# Battery and wire

1. Dec 5, 2013

### chimay

There is a battery connected to a non ideal circular conductor wire.
What is the specific mechanism by which one electron inside the wire “feel” the presence of the battery because of electric field?

2. Dec 5, 2013

### kended

In short the battery creates an electric field. Since an electrical conductor has "free" electrons available for conduction (that is negative charges), those charges gets pulled towards the positively charge side of the battery while the negative sides pushes them. By definition, ANY charge will "feel" an electric field by either being repulsed or attracted by it.

3. Dec 5, 2013

### wreckemtech

From what I understand, it's electrostatic repulsion. The battery will emit electrons at one terminal, and those electrons will "push" on other electrons in the wire and so on down the length of the wire. The other terminal of the battery will also take in electrons at the other end.As a result there is an electrostatic potential gradient (voltage) acting on the circuit. But at the level of the individual electron, it is the repulsion between electrons pushing it through the circuit.

4. Dec 5, 2013

### thegreenlaser

My understanding is that the battery creates an electric field everywhere around it, not just inside the wire. All the electrons and protons around the battery (in the air and the wire) do feel that force, it's just that the force isn't strong enough to do much to most of them because they're bound in place. The electrons in the wire, however, are able to move (because of the low resistance) and so they start to move under the influence of the electric field. So the electric field is everywhere, it's just that the electrons in the wire are the only thing that really respond to it in a noticeable way.

On a related note, if you took away the wire (so you just have a lonely battery) and you somehow cranked up the voltage of the battery, you would eventually see a spark between the two terminals. This is because you've made the electric field strong enough that the air molecules actually respond to it (the process is called dielectric breakdown).

5. Dec 5, 2013

### chimay

Wreckemtech and thegreenlaser, you proposed as answer just my doubts.
As for as Wreckemtech’s reply it’s like the charges inside the battery interacts by means of the electric field they create (electrostatic repulsion) and the action of the battery is confined inside itself. I discovered this point of view while reading “Introduction to electrodynamics” by David J.Griffiths and I found it very strange. In particular it was this that made me thing about this topic.
Thegreenlaser refers to something different, since battery creates the electric field and electrons feels it “passively”. This was my original thought before reading David J.Griffiths.
I would appreciate a sort of comparison between you, since I find the topic quite tricky.

English is not my mother tongue, so I hope you will excuse me for my mistakes.

6. Dec 5, 2013

### wreckemtech

The model I proposed is pretty much the classical way electric currents in conductors are modeled. It's incomplete, but it does explain some things. If conduction were completely due to the charges interacting with a field, then why don't batteries short-circuit more easily? If a field in and of itself is enough to move the electrons from one terminal to the other?

The dielectric potential of a vacuum is extremely high. To me, the fact that you "need" a conductor to move charge at all at lower voltages is pretty strong evidence for some type of "transmission" mechanism for electrons pushing one another through the conductive medium.

Or they could just be surfing an EM wave through the conductor which serves to focus the wave along its axis.

There are more than two possible explanations.

7. Dec 5, 2013

### thegreenlaser

So the argument seems to be that the electric field is zero (or close to zero) outside of the conductor yet somehow non-zero inside the conductor? This doesn't make sense to me... we're dealing with near-static fields, so we would expect the electric field to be close to conservative. That means if there's a 1.5 V drop between the terminals, the path integral of the electric field should be 1.5 V regardless of whether you take a path along the conductor or a path outside the conductor. If the electric field is zero outside of the conductor, then the path integral of the electric field along a path outside the conductor would have to be zero, which contradicts the fact that the electric field is conservative. So I would expect the electric field to be non-zero in the air as well.

We wouldn't expect current to flow through air/vacuum just because there's an electric field present, because there aren't any charges which can move to create that current. Suppose there's no conductor and you just have a battery sitting there producing an electric field in the air around it. There are certainly charges in the air and in the battery which are experiencing forces because of that electric field, but those charges are all bound in place, so they don't move. Once you add a conductor, you suddenly have charges which aren't bound in place, so they move under the influence of the electric field.

Which edition of the book do you have, and what section was it? I have the book, so I'd like to read this myself.

8. Dec 5, 2013

### thegreenlaser

Okay, I think I found the chapter in Griffiths that you're talking about (7.1.2). I think what I'm saying is consistent with what he's saying. He's just saying that when you connect the wire, there's a brief period of time where charges are building up in various locations on the wire to ensure that the current is constant around the wire. That's not the same as saying that current flows because one electron pushes on the next electron which pushes on the next and so on. Once the charge he's referring to is built up, then it just sits there, altering the electric field in such a way that the current flow is even all the way around the wire.

Once you're in that equilibrium state, then I believe the situation is as I described it: you have electric fields both inside and outside the wire. That electric field is generated both by the battery and by the static charge that's built up at various points on the wire. Inside the wire, the electric field pushes on conduction electrons causing current flow (via the relationship J=σE). Outside the wire, there are no free electrons for the electric field to push on, so there is no current flow.

EDIT: I think maybe you're getting tripped up by the source force fs that Griffiths refers to, which he says is confined to one portion of the loop (like the battery). What he's referring to, though, is not the electric force which pushes the current in the wire, but rather the force which allows the battery to generate a voltage in the first place. In a battery, fs is what causes positive charge to build up on one terminal and negative charges to build up on the other terminal (the positive charges attract the negative ones, so you couldn't separate them like that without some sort of source force fs applied). That particular force is indeed contained inside the battery. But then the separation of charges induced by fs creates an electric field all around the battery, which in turn induces a current flow in the wire.

Admittedly I'm not 100% confident that what I'm saying is correct, but I fail to see where the mistake is if there is one.

Last edited: Dec 5, 2013
9. Dec 5, 2013

### sophiecentaur

I would suggest that using the 'Field due to the battery' is not very fruitful. Field is the gradient of Potential and, as the actual distances involved are not specified, you cannot know the gradient but you do know the (say 12V?) potential. A long or short 12V battery will produce exactly the same effect and yet the fields in each case would be very different. Amazingly, circuits work more or less the same, whatever the layout or the lengths of connecting wires (assuming they are thick enough and sufficiently low resistance) so why not stick to Potential in your arguments? There is nothing more and nothing less fundamental about Field or Potential so why not choose the one that fits the situation better. You can say that Charges will move to places with lower potential. The path that they will take will be just that of the lowest potential. To stray outside the wire would involve a massive rise in potential so they do not go there.The mutual repulsion of (in this case) any surplus of electrons in any location will cause a force to move them a very small distance along the wire.
Can we not aspire to something higher than what they, sort of, taught you at school, when they probably never thought about it properly?

10. Dec 5, 2013

### wreckemtech

Okay. So he's saying that as a result of exposure to the voltage, the wire is developing a field of its' own. I don't disagree with this, per se as an explanation for the bulk movement of electrons. It can be demonstrated mathematically using Gauss' equations that there is an axial EM field being generated.

The problem I see with this being a complete description of the phenomenon of conduction is that if you place a positive and negative terminal in close proximity in a vacuum, there should be a strong potential between the two, and "free" electrons at the negative terminal available to jump across to the other terminal. Yet the conduction of a vacuum is nearly nil (permittivity of free space). Also, I think one would be hard pressed to find an explanation of how a negatively charged particle would not repel a similarly charged particle. So if one electron in a conductor moves (in the electrostatic model, the ones nearest the terminals would begin migrating first) and the other electrons are also free to move, they will. The electrons are indeed flowing through the medium (though not in linear fashion) and they do repel one another.

So possibly electrostatic repulsion explains how the required "charge" builds and equalizes while the field develops and stabilizes along the length of the conductor? I really think that both come into play, and that's possibly the reason that radiant EM waves travel just fine through a vacuum, but electricity finds it very difficult to do so.

I will take a look at the reading the two of you mentioned sometime in the next couple weeks. Things are very busy for me at the moment. Would you mind if we continued the discussion then?

11. Dec 6, 2013

### CWatters

Yes and no. Thinkl about what happens when you change the set up physically rather than electrically..

Clearly if you bend or twist the wire (without adding any shorts) the current won't significantly change. This suggests that the "like charges" issue dominates.

However if you consider a battery on it's own (in a vacuum) the positive and negative terminals do generate an electric field between them. In theory this could effect an external circuit but the impedance would have to be very high for that to be significant.

12. Dec 6, 2013

### sophiecentaur

Do you have any problem with the approach that involves Potential, rather than Field? Is it not intuitive enough?

13. Dec 6, 2013

### thegreenlaser

At a high level, the potential approach certainly provides an intuitive answer to the question "why does current flow in the wire?" But the question here seems to be "is the force that an electron feels directly due to the battery, or is it completely due to adjacent electrons in the wire?" Whether we talk in terms of potential or force, we still need to answer the question of whether the battery directly influences an electron part-way along the wire or whether the battery only influences the electrons nearest to it, which push the next electrons, which push the next electrons and so on.

It may be that the potential approach is better, but I think what's missing is an explanation of what exactly sets up that potential. Based on wreckemtech's post, for example, I think it's safe to say that it's not immediately obvious to everyone why an electron leaving the metal surfaces to travel through the vacuum/air would have to undergo a massive rise in potential. And there definitely is some confusion over whether there is a constant potential drop along the wire which is set up by the battery, or whether there's some sort of fluctuating potential gradient set up at each location by the electrons immediately adjacent to that location.

Edit: Since the question seems to be focused on where the force comes from, wouldn't a field approach be more appropriate in this case?

Last edited: Dec 6, 2013
14. Dec 6, 2013

### sophiecentaur

Here's a thought experiment, then: Take a resistor, attached to the terminals of a battery by two parallel wires. Consider an electron half way down this resistor. It has a force on it which will cause it to move. Put a 180° twist in the wire so the +- terminals of the battery are the opposite way round. Would the electron would experience a different force? The field 'due to the terminals of the battery' has reversed with respect to the resistor. If the cells in the battery were physically a bit wider or narrower, would the force be different? Isn't it due to the whole system / circuit? An electron can only respond to the field in its immediate vicinity.

A similar question, in a totally different context. Imagine a bicycle chain, driving the sprocket round. Is it the action of the chain links that happen to be in contact with the sprocket or is it due to the links on the chain wheel - or is it due to the foot pressing on the pedal? There is tension everywhere on the chain but it's higher on the top section than the bottom section. There is work done as the bike is propelled up a hill. Work is done ON the chain and BY the chain, depending on which bit you are considering.

Doesn't the gradient of the potential give you the force per unit charge?

15. Dec 6, 2013

### chimay

Thanks to you all for your help.

I and thegreenlaser share just the same uncertainty. This was what I was going to write before reading his post.
I red all the replies and they made me less confident about the " field generated by the battery" point of view . One point in particular is the shape of the wire. In fact if we accept that the field is due to the battery, then we have to accept that it creates a field with a direction wich depends upon the wire. ( intuition drives us to imagine a field parallel to the wire in each point )

16. Dec 6, 2013

### jartsa

Inside a thin wire an electron feels an electric field that depends on where the nearby electrons are.

Inside a thicker wire an electron feels an electric field that depends on where the electrons are in the larger neighborhood.

Inside a ridiculously thick wire an electron feels an electric field that depends on where all the electrons are.

Near a thin wire electric field points away from the wire or towards the wire depending on which electrode is the closest one. (measured along the wire) (we assume the wire is neutral on the average)

The ridiculously thick wire is equivalent to a really bad plate capacitor.

Last edited: Dec 6, 2013
17. Dec 6, 2013

### sophiecentaur

If there is no potential drop along the wire then any field will be at right angles to it, surely. The (external) Volts per Metre will vary with the spacing along the wire but there will be a vanishingly small field inside the metal. (As mentioned elsewhere, we're talking in terms of Capacitance). When the resistivity is low, the electrons need very small force to 'move them along' through the conductor on average. The only force of any consequence will be the 'compressive force' between the ends - OMG this is such a naff model; no wonder it's not used in any analysis I have come across. It's another one of those "what's really happening" discussions and the answer depends entirely on what part of the system you are dealing with and to what level you want to take it.

18. Dec 6, 2013

### wreckemtech

Sorry to jump in here and muddy things up again, but there will be a magnetic field with flux lines that wrap around the wire axially as soon as the electrons start to move in a certain direction. (Think about right-hand-rule in regards to moving charges and magnetic fields.) So perhaps that is part of the "field" effect?

This is a great discussion. Just wish I had more time for background reading right now...

19. Dec 7, 2013

### jartsa

I mean: Inside a thick copper wire an imaginary electron with eyes could look into the distance, and see large 3-dimensional chunks of copper in the distance, and feel the forces emanating from the distance.

Inside a thin wire that is not possible.