# Capacitors in series

1. Sep 12, 2008

### 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.

2. Sep 12, 2008

### 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.

3. Sep 12, 2008

### 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.

4. Sep 12, 2008

### Staff: Mentor

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?

5. Sep 12, 2008

### 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.

6. Sep 12, 2008

### atyy

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.

7. Sep 12, 2008

### 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.

8. Sep 12, 2008

### 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.

Last edited: Sep 12, 2008
9. Sep 12, 2008

### Staff: Mentor

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.

10. Sep 12, 2008

### Phrak

I am assuming no net charge.

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?

Last edited: Sep 12, 2008
11. Sep 12, 2008

### atyy

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=CTdVT/dt=CT[d(V1+V2)/dt]=CT[I/C1+I/C2]

Take the first and last terms in the above equalities and rearrange:
CT=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:tongue:).

Last edited: Sep 12, 2008
12. Sep 13, 2008

### 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.

13. Sep 13, 2008

### atyy

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.

Last edited: Sep 13, 2008
14. Sep 13, 2008

### ibc

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.

15. Sep 13, 2008

### Defennder

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

16. Sep 13, 2008

### atyy

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)

17. Sep 13, 2008

### atyy

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.

18. Sep 13, 2008

### 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...?

19. Sep 13, 2008

### atyy

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.

20. Sep 13, 2008

### 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.