# Charge between two isolated capacitors

Hello all,

I have been boggled by this problem the whole day today. Lacking a charge electrometer, I am limited in the ways I can test my theories.

I'll explain the problem in small steps:
Suppose you have two capacitors, each with the same capacitance. Lets say, 1 farad each.

You apply 5 Volts to each capacitor individually, thus having them store 5 Coulombs of charge each. (Charge / Capacitance = Voltage).

From my best understanding of charge storage on capacitors, (based on wikipedia), 5 coulombs of charge is on one plate the capacitor and -5 coulombs of charge is on the other plate. This applies for both of the capacitors.

To make the explanation easier, label the first capacitor's positive plate A+, its negative plate A-. Label the second capacitor's positive plate B+, its negative plate B-.

If I connect A- and B+ to my precise voltmeter, I get 0 V.
While the voltmeter is connected, if I connect A+ to B-, then I get 10 V on my voltmeter.

In classical terms of electromagnetics, this makes perfect sense. By connecting A+ and B-, the two capacitors are essentially in series, both with +5 V. Thus you get +10 V on the voltmeter.

This is, however an issue that I cannot get around:
If you have the two isolated capacitors, and connect A- to B+ with a very low resistance wire, apparently (as far as I was able to tell) there is absolutely no charge transfer between the two points. Let me restate this: There is 5 Coulombs on one end of a wire and -5 Coulombs on the other end of the wire and there is no movement at all of these charges.
If the two charges were isolated from the capacitors and connected by the wire, they would neutralize.. probably with a loud bang because 5 Coulombs is a lot of charge.

At first I tried to argue against this logic with the fundamental idea of a capacitor:
A- is attracted to A+. B- is attracted to B+. In order for charge to both decrease in A- and B+, (by decrease I mean neutralize.. charge is conserved), it must decrease in A+ and B-. Otherwise there would be excess charge in A+ and B-.
This means that the force in the capacitors between A+ and A- is a LOT more powerful than A- and B+, even though the charge of A+ = B+.
The force between A+ and A- is based on the capacitance. (Distance between the plates and the area of the plates.) However what is the force between A- and B+? ...assume 0 resistance wire.
...Maybe it's not called "Force" but something else.. either way, the logic stays the same.
The "pull" between A+ and A- is apparently much stronger than the "pull" between A- and B+ because there is no interaction when B+ and A- are connected.

I would leave is problem to rest except that it shows a small issue:
Even if A+ and A- attract each other much more than A- and B+ does, you would expect that at least an extremely small amount of charge would move between A- and B+ when they are connected.
This should in theory make A+ have more charge than A-.
Likewise B- would have more charge than B+.
However I have never seen this in practice.

I have searched for a solution for some time, and remembered an interesting phenomena:
Emission mechanism (for example thermionic and cold emission) has negative charge basically coming out of the conductor, into the air (or wherever.. could be a vacuum). Apparently high voltage, or high temperature is enough to overcome the attraction between A+ and A-.

Well then... Why doesn't the attraction between A- and B+ not give any noticeable charge movement at all?? Even though the voltage is only 5 or 10 Volts, 5 and -5 Coulombs on both ends should amount for something, no?
In mathematical terms, what is the force between A- and B+? ..regardless of this force, why isn't there any noticeable charge movement at all?

Thank you very much for reading. I have given much thought to this problem.
I hope you have something useful to share.

(Note - For those of you looking for an error in what I have written, I have simplified the problem to make it easier to understand. The total "pull" is (A- and B+) vs (A+ and A-, and B+ and B-). Meaning 2 times the "pull" of (A+ and A-) vs (A- and B+). Either way there should still be a little charge transfer between A- and B+ in theory.. Unless I'm missing something.)

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Electrons move and protons don't so they have been I guess you would say repelled by the negative charge set up across the other side of the capacitor and attracted to the E field on the other side. After the fields are all set up the situation is rather stable because the E field across the capacitor is much stronger than that on the other side of the wire, assuming the wire is much longer than the distance between the capacitor plates, than that in the wire itself, even the the potential difference across the 2 distances is the same. Thus the net force on some small charge dq in the negative plate remains in the direction of the positive plate but of course they can't jump across because of the insulator between them, but you see no current in the wire.

The force between A- and B+ is to a rough approximation Q *(Voltage/length of wire), meanwhile the force between the the two plates A- and A+ is Q * (Voltage/distance between capacitor plates). Since the distance between the plates is probably much smaller than the wire length the capacitor voltage wins in a sense.

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The force between A- and B+ is to a rough approximation Q *(Voltage/length of wire),

Where did you get this information?

sophiecentaur
Gold Member
I think that a cool person of light could be shed on this by considering what happens when this tiny current flows between the connected plates. You have to assume some finite resistance - else there will be a continuous oscillation set up. This resistance will cause some energy to be dissipated which, I think, will reduce in a drop in overall PD but with the same charge of 5C available across the outer terminals because there is no net loss of charge.

Another (more idealised) idea would be to consider what would happen if the two middle plates (without connecting wires) are slowly brought closer and closer together. The capacity between them will increase and increase without limit and, consequently, the PD will decrease (Q=CV) and decrease until it is zero when the plates touch. (Perfectly flat, parallel plates) No current will then flow and the 'outside' plates will have acquired a PD of 10V across them.

I think that a cool person of light could be shed on this by considering what happens when this tiny current flows between the connected plates. You have to assume some finite resistance - else there will be a continuous oscillation set up. This resistance will cause some energy to be dissipated which, I think, will reduce in a drop in overall PD but with the same charge of 5C available across the outer terminals because there is no net loss of charge.
Unfortunately, I've never detected charge going through this finite resistance.
Indeed, I expected for 5 Coulombs to remain on the ends of the capacitors, and have a slightly smaller charge on the inside plates.. but so far, I haven't seen such a thing happening.

Another (more idealised) idea would be to consider what would happen if the two middle plates (without connecting wires) are slowly brought closer and closer together. The capacity between them will increase and increase without limit and, consequently, the PD will decrease (Q=CV) and decrease until it is zero when the plates touch. (Perfectly flat, parallel plates) No current will then flow and the 'outside' plates will have acquired a PD of 10V across them.
What a great way of looking at the problem! You got around the issue with figuring out how much force there is between two charges in a wire. All we have to use is the capacitances.
Your explanation can be turned into a circuit that can be easily built:
Have 2 capacitors of the same capacitance and charge (as before).
Have a third capacitor between the two, connecting to A- and B+. Make sure this middle capacitor's capacitance is much larger than the first two. Also before connecting it, make sure that the middle capacitor initially has no charge.
In theory, based on the idea of capacitor charge, from what I understand, the charge at A- is attracted to the charge B+. The force between A- and B+ is more powerful than A+ and A- because the capacitor between A- and B+ is much more powerful than the capacitor between A+ and A-. The same goes for B+ and B-.
So in theory, by hooking up the super capacitor in the middle, you should at least get some amount of charge in the middle capacitor, and excess charge on the side plates of capacitors.

...however, this doesn't happen.
Just now I had two 2200 uF capacitors charged to 5.02 Volts.
I hooked up an uncharged 1 F supercapacitor in between the two.

Nothing happened.

How can this be explained?

Note in addition, that if you hook up a voltmeter up to A+ and B- while the middle capacitor is connected, you get a voltage between 0 and 10 volts, depending on the middle capacitor's capacitance. (Higher capacitance = Higher voltage (10 V) because less resistance is present when capacitor is not charged.)

Now this is making even less sense. :P

Anybody know what is happening here?

Clearly the middle capacitor has a higher force between its two plates, no?

Thank you sophiecentaur for posting your idea.

sophiecentaur
Gold Member
V. interesting.
What voltage (including sign) is across the centre capacitor? I suggest that it may be in the opposite sense - as, indeed, it is when the two capacitors are miles away from each other. I suggest that the vector sum of the PDs will actually be 10V (forgetting about any actual energy loss).
Try it.
;)

sophiecentaur,

I'm not very sure how you mean.
To answer your question, the center capacitor is 0 volts at all times. If you are to connect A+ and B-, then completing the circuit, you are feeding 10 Volts from the two side capacitors to the initially uncharged 1 Farad capacitor.

I have thought what you posted just now though.. regardless of the middle capacitor's capacitance, the voltage between A+ and B- should be 10 V. The reason my voltmeter is giving smaller voltages for smaller capacitances is that my voltmeter works by measuring an extremely small amount of current. If you have a capacitance that is a little too small, there isn't enough current to go around..
Unfortunately, I don't have an electrostatic voltmeter that would check the voltage not by current, but by charge.

Hm.. Absolutely no current, but plenty of voltage.
Okay.....
Let's say that my voltmeter is not capable of showing the increase in voltage present in A+ and B- because it is based on current, and there is no current which is why I am only getting 0 - 10V.

The middle capacitor would still gain charge from the two side capacitors, right?
In order for the middle capacitor to gain charge from the two side capacitors, charge would need to move from one plate to another. This should happen because the "pull" between the middle capacitor's plates are much more powerful than the "pull" between A+ and A, and B+ and B-.
This would cause the charges in A+ to not be so attracted to A- and the charges in B- to be not so attracted to B+.
Now we are assuming that my voltmeter is not capable of measuring between A+ and B-...
However, this would still mean that the center capacitor would still have some charge in it.
If you disconnect the center capacitor from the other two, you should have a nice big reading. Since it's a 1 F capacitor, the small current taken by the voltmeter should have almost no effect on the amount of charge left in the capacitor.
Thus you should see a high reading on the middle capacitor.
...and I saw 0 Volts on the capacitor... :grumpy:

Looks like I need to figure out a new method of looking at how charge is stored on a capacitor.

The best explanation I can come up with right now, is that for some odd reason, charge refuses to build up on one side of a capacitor and not the other.

Note that the charge build up can actually be view very simply in another type of experiment:
Take two parallel plates.. put them close to each other.. perhaps place some dielectric in between them.
Set the plates voltage to around 5000 Volts. Make sure the plates don't spark with each other.
Now with some insulator, take the two plates apart. Measure the voltage between the two plates with an electrostatic voltmeter.
It the voltage will be more like 50,000 Volts now.. depending on the distance you took apart the two plates.
The reason for this is understandable.. The capacitor has a constant charge, you change the capacitance.. based on:
Charge / Capacitance = Voltage
The Voltage must increase.

But.. for some reason this effect does not happen if you try and take the charge from A- and B+, leaving A+ and B- exposed...
I do not know why this is happening yet. If anybody knows, it would be nice if you could share.
As I said before, it seems I have to change my understanding of how charge is stored.

sophiecentaur
Gold Member
You may not be able to measure it but the PD between the + plate of one capacitor and the - plate of the other should be (is) 5V, if they were charged on the same device and held in perfectly insulating mounts. It is hard to measure because the capacity between them is so low. As they become more and more 'connected' that voltage will become 10V because they will float above and below the mean potential of the two joined terminals.
I think that the following is the 'nub' of the problem:-
Why should the middle capacitor 'gain charge' as you say? All the surplus electrons on the negative side of the 'top' capacitor are crowding on the - plate and all the 'holes' on the + plate of the lower one are on the lower plate. They are very strongly attracted to each other. There just isn't much charge 'outside' the devices because the charges are all held inside.
This is a bit like the fact that all the charges on a Van der Graaf generator ball or Faraday Cage are on the outside and none on the inside.

When you insert your third capacitor, you are just doing the equivalent of my bringing the backs of the plates together - the small Q and the C(added) will produce only a tiny V. The net charge of the two plates which were joined together will be zero.

Do you not have an electronic voltmeter? Their input impedance is very high and you appear to be using fairly beefy capacitors. But I guess the effective capacity between the two devices is only in the region of a very few pF to start with. Simply putting a high value C in parallel with this capacity will reduce, rather than increase, the voltage between them.

When you actually put a capacitor in circuit, the situation will change because the charges can flow away from between its plates. In this case, the other plate of each capacitor is undisturbed and still charged.

This "pull" you refer to relates, I assume, to the field between them times the charge. It will be considerable in Volts per metre because the spacing is sub-millimetre.

Boy, this is a fascinating question. I think I am beginning to get it though.

I think that the following is the 'nub' of the problem:-
Why should the middle capacitor 'gain charge' as you say? All the surplus electrons on the negative side of the 'top' capacitor are crowding on the - plate and all the 'holes' on the + plate of the lower one are on the lower plate. They are very strongly attracted to each other. There just isn't much charge 'outside' the devices because the charges are all held inside.
This is a bit like the fact that all the charges on a Van der Graaf generator ball or Faraday Cage are on the outside and none on the inside.

Although I tried not to, it seems like I have caused some confusion.

Charge moves from A- to one plate of the middle capacitor.
Charge moves from B+ to the other plate of the middle capacitor.
Charge is conserved.
A+ and B- contains the same amount of charge as it did before.

The "pull" I was referring to was what I call "force", but I am not sure if that is the correct way to call it.
Basically, the force between A+ and A-, and between B+ and B- is less than the force between A- and B+. The reason is that the capacitance between A- and B+ is higher than between A+ and A- and between B+ and B-.
Because the force is higher in the middle capacitor, the charge moves towards the middle capacitor.

Do you not have an electronic voltmeter? Their input impedance is very high and you appear to be using fairly beefy capacitors. But I guess the effective capacity between the two devices is only in the region of a very few pF to start with. Simply putting a high value C in parallel with this capacity will reduce, rather than increase, the voltage between them.
I have an electronic voltmeter. I do not have an electrostatic voltmeter. An electrostatic voltmeter leaves the charge alone. A regular voltmeter takes a sample of the charge. Even if the impedance is high, it may not be good enough.. this happens to me frequently when checking the voltage of a capacitor less than 10 uF or checking the voltage of a charge pump (voltage multiplier).

sophiecentaur
Gold Member
It struck me that a charged capacitor is analogous to an atom or, particularly a stable molecule, containing an equal number of positive and negative charges. It is electrically neutral and will only interact, electrically / chemically, with another object under extreme conditions. The equivalent experiment on two capacitors would be to slide one spiral wrapped pair of foils inside the other. Just touching their outsides really doesn't affect the internal charge situation much at all.

This is, however an issue that I cannot get around:
If you have the two isolated capacitors, and connect A- to B+ with a very low resistance wire, apparently (as far as I was able to tell) there is absolutely no charge transfer between the two points. Let me restate this: There is 5 Coulombs on one end of a wire and -5 Coulombs on the other end of the wire and there is no movement at all of these charges.
If the two charges were isolated from the capacitors and connected by the wire, they would neutralize.. probably with a loud bang because 5 Coulombs is a lot of charge.
When you go back to the first chapter of electrostatics you probably seen the experiment with a charged glass rod or some other insulating charged material held close to a thin conducting sheet. You probably easily accepted that when say the rod was -ve charged that the conductor surface facing the rod had a +ve charge inducted on it and therefore the surface away from the rod became -ve charged.

But now have a new and more in depth look at this. Just look at the conducting sheet on its own. Here we have (I think) the situation you are asking for because:
1 Although the opposite charges are sitting very close opposite each other they are not moving at all towards each other.
2 Since we are dealing with a conductor the voltage difference between every point on this conductor must be zero.
3 There is no electric field inside the conductor.
So what’s going on? Why do the -ve charges stay put? You could perhaps claim there’s a repulsive coulomb force between rod and electrons? But that’s not possible because the coulomb force field cannot exist inside a conductor.

The answer is that the field of a charge must always end somewhere on an opposite charge. If there are no objects anywhere nearby we just say that they end somewhere far away. And that solves the problem because here the electrons on the sheet are now being pulled by a field going off in the distance. And this field is generated by the –ve charges on the rod and the far away +ve charges.

The sheet only “used up” some field, meaning that where the sheet is there’s now no field.
Since there’s energy associated with this field it must mean that the sheet used up some energy. This energy is used up in 2 ways. First there must have been a current separating the charges and since the sheet has electrical resistance some power is consumed. Second there was a small pull on the sheet which was converted to heat when the moving sheet came to a stop. Both ways energy ends up being converted to heating up of the sheet.

sophiecentaur
Gold Member
I really don't see your problem here.
The charges stay one side of a conductor because they are attracted or repelled by something. There will be a high field 'out there' which has displaced them. There is a low field in the conductor because there is a vanishingly small PD across it. Just a hint of a field inside is enough to shift charges around. The amount of charge displaced during polarisation will be limited by the external fields; in just the same way, in a potential divider circuit, there is a PD across the upper resistor, zero PD across the connecting wire and some more PD across the lower resistor. The charges / currents will balance out when the fields / PDs are appropriate for the situation.
Per Oni
I don't know what you mean by "the sheet only used up some field". That is so far from a conventional description that its meaning is not clear.
I think there is some confusion between 'field' and 'field lines'. You can work out the force on a charge q in a field E due to some other charge or set of charges.
F=qE

or between two charges q1 and q2
F = keq1q2/r2 (Coulomb's law)

but 'field lines' are best used, I think, in graphical descriptions.

And Point 3: "why do the electrons stay put?"
I have given this some more thought. It doesn't need to be relevant to be original problem but:
OF course, there is a surplus of electrons on the negatively charged plate: they were put there initially. The charge distribution in the plate will be set by the pull of the electrons towards the external charge, balanced by the pull of the depleted atoms in the metal. Just because the field inside the metal is low doesn't mean that all the electrons have sloshed right over to one surface; they will occupy a finite depth, limited by forces of attraction and mutual repulsion. Because of the Coulomb forces, there can't really be an 'empty region' of metal at the back of the plate. I guess the depth of the skin of surplus electrons must be related to the PD across the capacitor and the spacing between the plates. The nearer electrons will, effectively, be screening those behind from the full force of the charges on the opposite plate. Are we looking at some exponential distribution here? Does this argument apply with or without resistance in the metal (resistance will just affect the time taken to reach equilibrium).

Per Oni
I don't know what you mean by "the sheet only used up some field". That is so far from a conventional description that its meaning is not clear.
You’r quite right. What I mean is that there’s potential energy locked up in an electric field. When the sheet takes up the space first occupied by the field then this energy gets transferred to the sheet.
Just because the field inside the metal is low doesn't mean that all the electrons have sloshed right over to one surface;
That’s true, not all free charges have travelled across. How much charge has flown? Well say the plate did not alter the original field. Then D=E*epsilon zero, and q=D*A. And N=q/e. If you would start with an realistic value for E say 100v/m then you come to the conclusion that only a tiny amount of free electrons have moved to the surface.
But I disagree about there still being some field inside the metal.
they will occupy a finite depth, limited by forces of attraction and mutual repulsion. Because of the Coulomb forces, there can't really be an 'empty region' of metal at the back of the plate. I guess the depth of the skin of surplus electrons must be related to the PD across the capacitor and the spacing between the plates. The nearer electrons will, effectively, be screening those behind from the full force of the charges on the opposite plate. Are we looking at some exponential distribution here? Does this argument apply with or without resistance in the metal (resistance will just affect the time taken to reach equilibrium).
Your train of thought is no doubt very useful and I can only admire you, but they are going quite a bit further then the OP was intended.

sophiecentaur
Gold Member
"But I disagree about there still being some field inside the metal. "
As far as I can see, the only situation i which there could be NO field in the metal would have to be when there was no imbalance of charge in it. That would be when it had no net charge and all electrons were evenly spread out and neutralising the nearby positive ion cores or, possibly, when all extra electrons were spread out around the outside by mutual repulsion (and the converse with positive charges) - but would this apply to any shape in general and not just a sphere?

As for the energy involved in moving the sheet about in the presence of a field, even with a perfect conductor, I think there would still be an energy transfer with Potential Energy being stored in the redistributed charges in the metal (polarisation).

But, as you say, the OP started off differently, discussing two real capacitors connected in the centre. There would, in that case, be a small amount of energy transfered as the + and - terminals were brought together to leave zero PD between them and twice the original PD appearing across the two in series. This energy would, presumably, be related to the original capacity between the two capacitors and the 10V difference between the leads (half C V squared) so not a lot!

I note some really nice capacitor sketch graphics on this site:

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm

I hope someone will study this link and then provide a sketch of this problem showing the charge regions at the plates, and at the dielectric interface, which holds a charge opposite to the plates.

I suspect there is no closed path for charge to flow when the wire is coupled. A KVL loop should also show zero voltage drop around any open or closed circuit path in space. A sketch should help clarify the theoretical study.

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The net charge at the surface of each plate contains two components.

A surface charge is induced on the metal conductors, which can't support an internal electric field.

A surface charge is induced in the dielectric right up against the metal conductors. But charge does not cross the metal/dielectric boundary.

The + and - charge is due to action at a small distance across the boundary.

The dipoles in the dielectric are like tiny compass needles fixed in space but able to rotate inside an electric field. They wiggle in the presence of an alternating current, so the device can pass AC, but they line up on the metal surfaces in the presence of a direct current source. After the source is removed, an ideal capacitor holds its charged state.

I'm fairly certain the voltage between two nodes must count the net charge on the plate and just inside the dielctric. I think this explains your voltmeter measurements but have not thought it all the way through.

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Excellent reference on basic electrostatic field problems and parallel plate capacitor:

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/ElectricForce/FlatSheet.html [Broken]

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Capacitors/ParallCap.html [Broken]

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SystemTheory,

I really like your attached picture. May I ask you where you got this illustration? It shows the pretty well the dielectric polarity that I have over looked. I originally ignored the dielectric because I thought that modern capacitors are mainly influenced by the area and distance between the two plates - that the dielectric is so thin, most of the charge is stored in the plates. Apparently I was wrong. Indeed, over the past few days I have been teaching myself a bit more about dielectric fields.

Thank you everyone for responding,
LiteHacker

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LiteHacker,

I had an excellent fields and waves professor 20 years back. Now I need to refresh most of the knowledge ...

You need to look also at the properties of a conductor, and how a surface charge is induced while it remains an equipotential surface.

There may be a free charge flow internal to the conductor wire that you're attaching to the precharged capacitors. This flow redistributes the surface charge on the wire, without changing the charge on the capacitors, such that the conductor acts as an equipotential surface under static equilibrium.

You can find techical diagrams rapidly online usually with a google image search:

The trouble is you need to have some sense of what you're looking for in advance, since a bunch of junk turns up too.

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Please note my analysis showing q = 0 in the figure at post #16 is incorrect. The charge in the dielectric at a plate surface is always smaller than the opposite charge on the metal plate.

Please refer to these two references and/or a good physics/electrostatics textbook for proper analysis:

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Capacitors/ParallCap.html [Broken]

The edit button does not appear at post #16 so I am unable to note the error there.

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