Capacitors in series and Kirchoff's law

[ibc]

I seems very simple, and I'm sure I understood it before, but for some reason I suddenly started to get confused with it

when we want to know the capacitance of 2 capacitors in series, we assume they both have the same charge on them, so I just want an explanation to why does it have to be this way.

 

[atyy]

Because of Kirchoff's law that charge doesn't accumulate at any junction in a circuit. The point between two capacitors is a junction, and so the total charge there should be zero.

 

[ibc]

Well, do you have something better than Kirchoff's law =X?
I mean Kirchoff's laws aren't that... absolute, more like a way to make electric circuits calculations simpler.
What I'm looking for is an explanation by the electric forces why the charge must be equal in both

lets say there is a battery connected to 2 capacitors in series, and we wait a while and the whole thing comes to it's equilibrium, I don't see anything wrong with 1 capacitor having +q charge on one plate and -q on the other, and the other one having +Q, -Q, besides, it seems more reasonable that their potentials will be equal, since they are connected by a conductive wire.

 

[jtbell]

Consider the segment of wire that connects the two capacitors directly (i.e. not through the battery). When the capacitors are uncharged, the total net charge in the wire and on the plates that it is connected to (one plate on each capacitor) is zero.

Suppose you were able to charge the capacitors in such a way that one has twice the charge of the other. Then the plate that is connected to one end of the connecting wire would have (say) +Q whereas the plate (on the other capacitor) connected to the other end would have -2Q. The connecting wire plus the two plates would now have a net charge of -Q instead of zero. Where would that charge come from?

 

[ibc]

conservation of charge still holds, since you still have the other 2 plates, which are not connected to the wire, one with charge +2Q, and one with charge -Q, the total amount of charge is still zero.

 

[atyy]

it seems more reasonable that their potentials will be equal, since they are connected by a conductive wire.

Yes, that is much more fundamental. The statement requires the assumption of identical geometry of both capacitors - or whatever the normal circuit theory approximations are. Another one is that energy is conserved in the circuit when a capacitor is charged or there is AC current. This cannot be fundamentally true, since you are changing the electric field of the capacitor, which should create a changing electromagnetic field and radiate energy away. These deviations go under the name of stray capacitance, stray inductance etc.

 

[atyy]

Actually, capacitance, as a property of the conductor, say as opposed to being a property of the conductor and field, doesn't even exist in the most general situation in classical electromagentism. C=Q/V, but V is a full description of the electric field only in electrostatics.

 

[Phrak]

Two capacitors in series don't have to have equal charge.

In the ideal world of circuit analysis it is implicitly assumed that the initial charge on each is equal (and usually zero) for two capacitors in series. Charge conservation then requires that the charge remains equal over time for an ideal capacitor (no parallel resistance or even sources of emf like in the real world). That charge that leave the plate of one ideal capacitor accumulates on the next.

 

[Dale]

Two capacitors in series must have equal charge by conservation of mass (and the assumption of a circuit with no net charge).

I understand what you are saying Phrak, but I think it is not helpful and will only confuse the OP.

 

[Phrak]

Two capacitors in series must have equal charge by conservation of mass (and the assumption of a circuit with no net charge).

I am assuming no net charge.

I understand what you are saying Phrak, but I think it is not helpful and will only confuse the OP.

Correct me if I'm wrong. I charge a 1 Farad capacitor to one volt. On each plate is plus or munus one Coulomb of charge. I connect it to another capacitor that is discharged--call it one Farad capacitance for definiteness. No charge moves until the circuit is complete.

Compete the circuit with a 1 volt battery. The charge distribution stays the same.

How is this confusing or wrong?

 

[atyy]

when we want to know the capacitance of 2 capacitors in series, we assume they both have the same charge on them, so I just want an explanation to why does it have to be this way.

Ahhh, I am now having exactly the same trouble you have - how do we state precisely the assumptions being made here.

It has something to do with the idea that for an isolated conductor in an electrostatic situation, the capacitance is a property of the body, independent of charge distribution and electric field strength, ie. the capacitance sums up everything we know about the geometry of a conductor.

And there's also an additional assumption that in circuit theory that the elements are independent. So for example, two resistors at different parts of the circuit are two conductors at different potentials, and do in principle form a capacitance. But we ignore this and treat the resistors as always being pure resistors.

One way to "avoid" the equal charge assumption is to think of Kirchoff's current laws as fundamentally true (in the realm of circuit theory, don't imagine that the junction law is enforcing charge conservation).

C=Q/V
Q=CV
I=C(dV/dt). This step is illegal, because the definition of capacitance only holds under electrostatic conditions, and we need the elecromagnetic four-potential, not just the scalar potential, in an electrodynamic situation. Nonetheless, for circuit theory, let's take this as our fundamental definition of capacitance.

In series:
I=C₁(dV₁/dt)=C₂(dV₂/dt)

Define total capacitance by:
I=C_TdV_T/dt=C_T[d(V₁+V₂)/dt]=C_T[I/C₁+I/C₂]

Take the first and last terms in the above equalities and rearrange:
C_T=I/[I/C₁+I/C₂]=C₁C₂/(C₁+C₂)

But of course, it's dangerous, engineers and experimentalists all have to be careful about when the assumptions of circuit theory fail (did I just imply something about theorists).

 

[ibc]

I understand that atyy, that's why I asked for the specific case when the ciruit reaches equilibrium, then we have no current, and we have to assume the charge is equal in both to get that result.
I guess the question is: if I connect a battery to 2 capacitors in series (and perhaps a resistor aswell), after some time it reaches equilibrium when V=V1+V2, and there's no current anymore. So I have 1 equation, but in order to know the charge or the voltage on each capacitor, I need another equation, which is (as I know it) that they have equal charge, but I don't understand why does it have to be this way.
DaleSpam: "Two capacitors in series must have equal charge by conservation of mass " - my question is: why does it HAVE TO be this way?
Beside that, I didn't quite understand why is it also dependent of the initial charge of the capacitors, since we are talking about when the circuit reaches equilibrium, I don't see how the equilibrium depends on the initial conditions (by the definition of equilibrium)
Though I think by that I start to understand, you all see the charge distribution as something that flows through the circuit and therefore must be distributed equally in all in-series capactiros, but what I'm looking for is an explanation to why in the steady equilibrium state, it has to be that way, regardless of what the circuit has been through, what holds the charges in each capacitors and makes them be equally charged, instead of making their voltage equal, as usualy happens when 2 charged objects are connected by a conductive wire (atyy said it depends on the capacitors geometry, though (regardless of if circuit or not) if you connect 2 charged object with no matter what geometry, what will happen is equalization of their potentials)

And I understand all what you are saying about simplifying unrealistic assumptions, but I am talking about simlpe classic circuit.

 

[atyy]

I understand that atyy, that's why I asked for the specific case when the ciruit reaches equilibrium, then we have no current, and we have to assume the charge is equal in both to get that result.
I guess the question is: if I connect a battery to 2 capacitors in series (and perhaps a resistor aswell), after some time it reaches equilibrium when V=V1+V2, and there's no current anymore. So I have 1 equation, but in order to know the charge or the voltage on each capacitor, I need another equation, which is (as I know it) that they have equal charge, but I don't understand why does it have to be this way.
DaleSpam: "Two capacitors in series must have equal charge by conservation of mass " - my question is: why does it HAVE TO be this way?
Beside that, I didn't quite understand why is it also dependent of the initial charge of the capacitors, since we are talking about when the circuit reaches equilibrium, I don't see how the equilibrium depends on the initial conditions (by the definition of equilibrium)
Though I think by that I start to understand, you all see the charge distribution as something that flows through the circuit and therefore must be distributed equally in all in-series capactiros, but what I'm looking for is an explanation to why in the steady equilibrium state, it has to be that way, regardless of what the circuit has been through, what holds the charges in each capacitors and makes them be equally charged, instead of making their voltage equal, as usualy happens when 2 charged objects are connected by a conductive wire (atyy said it depends on the capacitors geometry, though (regardless of if circuit or not) if you connect 2 charged object with no matter what geometry, what will happen is equalization of their potentials)

And I understand all what you are saying about simplifying unrealistic assumptions, but I am talking about simlpe classic circuit.

The derivation I gave says nothing about steady state, so it should hold it steady state too.

And the derivation also shows that the "simple classic circuit" is full of simplifying unrealistic assumptions.

I agree if you take one big capacitor with charge Q and one small capacitor with charge q, and connect them, they will not have equal charge at steady state. The connected plates will be equipotential, and the charge distribution depends on the geometry and relative spatial positions of the plates.

 

[ibc]

The derivation I gave says nothing about steady state, so it should hold it steady state too.

And the derivation also shows that the "simple classic circuit" is full of simplifying unrealistic assumptions.

I agree if you take one big capacitor with charge Q and one small capacitor with charge q, and connect them, they will not have equal charge at steady state.

So what is the difference in both cases?
How come in the circuit case we don't get equal potentials?

And I understand that we can take your derivation and generalize it to steady state, though as I said, I am looking for a steady state explanation, by forces / potentials / etc. which will prove that it must be that way.

 

[Defennder]

I agree if you take one big capacitor with charge Q and one small capacitor with charge q, and connect them, they will not have equal charge at steady state.

Why wouldn't they have the same charge at steady state? They're in series, are they not?

 

[atyy]

So what is the difference in both cases?
How come in the circuit case we don't get equal potentials?

And I understand that we can take your derivation and generalize it to steady state, though as I said, I am looking for a steady state explanation, by forces / potentials / etc. which will prove that it must be that way.

The circuit case is different because of the simplifying assumptions made in the derivation. Since we apply the derived equation for arbitrary time-varying currents and voltages, including all the currents until the time steady state is reached, the simplifying assumptions are also enforced at steady state.

It appears that the simplifying assumption that is "at fault" is that circuit elements remain "independent". The charge distribution on a system of conductors in general depends on the geometry and placement of conductors throughout all space. An example is when you tune your radio (which is adjusting a capacitor), and then the moment you walk away, it becomes slightly untuned. This is because you are a conductor, and your exact position in the room, determines the charge on the capacitor in the radio, even though you are not connected "in a circuit". (I think - Marten, tiny-tim and I had a discussion about this a few weeks ago, and we were all quite confused at some point, but I think we got it sorted out)

 

[atyy]

And I understand that we can take your derivation and generalize it to steady state, though as I said, I am looking for a steady state explanation, by forces / potentials / etc. which will prove that it must be that way.

Maybe such an explanation doesn't exist. The "equal charge" statement is not used to prove the formula for capacitors in series. Rather, the formula for capacitors in series is derived under the simplifying assumptions, and then assumed to hold for all currents and voltages, even steady state. And it is the formula for capacitors in series that leads to the "equal charge" statement.

 

[ibc]

So you're saying that in reality the capacitors do have equal potentials and not equal charge, and it's just a mistake in our calculations?

But how can it be that our simplifying assumptions gives a solutions which can be extremely far from reality, in that case wouldn't we use different assumptions...?

 

[atyy]

So you're saying that in reality the capacitors do have equal potentials and not equal charge, and it's just a mistake in our calculations?

But how can it be that our simplifying assumptions gives a solutions which can be extremely far from reality, in that case wouldn't we use different assumptions...?

Yes, I've always wondered why circuit theory works, and gives no sign that we are doing something wrong. For example, in charging a capacitor, we are obviously losing energy through electromagnetic radiation, and circuit theory must be wrong. But circuit theory itself does not betray any logical inconsistency. We only discover that it is wrong when we compare it against the full Maxwell equations.

But it is not that uncommon. Newtonian gravity is wrong, but it is a beautifully mathematically consistent theory.

 

[ibc]

I think there is a difference, Newtonian gravity starts to give bad results we approaching the speed of light or strong gravitational fields, so you can stay that cicuit theory is wrong when you have rapidly changing electric fields, which causes strong magnetic fields.
But in our case it's not about that, in both calculations, the one which determines the charges will be equal and the one which determines the potentials will be equal, are based on the same simple non-magnetic assumptions.

 

[jtbell]

conservation of charge still holds, since you still have the other 2 plates, which are not connected to the wire, one with charge +2Q, and one with charge -Q, the total amount of charge is still zero.

The only way to get this situation (given my initial setup) is for charge to flow through the capacitors, from one plate to the other, across the dielectric. But dielectrics are insulators! If a dielectric does conduct charge, which can happen if we make the potential difference large enough, the capacitor has broken down and is no longer a capacitor.

 

[atyy]

But in our case it's not about that, in both calculations, the one which determines the charges will be equal and the one which determines the potentials will be equal, are based on the same simple non-magnetic assumptions.

Yes, I think it is the "independence" assumption that is at fault.

In both the full electrostatic and circuit theory calculation, the connected plates are equipotential, so there's no problem there. The problem seems to be the assumption that the total capacitance is independent of the exact position of every conductor.

 

[ibc]

Yes, I think it is the "independence" assumption that is at fault.

In both the full electrostatic and circuit theory calculation, the connected plates are equipotential, so there's no problem there. The problem seems to be the assumption that the total capacitance is independent of the exact position of every conductor.

But after this independent assumptions, the connected capacitors are no longer equipotential.
So again, why would you make such assumption, if it's clear that it can give harshly wrong results, and there's a better assumption, which is just as simple, and gives good results.
As far as I know it it's a world-wide accepted theory, seems kind of weird that no one cares that it's not true... (and again, still not regarding magnetic effects and such things that make it less accurate, just the simple theory)

And just to make sure again that I understand correctly:
You are saying that in reality these capacitors WILL BE EQUIPOTENTIAL (regardless of magnetic effects and stuff), and therefore they WILL NOT HAVE THE SAME CHARGE ON THEM (as long as their capacities are different of course), and all these attempts above to explain why they can't have different charges are not true?

 

[Phrak]

this thread seems to be driven by misconceptions.

ibc, you do realize that the net charge on a capacitor is zero, don't you?

 

[Defennder]

I got lost after the first page. Now I have no idea whether they even accept the ideal condition assumptions of circuit theory when explaining why capacitors in series have the same charge.

 

[atyy]

And just to make sure again that I understand correctly:
You are saying that in reality these capacitors WILL BE EQUIPOTENTIAL (regardless of magnetic effects and stuff), and therefore they WILL NOT HAVE THE SAME CHARGE ON THEM (as long as their capacities are different of course), and all these attempts above to explain why they can't have different charges are not true?

Yes.

The attempts to explain why in circuit theory they have the same charge must lie in the approximations made for circuit theory.

I believe the approximation that is "at fault" is: the electric field does not depend on the exact position on all conductors in space, and that we can move the circuit elements independently, without affecting the electric fields in the vicinity of other circuit elements.

Edit: Note that in both circuit theory and the full electrostatic theory the connected plates are equipotential, so that is not the problem.

 

[atyy]

Some more comments about the assumptions:

C=Q/V
Q=CV
I=C(dV/dt). This step is illegal, because the definition of capacitance only holds under electrostatic conditions, and we need the elecromagnetic four-potential, not just the scalar potential, in an electrodynamic situation. Nonetheless, for circuit theory, let's take this as our fundamental definition of capacitance.

Tha above illegal step also contains the "circuit assumption" that Phrak and DaleSpam mentioned, and provides the interface to the assumptions of Kirchoff's laws. In addition to the illegal moves I already mentioned, another illegal move here was that we have made a current go across the capacitor, which of course doesn't happen. It only appears that way in a closed circuit (with changing potential difference). So this is a very crucial illegal step.

In series:
I=C₁(dV₁/dt)=C₂(dV₂/dt)

I believe this is where the "independence" assumption is taken, and we have treated the capacitance (and the surrounding electric field) of each capacitor as if no other conductors were present.

 

[Defennder]

I don't understand why you are complicating the explanation unnecessarily. I looked up the explanation in a freshman physics text and the one they gave was that the +Q charge induced on one plate would attract -Q charge on the other side of the plate. This happens on both sides of the capacitor plates on the extreme ends of the series arrangement and would in turn cause the charges on all the capacitors sandwiched between the two to be the same.

Instead, what you guys appeared to do was to drag in a lot of secondary (and I should say tertiary) factors to argue how the charges on series caps might not be equal and how factors such as distances between different capacitor plates and other conductors as well as other circuit elements might make it unequal. My first thought was that if want to consider all those irrelevant factors you might as well drag in the whole theory of transmission lines and extend this discussion to a few more pages to show how it might not be equal.

My first impression was that the question has already been answered by jtbell, and before he started delving into secondary factors, atyy himself. The question is, what level of complexity do you guys want to take this discussion to? Just how many more hundreds of other factors would you want to consider just to show that the charges may not be roughly equal?

 

[Dale]

Wow, this thread has really degenerated into a morass of confusion, complication, and misinformation.

First, circuit analysis is founded on a bunch of assumptions and conventions. The assumptions include that the size of the circuit is small, ideal components, etc. When these assumptions are combined with conservation of mass you get Kirchoffs Current Law and node voltage analysis, and when combined with conservation of energy you get Kirchoffs Voltage Law and mesh current analysis. One convention is that all circuit elements begin in their "ground" state unless explicitly indicated.

Phrak's comments are explicitly considering violations of the standard conventions and atyy's comments are explicitly considering violations of the standard assumptions. So IMO they are just adding unnecessary confusion to a simple situation. E.g. if one capacitor is initially charged and the other is not, then it is probably not be reasonable to consider them as a single equivalent series capacitor (the equivalent circuit would have a switch and voltage source in parallel to the charged capacitor).

Second, when you want to determine the equivalent capacitance of a series combination of capacitors you are free to analyze them in whatever circuit you choose. The natural circuit to perform the analysis is the one that begins with both capacitors in the conventional uncharged ground state. In this situation it is easy to see that the charge across each capacitor is equal by applying KCL at the node between the two capacitors. Current leaving one side of the node must enter the other side of the node.

Finally, it is most definitely not correct that the voltage across two capacitors in series is the same. Remember, in a capacitor i = C dv/dt. Two circuit elements in series share the same current, not the same voltage. So if the capacitors are different then the dv/dt will be different and, over time, will integrate to a different voltage.

 

[atyy]

Finally, it is most definitely not correct that the voltage across two capacitors in series is the same.

We did not come to this conclusion.

I agree the discussion is confusing. It is standard, but confusing, terminology to agree that a "charged capacitor" also "has no charge", the difference being whether one considers one plate or both plates. Unfortunately, similarly confusing terminology was at work in this discussion.

The discussion concerned why the standard assumptions of circuit theory lead to a result that violate Maxwell's equations.

if one capacitor is initially charged and the other is not, then it is probably not be reasonable to consider them as a single equivalent series capacitor (the equivalent circuit would have a switch and voltage source in parallel to the charged capacitor).

One point of discussion was indeed why two capacitors in series, one initially charged and the other not, may be considered a single equivalent capacitor, and may be considered so until steady state is reached.

 

[Dale]

We did not come to this conclusion.

Your conclusion was wrong.

Two circuit elements in parallel have the same voltage across them. Two circuit elements in series have the same current through them.

 

[atyy]

Your conclusion was wrong.

Two circuit elements in parallel have the same voltage across them. Two circuit elements in series have the same current through them.

I should have said, "We did not come to the conclusion that the voltage across two capacitors in series is the same".

 

[Dale]

I should have said, "We did not come to the conclusion that the voltage across two capacitors in series is the same".

Unfortunately your comments have been so confusing and complicated that this is exactly the wrong conclusion that you have mislead ibc to.

You are saying that in reality these capacitors WILL BE EQUIPOTENTIAL (regardless of magnetic effects and stuff), and therefore they WILL NOT HAVE THE SAME CHARGE ON THEM (as long as their capacities are different of course), and all these attempts above to explain why they can't have different charges are not true?

 

[atyy]

Edit: Note that in both circuit theory and the full electrostatic theory the connected plates are equipotential, so that is not the problem.

I should have said, "We did not come to the conclusion that the voltage across two capacitors in series is the same".

Unfortunately your comments have been so confusing and complicated that this is exactly the wrong conclusion that you have mislead ibc to.

The two quotes of mine above are not contradictory, but I agree they may appear to be. ibc and I were not talking about the potential difference ('voltage') across a capacitor.

 

[atyy]

Two capacitors in series don't have to have equal charge.

In the ideal world of circuit analysis it is implicitly assumed that the initial charge on each is equal (and usually zero) for two capacitors in series. Charge conservation then requires that the charge remains equal over time for an ideal capacitor (no parallel resistance or even sources of emf like in the real world). That charge that leave the plate of one ideal capacitor accumulates on the next.

E.g. if one capacitor is initially charged and the other is not, then it is probably not be reasonable to consider them as a single equivalent series capacitor (the equivalent circuit would have a switch and voltage source in parallel to the charged capacitor).

ibc, I think you are right that the independence assumption is not primarily at fault, but more the initial conditions. The derivation of the formula for capacitors in series in #11 only contains the current (dQ/dt) and the change in voltage (dV/dt). Hence the actual charge and voltage probably cannot be uniquely determined from that formula alone, and presumably require the specification of initial conditions. Thus the formula for capacitors in series does not erroneously indicate the "equal charge statement", contrary to what I previously stated. Instead the "equal charge statement" presumably holds only under certain initial conditions as Phrak and DaleSpam have indicated.

 

[Dale]

The equal charge does only hold under certian assumptions, including the standard circuit analysis assumptions, but the formula for two capacitors in series can be derived without it.

i = C_1 \: \frac{dv_1}{dt} (1)

i = C_2 \: \frac{dv_2}{dt} (2)

v = v_1 + v_2 (3)

Solving (1), (2), and (3) for i in terms of v we obtain

i = \frac{C_1 \: C_2}{C_1 + C_2} \: \frac{dv}{dt} (4)

which is the formula for a single capacitor of capacitance

C = \frac{C_1 \: C_2}{C_1 + C_2} (5)

 

[schroder]

I seems very simple, and I'm sure I understood it before, but for some reason I suddenly started to get confused with it

when we want to know the capacitance of 2 capacitors in series, we assume they both have the same charge on them, so I just want an explanation to why does it have to be this way.

I think the original question was answered quite well by jtbell, way back in post number 5 on the first page. That is not a criticism of the rest of the posts, as that is what good science is all about: asking questions, especially “Why” and then answering that. Why the charge has to be equal on two or more capacitors connected in series may be best answered by turning the question around and asking “Is it possible to have different charges on two capacitors connected in series, in a complete circuit?” The answer to that question is “No”. When you connect two caps in series, the plates that are connected via the conducting leads become One plate, electrically speaking, and that is the language we are speaking here today. Capacitance is the capacity to hold charge. If you now push a current through the series connection and fill that one plate with a charge of one coulomb, for example, then one coulomb of charge exists across both capacitors by virtue of sharing that one plate. For example, if you place a 0.1 farad cap in series with a 0.01 farad cap and charge the center plate to one coulomb, then each cap will have a charge of one coulomb across it. The voltage each cap develops is determined by V = q/C, so the 0.1 f cap develops 10 Volts across it and the 0.01 cap develops 100 Volts. As you can see, the voltage is not equal but the charge is equal. You can extend this model to include any number of caps you wish keeping in mind that where the leads are connected together, there is effectively only one plate in play.

 

[Phrak]

The hydrolic model for electric circuits lends itself very well to simple LRC curcuits.

With it, all this confusion would vanish. Voltage is pressure, Charge is a volume of water. Wires are pipes. Capacitors are rubber membranes stretched across a wide spot in the pipe. Switches are valves.

The charge on a capacitor is the volume of water displacing a membrane.

Now it becomes of a simple question of what happens to the water trapped between two membranes.

With this simple model applied it becomes obvious, Schroder that the answer is "no" in the most naive initial conditions where both membranes are initially at neutral, but in general, the answer is otherwise.

Dale. I don't agee that these minor swindles while teaching a subject are all that wonderful. There have been at least 4 other times on this forum I recall running into this sort of misinformation become a widely held belief widely defended.

 

[Dale]

Dale. I don't agee that these minor swindles while teaching a subject are all that wonderful.

Personally, I think simplifying assumptions and conventions are "all that wonderful" and I don't know why you would characterize them as "swindles".

 

[ibc]

I agree that this whole thread is filled with confusion and misunderstanding, so I will try to organize some things and clear some mess.

1) As I mentioned before, although there were several posts regarding electromagnetic effects which are not part of the circuit theory assumptions and simplifications, my questions did not regard these effects, neither atyy and myself's discussion about the theory's mismatch to reality, we were only using simple electricity calculations.
2) Most of that discussion was about that the 2 capacitors being equipotential, since they are connected by a conducting wire, but now I think this statement is a bit problematic, since only one plate from each capacitor is connected, and it seems like talking about their optential is somewhat meaningless, because we assume both plates are in a uniform field, which only exists inside the capacitor.
3) What I kept asking for, is an explanation to why 2 capactiors, when being connected by a conducting wire, and the example I gave is that both capactiors are far away.
I now understand that one of the circuit assumptions (as one mentioned) is that it must be small, or in that case that the capacitors are close enough together, and indeed we see that if we get the 2 capactiors close enough together, until their plates converge, they act as one capacitor, with capacitance as mentioned in the calculations.

4) So even though this small-distance assumption leads to a certain total capacitance, which we can get under the assumption the 2 capacitors have the same charge (under certain initial conditions), suggests that the second part is true, and that there is a 3rd way to get this same conclusion, which is the calculation many of you made, with the corrents and V=V1+V2, which gives more reason to believe it's true, you are missing the point of my initial question, I know and understand that current calculation, but I asked:
Assume we have a circuit with 1 battery and 2 capacitors in series (both capacitors starts with zero charge), after a while we get V=V1+V2, and therefore the circuit comes to stationary condition. I want a proof that in and by that stationary condition the charge is equal in both capacitors (there is no current in the circuit then, so you can't use the current calculation and say the same current is running through both of them).

I hope we can have a more organized and less confused discussion now =)

 

[Phrak]

Personally, I think simplifying assumptions and conventions are "all that wonderful" and I don't know why you would characterize them as "swindles".

OK. So the convensions are those we encounter in elementry circuit analysis.

What are the simplifying assumptions required for equality of charge?

I could scroll through this thread and accumulate the various wrong simplifying assumptions to justify the expected result...

I'm assuming that you do realize that the charge does not have to be equal with two capacitors in seriew within the conventions of elementry circuit analysis. You haven't said as much, so I don't really know.

 

[Dale]

OK. So the convensions are those we encounter in elementry circuit analysis.

What are the simplifying assumptions required for equality of charge?

Initially same charge (e.g. both initially uncharged).

I'm assuming that you do realize that the charge does not have to be equal with two capacitors in seriew within the conventions of elementry circuit analysis. You haven't said as much, so I don't really know.

I understand what you have been saying, but when someone says "two capacitors in series" they usually mean the circuit on the left, not something like the circuit on the right. You can call the circuit on the right "two capacitors in series", but it is definitely not conventional, and therefore not helpful to someone learning.

 

[Dale]

Assume we have a circuit with 1 battery and 2 capacitors in series (both capacitors starts with zero charge), after a while we get V=V1+V2, and therefore the circuit comes to stationary condition. I want a proof that in and by that stationary condition the charge is equal in both capacitors (there is no current in the circuit then, so you can't use the current calculation and say the same current is running through both of them).

Consider the attached circuit. This circuit has the following equations:

\begin{array}{l}<br /> \frac{V_B(t)-V}{R}+C_1<br /> \left(V_B&#039;(t)-V_A&#039;(t)\right)=0 \\<br /> C_2 V_A&#039;(t)+C_1<br /> \left(V_A&#039;(t)-V_B&#039;(t)\right)=0 \\<br /> V_A(0)=0 \\<br /> V_B(0)=0<br /> \end{array} (1)

Which has the solution:

\begin{array}{l}<br /> V_A(t) = \frac{C_1 \left(1-e^{-\frac{(C_1+C_2) t}{C_1<br /> C_2 R}}\right) V}{C_1+C_2} \\<br /> V_B(t) = V-e^{-\frac{(C_1+C_2) t}{C_1 C_2 R}} V<br /> \end{array} (2)

Which in the limit as t -> infinity goes to:
\begin{array}{l}<br /> V_A(\infty) = \frac{C_1 V}{C_1 + C_2} \\<br /> V_B(\infty) = V<br /> \end{array} (3)

So

\begin{array}{l}<br /> Q_1(\infty) = C_1 (V_B(\infty) - V_A(\infty)) = \frac{C_1 C_2 V}{C_1 + C_2} \\<br /> Q_2(\infty) = C_2 V_A(\infty) = \frac{C_1 C_2 V}{C_1 + C_2}<br /> \end{array} (4)

And therefore the charges are equal in the steady-state.

 

[atyy]

2) Most of that discussion was about that the 2 capacitors being equipotential, since they are connected by a conducting wire, but now I think this statement is a bit problematic, since only one plate from each capacitor is connected, and it seems like talking about their optential is somewhat meaningless, because we assume both plates are in a uniform field, which only exists inside the capacitor.

Ah, so everyone was right that we were not talking about the same thing at all! I was talking about the connected plates having the same potential (equipotential), not the two capacitors having the same potential difference across them. In both circuit theory and the full electrostatic theory, the connected plates have the same potential, but the capacitors have different potential differences. Note that V in C=Q/V is the potential difference, not the potential.

I don't understand why you are complicating the explanation unnecessarily. I looked up the explanation in a freshman physics text and the one they gave was that the +Q charge induced on one plate would attract -Q charge on the other side of the plate. This happens on both sides of the capacitor plates on the extreme ends of the series arrangement and would in turn cause the charges on all the capacitors sandwiched between the two to be the same.

As stated completely correctly by Defennder, in the simple case of an uncharged node, the excess charge transferred to one capacitor is equal to the deficit of charge from the other capacitor because of charge conservation at that node. It is important to realize that Kirchoff's laws are not just calculating conveniences. The node law is the statement of charge conservation, which is the answer to your question. The loop law is the assumption of a conservative electric field.

 

[Phrak]

Initially same charge (e.g. both initially uncharged).

I understand what you have been saying, but when someone says "two capacitors in series" they usually mean the circuit on the left, not something like the circuit on the right. You can call the circuit on the right "two capacitors in series", but it is definitely not conventional, and therefore not helpful to someone learning.

Your assumptions are that both the common node doesn't include other elements drawing current at any time, and that the initial conditions are q_1 = q_2.

 

[Phrak]

By the way, we put labels on the labels on the unterminated bubbles.

 

[ibc]

Consider the attached circuit. This circuit has the following equations:

\begin{array}{l}<br /> \frac{V_B(t)-V}{R}+C_1<br /> \left(V_B&#039;(t)-V_A&#039;(t)\right)=0 \\<br /> C_2 V_A&#039;(t)+C_1<br /> \left(V_A&#039;(t)-V_B&#039;(t)\right)=0 \\<br /> V_A(0)=0 \\<br /> V_B(0)=0<br /> \end{array} (1)

Which has the solution:

\begin{array}{l}<br /> V_A(t) = \frac{C_1 \left(1-e^{-\frac{(C_1+C_2) t}{C_1<br /> C_2 R}}\right) V}{C_1+C_2} \\<br /> V_B(t) = V-e^{-\frac{(C_1+C_2) t}{C_1 C_2 R}} V<br /> \end{array} (2)

Which in the limit as t -> infinity goes to:
\begin{array}{l}<br /> V_A(\infty) = \frac{C_1 V}{C_1 + C_2} \\<br /> V_B(\infty) = V<br /> \end{array} (3)

So

\begin{array}{l}<br /> Q_1(\infty) = C_1 (V_B(\infty) - V_A(\infty)) = \frac{C_1 C_2 V}{C_1 + C_2} \\<br /> Q_2(\infty) = C_2 V_A(\infty) = \frac{C_1 C_2 V}{C_1 + C_2}<br /> \end{array} (4)

And therefore the charges are equal in the steady-state.

Well thanks for finally answering my question =)
although I was thinking of a more simple circuit, and what I was looking for is an explanation without currents, I mean, for a reason that the charges will be equal, just by forces and potentials, but perhaps such reason does not exist.
but you did proove this point nicely
thanks

Ah, so everyone was right that we were not talking about the same thing at all! I was talking about the connected plates having the same potential (equipotential), not the two capacitors having the same potential difference across them. In both circuit theory and the full electrostatic theory, the connected plates have the same potential, but the capacitors have different potential differences. Note that V in C=Q/V is the potential difference, not the potential.

I did get a bit confused at first with the total potential and only one plate potential, though I wasn't talking about that, I just said that it's problematic to tak about these plates potentials, because we talk about uniform field that only exists inside the capacitor, and that it's problematic to choose a general point of perspective for both capacitors, to know the potential on each plate.

 

[Dale]

Well thanks for finally answering my question =)
although I was thinking of a more simple circuit, and what I was looking for is an explanation without currents, I mean, for a reason that the charges will be equal, just by forces and potentials, but perhaps such reason does not exist.
but you did proove this point nicely

You are welcome.

A more simple circuit does not exist. If you remove the resistor then you require an infinite current in order to instantaneously charge the capacitors when you throw the switch.

There is a non-current explanation: conservation of charge at the node between the capacitors. If the capacitors begin uncharged then any charge that is displaced on one side of the node must go to the other side of the node. The circuit example I posted arrives at the equal-charge result because the conservation of charge is built into KCL.

 

[epenguin]

(Supposed to represent a capacitor ---||--- with dielectric |)

---₁⁺|||^-₂---

Why Q₁ = - Q₂ cannot be the question, that is just conservation of charge.

Also for the same reason, below

---₁⁺|||^-₃---------------------------------------------------₄⁺|||^-₂---Q₃ = Q₄. But conservation of charge cannot be the reason why Q₃ = -Q₁. It is not immediately obvious, since we are debating it, what physical law would be broken if Q₄ were equal to –Q₃ but different from Q₁.
I think this was the original question of ibc??, couldn’t you have Q₁ = - Q₂ and Q₃ = -Q₄, but Q₄ something different from Q₁?

In a useful charged capacitor some work has been done (against the attractive force between opposite charges) to separate the charges. This has been made smaller by making the plate separation short and filling the gap with dielectric which diminishes the force between the separated charges, i.e. the work of charge separation. No matter if an electron really goes a long way round through wire and battery - the work is path-independent so is the same as transporting one through the dielectric from one plate to another .

In the case of 2 capacitors in series it is the same, this charge separation is equivalent to transport of a charges across both dielectrics. As long, that is, as each capacitor remains on balance neutral, i.e. charge on one plate equal and opposite to that on the other, i.e. for each electron moved from 1 to 2, one moves from 4 to 3. For charge inequality on the other hand, the work done is that of moving an electron from 1 to 2 without any compensation, through the air say rather than the dielectric, permitting practically the maximum back attraction over a relatively large distance, say of the order of centimeters instead of fraction of a millimetre. This is much more work that the previous case. The amount of charge imbalance for a given voltage this way is I think the same as that of the capacitor

---₁⁺|-------------------------air---------------------------------|^-₂--

under the same voltage – very little. That is, practically Q₁ = -Q₂ etc.

Well thanks for finally answering my question =)
although I was thinking of a more simple circuit, and what I was looking for is an explanation without currents, I mean, for a reason that the charges will be equal, just by forces and potentials, but perhaps such reason does not exist.

I think that also is a good question which I will leave to someone else. You actually want to do a lot of work in storing charges, i.e. you want capacitors to store a lot of energy. It may seem a bit paradoxical that you are storing more energy by making everything less intense, the charge density less by greater area, its separation less by small d, and the forces less by the dielectric.

 

[epenguin]

correction.
That sentence near the end I meant
"under the same voltage – very little. That is, practically Q₁ = -Q₃ etc."