might possess, then the values of the charges would
>not necessarily split up into exactly -Q and +Q on plates B and C but there
>would also be some partial charge distributed along the surface of the
>conducting connector between them.
>To solve for the final charge distribution
>in this case would entail, in general, solving Laplace's equation over all
>space with the requirement the the potential gradient at all metallic
>surfaces be locally perpendicular to the surface. This will zero out the
>electric field in the interior of the metal.

BTW, my explanatory framework has no difficulty let alone an "overwhelming difficulty in explaining the reasonable case of two caps of different C value, precharged to the same voltage." The reason that I did not happen to consider such a precharged case was that it was not a concern of the original questioner.

I'll forgo any further public sniping here since John Mallinckrodt did such a fine job of it in his previous post.

David Bowman

------------------------------

Its early in the morning, so this is with some hesitancy, but hey, I've already embarrassed myself.

David remarked, concerning the requirement that all field lines starting on plate A must terminate on plate B:

"Notice that this requirement of no net excess electric flux external to each of the capacitors is equivalent to a requirement of a vanishing field in the region between plates B and C."

A very minor quibble, they are not precisely equivalent, as an odd field line or two may leave plate A and terminate on plate B by first going around plate B and entering that region between plates B and C.

For ideal infinite parallel plate capacitors, they are precisely equivalent (as David, quite quickly and correctly, mentioned in pointing out one of my errors). And I presume that in most reasonable geometries (not extreme geometries) they will be very very close to being equivalent, as almost all field lines would be in the region between plate A and B.

Joel Rauber

------------------------------

> Care to try?
> Procedure:
> Connect cap 1 and 2 to battery 3.
> Leads B and C to pos terminal
> Leads A and D to neg terminal.
> Q's ???

One more qualification is necessary, if one wants to be "practical." The time during which the circuit is closed must be very short in comparison with R1*C1 and R2*C2. This was implicitly stated in the first message (leakage resistances R1 and R2 have no effect). The engineers know about the effect of leakages very well; their d.c. circuit rule is "not to use capacitors in series unless a parallel resistor is connected to each of them to control the distribution of charges."

But physics textbook writers let us believe that Q1 and Q2 are equal in the equilibrated (steady) state. We discussed this issue three years ago under the subject "a myth about capacitors in series." How can I resist reminding you?

The basic reference is: A.P. French, The Physics Teacher, May 1993. Also read E.D. Noll, "Capacitors in series: A laboratory activity to promote critical thinking," Physics Education, November 1996.

Ludwik Kowalski

P.S. In my opinion "public snipings" and "embarrassing comments" are perfectly OK as long as we know that everybody is motivated by the concerns which unite us on this list. We learn from each other and this is the most important.

------------------------------

Joel, you are quite correct about the stray fringe field lines. I was being somewhat sloppy in my writing. It's hard to say things in a way that is both lucid and fully qualified against all possible contingencies and complications. I was considering the infinite plate short plate separation distance approximation. I also did not include any perturbing effects on the charge distribution due to other possible charges and fields from other nearby objects. For instance, I did not include the perturbative effect of the field of the battery and of the induced surface charges on the wires that connect it to plates A and D either. (I also did not include the fact that charge is quantized and not infinitely divisible. Nor did I include leakage effects between the capacitor plates either.)

David Bowman

------------------------------

I see Ludwik finally checked into the discussion with his favorite objection to textbook treatments of series capacitors that give no hint that leakage effects drastically affect the steady state results in realistic (i.e. leaky) capacitors.

How did you ever hold your tongue (er, keyboard keys) for so long?

David Bowman

------------------------------

>In my opinion "public snipings" and "embarrassing comments" are
>perfectly OK as long as we know that everybody is motivated by the
>concerns which unite us on this list. We learn from each other and
>this is the most important.

Hear, hear! This is academic debate. Our style may not be correct "netiquette," but many of us are utterly fed up with artificial correctness. I'll join your party, Ludwik.

Leigh Palmer

------------------------------

Postscript: Since this series of postings, I have come across another paper which has an excellent discussion of this topic.

B.L. Illman and G.T. Carlson, "Equal plate charges on series capacitors?" TPT vol. 32, p. 77 (1994). Also see the follow-up letter by A.P. French, "Another in our series on capacitors," TPT vol. 32, p. 262 (1994).

Cheers, Carl Mungan