Capacitors in series and Kirchoff's law

[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:

$$\begin{array}{l} \frac{V_B(t)-V}{R}+C_1 \left(V_B'(t)-V_A'(t)\right)=0 \\ C_2 V_A'(t)+C_1 \left(V_A'(t)-V_B'(t)\right)=0 \\ V_A(0)=0 \\ V_B(0)=0 \end{array}$$ (1)

Which has the solution:

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

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

So

$$\begin{array}{l} Q_1(\infty) = C_1 (V_B(\infty) - V_A(\infty)) = \frac{C_1 C_2 V}{C_1 + C_2} \\ Q_2(\infty) = C_2 V_A(\infty) = \frac{C_1 C_2 V}{C_1 + C_2} \end{array}$$ (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:

$$\begin{array}{l} \frac{V_B(t)-V}{R}+C_1 \left(V_B'(t)-V_A'(t)\right)=0 \\ C_2 V_A'(t)+C_1 \left(V_A'(t)-V_B'(t)\right)=0 \\ V_A(0)=0 \\ V_B(0)=0 \end{array}$$ (1)

Which has the solution:

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

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

So

$$\begin{array}{l} Q_1(\infty) = C_1 (V_B(\infty) - V_A(\infty)) = \frac{C_1 C_2 V}{C_1 + C_2} \\ Q_2(\infty) = C_2 V_A(\infty) = \frac{C_1 C_2 V}{C_1 + C_2} \end{array}$$ (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."

 

[atyy]

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.

Yes, that was puzzling to me too. The equilibrium distribution is dependent on the initial charge distribution, but fundamentally, it seemed to me that equilibrium for charged nodes could still be handled by the condition of equilibrium, equipotentiality of connected plates, charge conservation at each node, and the definition of capacitance for each capacitor in series (which if you read below may not be quite right). But the equations I got were unduly messy for such a simple situation.

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.

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.

And schroder's comments also indicated a good trick for the equilibrium distribution, but I had trouble getting the trick right with more capacitors

Then I found this interesting comment on Wikipedia: C=Q/V does not apply when there are more than two charged plates, or when the net charge on the two plates is non-zero. To handle this case, Maxwell introduced his "coefficients of potential". If three plates are given charges Q1,Q2,Q3, then the voltage of plate 1 is given by V1 = p11Q1 + p12Q2 + p13Q3, and similarly for the other voltages.
http://en.wikipedia.org/wiki/Capacitance

 

[atyy]

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.

Then I found this interesting comment on Wikipedia: C=Q/V does not apply when there are more than two charged plates, or when the net charge on the two plates is non-zero. To handle this case, Maxwell introduced his "coefficients of potential".

The coefficients of potential appear to matter only if the independence assumption is violated. If it is not, I think Phrak gives the correct statement of the two necessary assumptions for applying the capacitors in series formula. The following articles contain an amusing discussion about the violation of the first assumption (which I had not appreciated in my derivation in post #11):

A myth about capacitors in series
L. Kowalski
Phys. Teach. 26, 286 (1988)

Are the textbook writers wrong about capacitors?
A. P. French
Phys. Teach. 31, 156 (1993)

Equal plate charges on series capacitors?
B. L. Illman and G. T. Carlson
Phys. Teach. 32, 77 (1994)

Another in our series on capacitors
A. P. French
Phys. Teach. 32, 262 (1994)

 

[ibc]

The following articles contain an amusing discussion about the violation of the first assumption (which I had not appreciated in my derivation in post #11):

A myth about capacitors in series
L. Kowalski
Phys. Teach. 26, 286 (1988)

Are the textbook writers wrong about capacitors?
A. P. French
Phys. Teach. 31, 156 (1993)

Equal plate charges on series capacitors?
B. L. Illman and G. T. Carlson
Phys. Teach. 32, 77 (1994)

Another in our series on capacitors
A. P. French
Phys. Teach. 32, 262 (1994)

Do you know perhaps where we can find them? (internet, link, or whatever, I'm just not very familiar with knowing where do find such things =x )

 

[atyy]

Do you know perhaps where we can find them? (internet, link, or whatever, I'm just not very familiar with knowing where do find such things =x )

The articles can be found at:
http://scitation.aip.org/tpt/

Unfortunately they require a subscription, and they aren't on arXiv. So I will just quickly describe the most instructive situation I found, from Kowalski's and French's articles.

A) Consider 2 capacitors C1, C2 in series with a constant emf V across X, Z (node Y between the capacitors is labelled for convenience):
X---C1---Y---C2---Z,

B) Then consider a modification of the above setup, still with C1,C2 and V across X,Z, but now with R1 parallel to C1, R2 parallel to C2:
X---C1---Y---C2---Z
X---R1---Y---R2---Z

We start off with no charged nodes and no current flowing in both situations. In both A and B, there is no current flowing through C1,C2 at steady states. But in B, there is current through R1,R2 at steady state. This causes the steady state voltage between XY and YZ to be different in situations A and B. When one thinks about it carefully, the capacitors are not in series when the circuit is being charged, in the sense that the same current is not flowing through them, which is why the derivation in post #11 fails. Ironically, they are "in series" at steady state when no current flows through C1 and C2! So if we require capacitors in series to be in series at all times, then perhaps the statement of Phrak's first assumption is not strictly necessary (equivalent to the definition of capacitors in series). However it is definitely helpful. Kowalski comments that real capacitors have leak currents, which are modeled by R1, R2, and so one must be careful about modelling a real situation with ideal elements.

So I guess the checklist is:
1) Are the real elements are equivalent to the ideal elements?
2) Is the circuit small enough to apply circuit theory?
3) Are the elements sufficiently small and the distance between the elements sufficiently large to take the independence assumption?
4) Are the capacitors in series? (Phrak's first assumption in post #45)
5) What is the initial charge distribution at the nodes?
- If it is zero, then, for capacitors in series, the charge on connected plates will be equal and opposite, by conservation of charge. (Phrak's second assumption in post #45)
- If the nodes are charged, the equilibrium condition is given by the equillibrium voltage across the series, the equipotentiality of connected plates, charge conservation at each node, and the definition of capacitance for each capacitor in series.