# Uncharged capacitors connected in series

• vcsharp2003

#### vcsharp2003

Homework Statement
When two uncharged capacitors are connected in series as shown then ultimately all the plates of capacitors end up with same magnitude of charges. Why would this happen?
Relevant Equations
##C = \dfrac {q} {V} ##, where C is capacitance of capacitor, q is magnitude of charge on each plate of capacitor and V is the potential difference between the plates of capacitor
I came across the following explanation from the famous book of Sears and Zemansky which I am unable to understand. I can get the initial part where a positive charge goes to the top plate of C1 since the point a is at a +ve potential causing free electrons to transfer from top plate of C1 to the point a. But after that part, it gets confusing especially the part that says "until all of the field lines that begin on the top plate end on the bottom plate. This requires that the bottom plate have charge -Q. ". Why does charge transfer stop when the bottom plate has acquired a -Q charge?

Two capacitors are connected in series (one after the other) by conducting wires between points a and b. Both capacitors are initially uncharged. When a constant positive potential
difference Vab is applied between points a and b, the capacitors become charged; the figure shows that the charge on all conducting plates has the same magnitude.
To see why, note first that the top plate of C1 acquires a positive charge Q. The electric field of this positive charge pulls negative charge up to the bottom plate of C1 until all of the field lines that begin on the top plate end on the bottom plate. This requires that the bottom plate have charge -Q. These negative charges had to come from the top plate of C2, which becomes positively charged with charge +Q. This positive charge then pulls negative charge -Q from the connection at point b onto the bottom plate of C2. The total charge on the lower
plate of C1 and the upper plate of C2 together must always be zero because these plates aren’t connected to anything except each other. Thu, in a series connection the magnitude of charge on all plates is the same. Last edited:
• Delta2
the part that says "until all of the field lines that begin on the top plate end on the bottom plate. This requires that the bottom plate have charge -Q. ".
That argument doesn't really work. Suppose instead of connecting the capacitors to a battery we were connect them to two separate identical conducting spheres, but one with charge Q and one with charge -2Q. We would not then expect all the capacitor plates to have the same charge magnitude.

Instead, start with a battery and one initially uncharged capacitor. Whatever charge the battery passes to one plate it must pass an equal and opposite charge to the other.
Now see if you can extend it to a series of capacitors.

We would not then expect all the capacitor plates to have the same charge magnitude.
Yes, that's correct since the two points a and b would then have +Q and -2Q charges.

That argument doesn't really work. Suppose instead of connecting the capacitors to a battery we were connect them to two separate identical conducting spheres, but one with charge Q and one with charge -2Q. We would not then expect all the capacitor plates to have the same charge magnitude.

Instead, start with a battery and one initially uncharged capacitor. Whatever charge the battery passes to one plate it must pass an equal and opposite charge to the other.
Now see if you can extend it to a series of capacitors.

The explanation appears to say that when the bottom plate of C1 acquires a -Q charge, then the electric lines of force starting from positive top plate will end in negative bottom plate, which seems to make sense since there would be no electric field outside the top capacitor's two plates i.e. for C1and therefore no force to move electrons towards the bottom plate from the adjacent C2 capacitor's plate.

So, then my doubt is why would there not be any electric lines of force below the bottom plate of C1 when the charges are +Q and -Q on top and bottom plates of C1?

why would there not be any electric lines of force below the bottom plate of C1 when the charges are +Q and -Q on top and bottom plates of C1?
You seem to have answered that for yourself in the first para.
If there's Q on one plate and -Q on the other than all lines from one plate terminate at the other.

• vcsharp2003
You seem to have answered that for yourself in the first para.
If there's Q on one plate and -Q on the other than all lines from one plate terminate at the other.

The same book by Sears and Zemansky shows the electric field for a charged capacitor as in figure below.
So, clearly the electric field extends beyond the plates i.e. above and below the top and bottom plates. The field below the bottom plate cannot have any electrons flow into it due to the electric field direction and thus electron flow into bottom plate would stop at this stage. That argument doesn't really work. Suppose instead of connecting the capacitors to a battery we were connect them to two separate identical conducting spheres, but one with charge Q and one with charge -2Q. We would not then expect all the capacitor plates to have the same charge magnitude.

Instead, start with a battery and one initially uncharged capacitor. Whatever charge the battery passes to one plate it must pass an equal and opposite charge to the other.
Now see if you can extend it to a series of capacitors.

I think the electric field approach makes sense since free electrons can only move if a force due to electric field acts on them. Once an equal and opposite charge accumulates on the other plate, then the resulting electric lines of force as shown in post#6 cannot exert a net force on free electrons to cause them to move. So, electron transfer i.e. charge transfer stops.

Basically what we have here is Kirchhoff's current law : The currents ##I_1=\frac{dq_1}{dt}## and ##I_2=\frac{dq_2}{dt}## at each of the capacitors are equal since from Kirchhoff's current law we can get that ##I_1-I_2=0## and from this last equation if we integrate w.r.t time t we can get that ##q_1(t)=q_2(t)+C## where the constant C will be zero if at t=0 we have ##q_1(0)=q_2(0)=0 ## that is initially uncharged capacitors.

Kirchhoff's laws are often taken to be the axioms of circuit theory. The book attempts to give a somewhat more detailed explanation of why Kirchhoff's current law is not exactly an axiom but can have a deeper explanation at least in this case of capacitors. The way I know it is that Kirchhoff's current law is a consequence of the continuity equation (or equivalently of the Maxwell-Ampere law from which we can derive the continuity equation) with the additional assumption that the charge density in every point of the circuit is constant in time.

• vcsharp2003
The currents I1=dq1dt and I2=dq2dt at each of the capacitors

Are these the currents flowing in wires connecting the ends of capacitors that are connected to the battery i.e. top plate of top capacitor connected to +ve terminal of battery and bottom plate of bottom capacitor connected to -ve terminal of battery? Point a is like +ve terminal while point b is like -ve terminal.

Are these the currents flowing in wires connecting the ends of capacitors that are connected to the battery i.e. top plate of top capacitor connected to +ve terminal of battery and bottom plate of bottom capacitor connected to -ve terminal of battery? Point a is like +ve terminal while point b is like -ve terminal.
Well yes for ##I_1## but for ##I_2## I was taking to be the current of the branch in between the plates of the two capacitors. Anyway if you take ##I_3## to be the current between the bottom plate of the bottom capacitor and the -ve terminal of battery you can prove using Kirchhoff's current law that all 3 currents are equal.

• vcsharp2003
Well yes for I1 but for I2 I was taking to be the current of the branch in between the plates of the two capacitors.

For applying Kirchoff's current law to ##I_1## and ##I_2## did you take the top capacitor as the node?

• Delta2
For applying Kirchoff's current law to ##I_1## and ##I_2## did you take the top capacitor as the node?
Well yes, and I can assume your next question will be if we are allowed to take a whole capacitor as a node. Well I am not sure to be honest.

• vcsharp2003
There is another problem I see as I said Kirchhoff's current law comes from continuity equation with the additional assumption that the charge density is constant, however the charge density isn't constant ,at one capacitor plate the charge density is increasing and at the other capacitor plate the charge density is decreasing.

My only hope to save this would be to bring in the concept of displacement current but I just don't find it a very good idea.

• vcsharp2003
Suppose instead of connecting the capacitors to a battery we were connect them to two separate identical conducting spheres, but one with charge Q and one with charge -2Q. We would not then expect all the capacitor plates to have the same charge magnitude.

IMO, despite that, the capacitor plates themselves should have equal and opposite charges to prevent the electric field from penetrating the conductor. If you model the capacitor as two parallel plates, the charge is induced in their surfaces to block the other's field.

• • vcsharp2003 and Delta2
Now that I recall there was another thread here in these forums, where we were arguing about Gauss's law and this subject. According to one line of thought, if one makes the assumption that the fringing field is negligible then using Gauss's law and taking as surface any surface that encloses the whole capacitor, the electric flux through this surface would be zero (since we make the assumption that the fringing field is negligible) and hence the total charge enclosed by the capacitor would be zero, or equivalently each plate carries equal and opposite charge.

• vcsharp2003
IMO, despite that, the capacitor plates themselves should have equal and opposite charges to prevent the electric field from penetrating the conductor. If you model the capacitor as two parallel plates, the charge is induced in their surfaces to block the other's field.
How is that going to work if the circuit consists of two identical parallel capacitors, the total charge of the system being nonzero?

How is that going to work if the circuit consists of two identical parallel capacitors, the total charge of the system being nonzero?
In the way you originally proposed the problem, you have two spheres connected to the plates.

In total, yes, you have two conductors with different charges, and the charge distribution will be complex. But the charge should be distributed in such a manner that, on the surface of the plates, the charge is the same in magnitude but opposite in sign. We can neglect parasitic capacitances between the plates and the spheres, which would broaden the concept of the "capacitor" of the problem and complicate the demonstration.

But even if you get two parallel plates and apply a +Q and -2Q charge to each of them, the charges would still rearrange in a way that, in the inner face of the plates, the charge is the same. I calculated that supposing each plate can have inner and outer surface charges and that the electric field inside the plate is zero. You get that the inner charges for the capacitor plates are 3Q/2 and -3Q/2, respectively.

But even if you get two parallel plates and apply a +Q and -2Q charge to each of them, the charges would still rearrange in a way that, in the inner face of the plates, the charge is the same. I calculated that supposing each plate can have inner and outer surface charges and that the electric field inside the plate is zero. You get that the inner charges for the capacitor plates are 3Q/2 and -3Q/2, respectively.
Sure, but the argument quoted in post #1 does not discriminate inner and outer surfaces. My point is that for the plates (taking each as a whole) to have equal and opposite charge you need the external source of charge (the battery) to be net zero. The quoted argument fails to use that fact.

Sure, but the argument quoted in post #1 does not discriminate inner and outer surfaces. My point is that for the plates (taking each as a whole) to have equal and opposite charge you need the external source of charge (the battery) to be net zero. The quoted argument fails to use that fact.

Is it possible that with equal and opposite charges on two plates of capacitor in scenario of my original question, the electric field becomes so that no electron motion is possible?