Electric field within a battery

In summary, the conversation discusses the concept of EMF in batteries and how it is related to the electric field inside the battery. The participants have different views on how to define EMF and the role of the electric field in producing it. One participant suggests that the electric field within the battery is not a useful concept, while another mentions the importance of understanding the details of the electric field in the battery's operation. The conversation also touches on the concept of charge separation and its relation to the conservative electric field within the battery. Overall, the participants agree that studying a specific type of battery in depth is the best way to gain understanding of its operation.
  • #36
I believe all of the equations Skilling quotes are correct, so long as you replace ##\mathbf{E}_s## with ##\mathbf{E}##, and ##\mathbf{E}_m## with ##\mathbf{f}_s##. I have no idea why he is treating the force of the battery as an electric field.

Furthermore, for reasons that @Dale notes in the comment thread of the Split Electric Fields insight, ##\mathbf{f}_s## is conservative (i.e. derivable from a chemical potential) and thus cannot be the non-conservative component of the electric field.

Whilst you can certainly always perform the Helmholtz decomposition of ##\mathbf{E} = \mathbf{E}_{cons} + \mathbf{E}_{n-cons}##, and come up with formulae like $$\nabla \times \mathbf{E}_{n-cons} = -\partial_t \mathbf{B}$$I don't see how this is ever too helpful. More importantly, such a decomposition is irrelevant in the example of a static battery where there is only one conservative electric field.
 
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  • #37
etotheipi said:
I am not sure about this. Griffiths says

There appears to be no constraint that this force per unit charge be an electric field. And indeed, most other sources I have consulted today echo Griffiths.
All the examples quoted by Griffith are effectively E fields. A "force per unit charge" IS the definition of an E field.

I have zero disagreement with Griffith.

etotheipi said:
I believe all of the equations Skilling quotes are correct, so long as you replace ##\mathbf{E}_s## with ##\mathbf{E}##, and ##\mathbf{E}_m## with ##\mathbf{f}_s##. I have no idea why he is treating the force of the battery as an electric field.
.

Yes I tried to have you see your E is my Es and your ##f_s## is my Em. As to why ##f_s## consitutes an E field, that is because otherwise the line integral of the E field around a closed path would be zero. It's not. It's the emf.

I suggest studying this matter further in Richard Feynman's Lectures on Physics vol II chapt. 23.

Dale said:
Oh my. This is not only wrong, it is obviously wrong.

Consider the usual open circuit measurement of a battery after all transients have ended. There is a non zero charge distribution that produces an E field in the battery. This E field is produced by a static charge distribution in the usual way so it is conservative and has an associated electrostatic potential.

There is also a chemical potential, usually called the EMF. Since the transients have all ended and the battery is in equilibrium then this chemical potential is equal and opposite the electric potential.

OK so far.
It is therefore also conservative.

Absolutely wrong. I can't imagine how you conclude this. You might also be advised to refer to Feynman's Lectures Vol II chapt. 22 eq. following eq. 22.3 where he shows that the circulation of the E field is not zero but actually the emf.

Maybe the problem is your assumption that a chemical potential is the same thing as an electrical potential. The two are completely different concepts. The unit of chemical potential is the joule, not the volt, just for starters.
 
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  • #38
rude man said:
All the examples quoted by Griffith are effectively E fields. A "force per unit charge" IS the definition of an E field. A "force per unit charge" IS the definition of an E field.

I strongly disagree with this; it is just not correct. If I take a ball with a charge of ##3C## and push it along a table with a force of ##6N##, does that mean that I constitute an electric field of ##2NC^{-1}##? Most certainly not. The electric field is the electric force per unit charge, and here we have a chemical force from the battery. From Wikipedia:

"The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point."

rude man said:
Yes I tried to have you see your E is my Es and your ##f_s## is my Em. As to why ##f_s## consitutes an E field, that is because otherwise the line integral of the E field around a closed path would be zero. It's not. It's the emf.

I suggest studying this matter further in Richard Feynman's Lectures on Physics vol II chapt. 23.

I will have a look at Feynman, but I don't think he will say anything like you are proposing. Even in the opening of your insights article you write
The first and admittedly controversial example is that of a chemical battery (Fig. 1a below). It’s controversial because most folks consider that there is only one E field in the battery, which corresponds to my Es. And, they are in good company. No less a luminary than Richard Feynman makes this statement: Arguing just as we did for the ideal capacitor, we see that the potential difference between the terminals a and b is just equal to the line integral of the electric field between the two plates when there is no longer any net diffusion of the ions.
 
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  • #39
By now you must see why I call the battery example controversial.

Feynman does not talk explicitly about a battery in chapt. 22. He takes a coil with current i then writes the circulation of E. There is no E field in the wire; it's a perfect conductor. There is an electrostatic field outside the wire. The circulation of E is thus L di/dt, not zero.

There is an irony here. Feynman gets the right answer for the circulation but does a no-no: he finds that the circulation of a conservative field is non-zero. The real rationale is that there are equal and opposite E fields in the wire: Em and Es. The circulation of the Es fields inside the wire and outside the coil gives zero; the circulation of Em is what gives the right answer which is L di/dt.
 
  • #40
rude man said:
There is an irony here. Feynman gets the right answer for the circulation but does a no-no: he finds that the circulation of a conservative field is non-zero. The real rationale is that there are equal and opposite E fields in the wire: Em and Es. The circulation of the Es fields inside the wire and outside the coil gives zero; the circulation of Em is what gives the right answer which is L di/dt.

I don't know how you conclude this. There are no faults with Feynman's analysis here.

He takes the line integral of the electric field ##\mathbf{E}## around a closed loop to obtain a non-zero EMF (i.e. ##\mathbf{E}## is not conservative here since we have a time-varying magnetic field inside the box!).

However he takes the integral for potential difference along a path outside the box where ##\mathbf{B}## is zero, and the electric field ##\mathbf{E}## is completely conservative, allowing us to define a potential function.
 
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  • #41
etotheipi said:
I don't know how you conclude this. There are no faults with Feynman's analysis here.

He takes the line integral of the electric field ##\mathbf{E}## around a closed loop to obtain a non-zero EMF (i.e. ##\mathbf{E}## is not conservative here since we have a time-varying magnetic field inside the box!).
His loop is thru the wire and outside the coil. The mag field is not part of this E loop. We are dealing with an E loop, not a B loop.

If the field in the wire were zero and the field outside the coil is conservative then the circulation of the E field would be zero. As he says, it's not, it's L di/dt.
 
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  • #42
rude man said:
If the field in the wire were zero and the field outside the coil is conservative then the circulation of the E field would be zero. As he says, it's not, it's L di/dt.

No! The line integral of a conservative field around a closed loop, if the vector field is conservative around the whole loop, is zero! The line integral of a conservative field from ##a## to ##b## can be non-zero (think of work done by any conservative force...)! Specifically, like he says, even though ##\mathbf{E}## is conservative on the path outside the box, the line integral along this path is $$\oint_a^b \mathbf{E} \cdot d\mathbf{l} = L\frac{dI}{dt}$$There is no contradiction here!
 
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  • #43
etotheipi said:
The line integral of a conservative field around a closed loop, if the vector field is conservative around the whole loop, is zero!
Right!
We have such a closed loop:
The integral of E inside the wire is zero.
The outside E field is conservative.
So there are only conservative (and zero) fields.
We have closed the loop and the only thing in it is a conservative E field.
But the integral of the loop is finite. That is impossible.

The dilemma is resolved if and only if you accept the existence of two equal and opposite E fields in the wire. As I explained.
 
  • #44
rude man said:
Right!
We have such a closed loop:
The integral of E inside the wire is zero.
The outside E field is conservative.
So there are only conservative (and zero) fields.
We have closed the loop and the only thing in it is a conservative E field.
But the integral of the loop is finite. That is impossible.

The dilemma is resolved if and only if you accept the existence of two equal and opposite E fields in the wire. As I explained.

No! Inside the box ##\nabla \times \mathbf{E} = -\partial_t \mathbf{B} \neq \mathbf{0}##. The electric field inside the wire is not conservative.

The line integral of ##\mathbf{E}## along the outer path is the voltage, the line integral of all forces per unit charge around the whole loop is non-zero and equals the EMF. There is no paradox and no weirdness here, it's just vector fields. As vanhees (and Feynman!) note, there is no need to perform any such decomposition of the electric field.
 
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  • #45
Adesh said:
I think that’s not correct, EMF is defined as the work done by all forces between the terminals except the electrostatic one. EMF is some kind of an internal mechanism that forces the positive charge to the positive terminal and negative charge to the negative terminal.
The EMF is DEFINED as (in SI units)
$$\mathcal{E}=\int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B}).$$
Here ##\vec{v}(t,\vec{x})## is the velocity along the boundary ##\partial F## of an arbitrary surface ##F##. It obeys the complete integral Faraday Law of induction,
$$\dot{\Phi}_{\vec{B}}=-\mathcal{E}.$$
This law, applied to "infinitesimal surfaces" and boundaries leads to the local differential form of Faraday's Law, which is one of Maxwell's homogeneous equations,
$$\partial_t \vec{B} + \vec{\nabla} \times \vec{E}=0.$$
There is one and only one electromagnetic field ##(\vec{E},\vec{B})##. There's no artificial split needed and also no Helmholtz decomposition is of any merit here. The Helmholtz decomposition of the fields is helpful in the static case, while in the general time-dependent case it leads to overly complicated apparently non-local expressions for the solution of Maxwell's equations which are of little to no use in finding these solutions.

The appropriate mathematical tool is Green's theorem in 4D (Minkowski space), which can be interpreted as a kind of generalized Helmholtz decomposition principle + the choice of retarded solutions over advanced or any superposition including the advanced solutions. You and up with the retarded potentials in the Lorenz gauge (first found by Lorenz not Lorentz) or, equivalently, the Jefimenko equations for the electromagnetic field which is a gauge invariant and thus physically interpretable expression. The conclusion is that the true causal sources of the em. field are the charge-current distributions, all obeying gauge-invariant local field equations.

Concerning a battery at rest with its terminal not connected you have a chemical electromotive force leading to charge separation and and electrostatic field. As any electrostatic field, it's a potential field, because in Faraday's Law ##\partial_t \vec{B}=0##.

For details see

https://doi.org/10.1119/1.19327
 
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  • #47
In order to clear up a few things, when you talk about electric fields in batteries then you could be talking about a number of things:

  • The electric field within the electrode particle.
  • The electric field within the electrolyte.
  • A general electric field within the anode or cathode, usually obtained by volume averaging or asymptotic homogenisation.

Generally in battery modelling, we don't consider magnetic fields at all, which allows the introduction of an electric potential.

The electric fields are set up by the charges on the lithium ions as they diffuse in and out of the electrode particles and get transported within the electrolyte. As for chemistry, that is dealt with via a boundary term for the diffusion of the lithium in/out of the electrodes, the equation used is called the Butler-Volmer equation.
 
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  • #48
However, when working in the battery industry and researching how batteries work, how they fail, and why, things like this become relevant. I was just looking for "electric field in a battery" in the hopes someone else has made the same observations as we have.

The generic dipole electric field lines (stretched a bit due to battery geometry) we can see with thermal imaging because potential causes current, and current causes heat. Heat kills battery, hence, interest in field lines. So far as I can tell nobody has seen this, and I would not normally think of it except something that looks like a field line is produced in some test cells and if we run them long enough, they burn up the same way.

hutchphd said:
It is certainly true that electromagnetic forces control every aspect of a common battery and that the work done on the electrons to provide them emf is all electromagnetic. But to talk about "an electric field" implies a well-described and static structure that does not comport with reality. The process is complicated and dynamic and not characterized by "an electric field". It involves many different electric fields over an ion's trajectory and so the statement should be discouraged.. What is well defined is the path integral for each electron which we call the potential difference.
The controlling electric fields develop at the interfaces between the ~liquid electrolyte and the metal electrodes. The flow of ionic constituents in the electrolyte, driven by chemical processes, is balanced by the contervailing flow of electrons in the attached wire to maintain a dynamic equilibrium. If the flow stops the chemistry stops. I've never found it useful to worry about the detailed fields inside the battery...for instance the voltage is essentially independent of geometry. The internal resistance is not. Of course there is an entire branch of chemistry devoted to such details!
So I would not describe the internal emf of a battery as being associated with "an electric field". The concept is not particularly useful and often confusing. There is a chemical process which supplies a fixed emf to the electrons. That's all you need to know.
If you are really interested in the gory details I suggest that you choose a particular type of battery (say lead acid) and learn everything you can about it. Then you can generalize that knowledge.
 
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  • #49
book_collector said:
However, when working in the battery industry and researching how batteries work, how they fail, and why, things like this become relevant. I was just looking for "electric field in a battery" in the hopes someone else has made the same observations as we have.

The generic dipole electric field lines (stretched a bit due to battery geometry) we can see with thermal imaging because potential causes current, and current causes heat. Heat kills battery, hence, interest in field lines. So far as I can tell nobody has seen this, and I would not normally think of it except something that looks like a field line is produced in some test cells and if we run them long enough, they burn up the same way.
I've had a lot of experience with thermal effects in batteries.
 
  • #50
book_collector said:
However, when working in the battery industry and researching how batteries work, how they fail, and why, things like this become relevant.
Interesting.
The fact that the patterns you see mimic field lines does not necessarilly indicate a direct relation to electric field. There are a variety "conserved flow" situations which generate a similar pattern. But maybe it is electric field driven. I am cerainly not an expert!
 
  • #51
book_collector said:
However, when working in the battery industry and researching how batteries work, how they fail, and why, things like this become relevant. I was just looking for "electric field in a battery" in the hopes someone else has made the same observations as we have.

The generic dipole electric field lines (stretched a bit due to battery geometry) we can see with thermal imaging because potential causes current, and current causes heat. Heat kills battery, hence, interest in field lines. So far as I can tell nobody has seen this, and I would not normally think of it except something that looks like a field line is produced in some test cells and if we run them long enough, they burn up the same way.
hutchphd said:
Interesting.
The fact that the patterns you see mimic field lines does not necessarilly indicate a direct relation to electric field. There are a variety "conserved flow" situations which generate a similar pattern. But maybe it is electric field driven. I am cerainly not an expert!
This is true of course. I am not an electrical engineer, but found myself working in the battery field where I examine battery chemistries and why they work (or don't work). I paid attention in school, and when I saw these anomolies it does in fact resemble electric field lines quite strongly, in fact... nearly identical. These were observed indirectly of course, through thermal signature, but if you pump enough current through a conductor, you generate heat due to resistance.

This of course i not total nonsense, but at the same time I lack the background in EE to really help me out. What I do know is electric field lines are generated by electric charges between a dipole. This is old textbook stuff going back many decades.

Consider a battery, an ion battery, where you have current flowing both ways - electricity in one direction, and positive ions in the other. The pattern is almost a spot-on match for the generic textbook illustration of field lines. Both the anode and cathode materials are good electrical conductors, so the charge migration would be mostly laminar in a layered battery, at some point crossing over to the opposing electrode. This should follow some sort of "path of least resistance" which in a current that is high enough could form waves with nodes and troughs in what is essentially a 2D space.

I was looking through the literature, then the internet for anything that may give me clues to previous work, and came across this thread. Which is full of experts, of course.
 
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  • #52
hutchphd said:
Interesting.
The fact that the patterns you see mimic field lines does not necessarilly indicate a direct relation to electric field. There are a variety "conserved flow" situations which generate a similar pattern. But maybe it is electric field driven. I am cerainly not an expert!
Since there is definitely a current flow, this lends to a field being created since it's a dipole with current flowing in one direction at a specified rate. It was an accidental obersvation but it was ironic because I predicted it as a possibility based on physical findings in the damaged parts of another battery of the same model.
 

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