Electric field within a battery

[etotheipi]

I've been reading through this paper to try and get a better understanding of how batteries work. The analysis there is fine (they consider a voltaic cell to charge a capacitor in order to derive $\Delta V=\varepsilon$, and go via an energy route), but it doesn't really touch upon the fields inside the battery. I assume that there are no non-conservative electric fields inside the battery arising from time varying magnetic fields (this could be wrong...).

If I take A to be the negative terminal and B to be the positive terminal, then the EMF from A to B is the line integral of all forces (per unit charge), $$\varepsilon = \int_A^B (\frac{1}{q} \vec{F}_{chemical} + \vec{E}) \cdot d\vec{r}$$ Whilst the change in potential from A to B is $$\Delta V = -\int_A^B \vec{E} \cdot d\vec{r}$$ I am not sure how we can show from here that these two quantities are equal. To make matters worse, Wikipedia says this:

Inside a source of emf that is open-circuited, the conservative electrostatic field created by separation of charge exactly cancels the forces producing the emf. Thus, the emf has the same value but opposite sign as the integral of the electric field aligned with an internal path between two terminals A and B of a source of emf in open-circuit condition (the path is taken from the negative terminal to the positive terminal to yield a positive emf, indicating work done on the electrons moving in the circuit).[13] Mathematically:

There are quite a few things I don't understand about this. First of all, they use a field $\mathbf{E}_{cs}$ - is that to imply that they are performing a decomposition into a conservative component and non-conservative component $\mathbf{E}_{ncs}$? And if so, why would the non-conservative component arise?

The only way I can make sense of Wikipedia's definition is to suppose that the electric field can be decomposed $\vec{E} = \vec{E}_{cs} + \vec{E}_{ncs}$, and then asserting that $\vec{E}=0$ and $\frac{1}{q}F_{chemical} = -\vec{E}_{cs}$. Though I'm not sure if this makes any sense at all.

So I wondered whether anyone could clear up what is actually going on? Thanks!

 

[hutchphd]

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.

 

[etotheipi]

I see your point, though I don't think all hope is lost for a simple model. We consider a voltaic cell comprised of two electrodes, and at each electrode there is an EMF across the double layer.

I'm aware that there are many complications (electrochemical energies, Volta/Galvani potentials? etc.) however is there a simplified way of describing the means by which an EMF is produced across the electrode-electrolyte interface?

Would you also say that the Wikipedia derivation is incorrect?

 

[Dale]

The cs probably means “charge separation”. Is there anything specific in the text that leads you to believe it means “conservative” instead?

 

[hutchphd]

This is the chemistry of half-cells (oxidation-reduction...) and if you want to dive in, OK, but please don't splash me. Truth be told I do not really understand thermodynamics and do not wish to talk about entropy when designing circuits! So I go with the simplest credible model.

 

[Delta2]

If I take A to be the negative terminal and B to be the positive terminal, then the EMF from A to B is the line integral of all forces (per unit charge),

ε=∫BA(1q→Fchemical+→E)⋅d→​

I think you doing a mistake here , considering the $F_{chemical}$ force in the definition of EMF. The EMF is simply defined as the line integral of the electromagnetic Lorentz force $F_L=(\mathbf{E}+\mathbf{v}\times\mathbf{B})q$ so the EMF is always defined as $$\mathcal{E}=\int_A^B(\mathbf{E}+\mathbf{v}\times\mathbf{B})d\vec{r}$$

I am not an expert in how batteries work but I think inside a battery chemical reactions happen that create separation of charges. We can macroscopically attribute the separation of charges to an $F_{chemical}$ force (though this is not really the case) and the result of separation of charges is a conservative electric field that cancels out $F_{chemical}$. I don't think there is non conservative component of the E-field inside a chemical battery at least not when the battery doesn't have a current through itself.

 

[etotheipi]

I think you doing a mistake here , considering the $F_{chemical}$ force in the definition of EMF. The EMF is simply defined as the line integral of the electromagnetic Lorentz force $F_L=(\mathbf{E}+\mathbf{v}\times\mathbf{B})q$ so the EMF is always defined as $$\mathcal{E}=\int_A^B(\mathbf{E}+\mathbf{v}\times\mathbf{B})d\vec{r}$$

In that case we would have (I'll just call $\mathbf{E}_{cs} = \mathbf{E}$ from here on)

$$\varepsilon = \int_A^B \mathbf{E} \cdot d\mathbf{r}$$

whilst the Wikipedia definition has a negative sign in it for a strange reason (perhaps they are just negating their result so it comes out positive?). However I thought we must also take into account the chemical forces in the definition of the EMF:

 

[Delta2]

However I thought we must also take into account the chemical forces in the definition of the EMF:

I think here Wikipedia goes wrong. If we take the sum, and having in mind that the conservative electric field inside the battery, cancels out $F_{chemical}$ then the EMF of a battery would be always zero.

 

[etotheipi]

I think here Wikipedia goes wrong. If we take the sum, and having in mind that the conservative electric field inside the battery, cancels out $F_{chemical}$ then the EMF of a battery would be always zero.

This makes a lot more sense, but it wouldn’t it then mean that the potential difference from A to B (defined with a negative sign before the integral of the electric field) would carry the opposite sign compared to the EMF from A to B (defined here as the integral of the electric field with no negative sign)?

That is the only thing that’s still causing a bit of doubt.

 

[Delta2]

This makes a lot more sense, but it wouldn’t it then mean that the potential difference from A to B (defined with a negative sign before the integral of the electric field) would carry the opposite sign compared to the EMF from A to B (defined here as the integral of the electric field with no negative sign)?

That is the only thing that’s still causing a bit of doubt.

This looks like a sign convention issue. However it causes me a headache when I try to resolve it. It is well known fact that inside the battery the E-field is doing negative work, it opposes the flow of charges. Outside the battery in the rest of the circuit the E-field does positive work. I think the line integral in the definition of V should be taken for a path outside the battery (the path along the circuit), so we would have $$-\int_{circuit path}\mathbf{E}\cdot\mathbf{dl}=\int_{battery path} \mathbf{E}\cdot \mathbf{dl}$$ or $$\oint\mathbf{E}\cdot \mathbf{dl}=0$$ as it is expected for a conservative e-field.

 

[anorlunda]

This is certainly a FAQ. It is very ripe for an Insights Article (hint hint hint).

There was an excellent post recently on exactly this question. But I can't find it. The Related posts list on the bottom of this page didn't show it.

But this post from an earlier thread might help.

There are two E fields within a battery. One is electrostatic and points from + to -, the other is the source of emf and points in the opposite direction, i.e - to +.

(Positive) charges are impelled by the emf field from the cathode to the anode and simultaneously repelled by the electrostatic field.towards the anode. The two fields are equal and opposite in direction. There is no net E field within a battery.

Dr. Shankar of Yale, in his excellent intro physics course which you can follow on youtube, likens the procress to a ski lift: there is the pull on the rope pulling you up and gravity resisting the pull. He completes the analogy with a battery-resistor circuit.

 

[etotheipi]

This is certainly a FAQ. It is very ripe for an Insights Article (hint hint hint).

There was an excellent post recently on exactly this question. But I can't find it. The Related posts list on the bottom of this page didn't show it.

But this post from an earlier thread might help.

Maybe you were referring to @vanhees71's excellent post here where he discusses $$\mathcal{E}=\int_C \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B})$$ On another note, from what I have read I am not sure if performing the Helmholtz decomposition of the electric field as is mentioned in the post you quote is particularly physical.

Although I'm particularly interested in this case just in evaluating the EMF of the battery. If we perform a line integral of the Lorentz force from the negative to the positive terminal then we obtain a negative EMF, which is oppositely signed to the potential difference in the same direction, which appears to be slightly odd.

I also found another treatment here (starting at end of page 5) which includes the work done by a chemical force from the source in the equation for EMF. However they perform the integration around the whole loop and it's not then obvious to me how this relates to the EMF across just the battery.

 

[etotheipi]

Griffiths gives a slightly different treatment. He says that the force per unit charge $\mathbf{f} = \mathbf{f}_s + \mathbf{E}$ where $\mathbf{f}_s$ is in this case a chemical force from the battery, and $\mathbf{f}_s = \mathbf{0}$ outside the battery.

Furthermore, he says that for an ideal battery the net force on a charge is zero, so $\mathbf{E} = -\mathbf{f}_s$. We can then write the p.d. from $a$ to $b$ as $$V = -\int_a^b \mathbf{E} \cdot d\mathbf{l} = \int_a^b \mathbf{f}_s \cdot d\mathbf{l} = \oint \mathbf{f}_s \cdot d\mathbf{l} = \mathcal{E}$$ where in the last step we have extended the bounds of integration to the whole loop since $\mathbf{f}_s=\mathbf{0}$ outside the battery, and we equate the final integral to $\mathcal{E}$ since $$\mathcal{E} = \oint \mathbf{f} \cdot d\mathbf{l} = \oint \mathbf{E} \cdot d\mathbf{l} + \oint \mathbf{f}_s \cdot d\mathbf{l} = \oint \mathbf{f}_s \cdot d\mathbf{l}$$ because the $\mathbf{E}$ field is conservative.

One thing I note is that the $\mathcal{E}$ is taken around the whole loop. This reasoning would indeed appear to imply that the EMF between the terminals of the cell is zero, since the net force is zero inside the battery. This is a little peculiar...

 

[rude man]

This is certainly a FAQ. It is very ripe for an Insights Article (hint hint hint).

There was an excellent post recently on exactly this question. But I can't find it. The Related posts list on the bottom of this page didn't show it.

But this post from an earlier thread might help.

Thank you @anorlunda for quoting my post. It may have been my earlier one on the subject.

Yes, there are two sources of E fields inside the battery, one I call $E_m$ which is non-conservative $(\nabla \times \bf E \neq 0)$ and the other electrostatic which of course is conservative which I call $E_s$.

I introduced this fact as an effort to show more generally that every emf is associated with a non-conservative E field. I gave several other examples which require understanding of the separate sources of $E_m$ and $E_s$ fields (cf. link below).

In a battery the chemical reaction generates an $E_m$ field which is used to move charge from the cathode to the anode against the $E_s$ field which builds up just as in a capacitor. Equilibrium between the two forces is reached when $E_m = -E_s$. $E_m$ points from cathode to anode; $E_s$ from anode to cathode. So the net E field in the battery is zero.

If anyone wants more of the same they can look at my blog at https://www.physicsforums.com/threads/how-to-recognize-split-electric-fields-comments.984872/

 

[Delta2]

One thing I note is that the EE\mathcal{E} is taken around the whole loop. This reasoning would indeed appear to imply that the EMF between the terminals of the cell is zero, since the net force is zero inside the battery. This is a little peculiar...

I think EMF should always be considered around closed loops, cause it then shows up which is the "real EMF" that is which force does the real work supplying with energy the circuit. In the case of the battery the conservative E-field does positive work in the circuit path, and negative work in the battery path so the total work in the total closed path is zero. The real work is done by the chemical force $\mathbf{f_s}$ the work of which around a closed path is not zero.
The same happens in the case of an AC-generator that has a rotating coil. There inside the coil the total force is again zero cause the $\mathbf{v}\times \mathbf{B}$ force is canceled by the force from a conservative E-field inside the coil. But around a closed loop the contribution from the conservative E-field vanishes and only the $\oint (\mathbf{v}\times \mathbf{B})\mathbf{dl}$ survives which is well known fact that it is the EMF of the rotating coil.

 

[rude man]

One thing I note is that the $\mathcal{E}$ is taken around the whole loop. This reasoning would indeed appear to imply that the EMF between the terminals of the cell is zero, since the net force is zero inside the battery. This is a little peculiar...

It's peculiar because it's wrong. I think you're misinterpreting Griffith. I could see nothing wrong in what he said, including every equation.

OK, by his annotation E is the conservative field (which I call $E_s$) and $f_s$ is the emf-generated, non-conservative field (which I call $E_m$).

Although numerically accurate I don't like equating E (meaning emf here) to V (which is p.d.). emf and p.d are separate entities and must be recognized as such even though they have the same units. An emf is always generated when some other form of energy is changed to electrical .$E_s$ fields are always generated by charges. The flux lines begin and end on charges. There are no charges associated with an $E_m$ field (unless current flows of course).

You need to take the circulation of fields $E$ and $f_s$ separately. The circulation of $E$ is zero since it's conservative. BTW E exists outside the battery body also, from anode to cathode, same as inside, same direction and magnitude. So if there's an external circuit like a series resistor you can take your closed contour around the resistor and either inside or outside the battery body. If there is just the battery then the closed E loop is taken inside and outside the battery terminals. The result is of course zero in both cases.

But $f_s$ is really the emf-generated electric field even though lots of folks don't seem to like calling it an electric field (they're wrong! ). In any case when you compute the circulation of $f_s$ you get the emf of the battery, not zero, just as Griffith does.

You might refer to my blog at https://www.physicsforums.com/threads/how-to-recognize-split-electric-fields-comments.984872/

 

[Delta2]

But fsfsf_s is really the emf-generated electric field even though lots of folks don't seem to like calling it an electric field (they're wrong! ). In any case when you compute the circulation of fsfsf_s you get the emf of the battery, not zero, just as Griffith does.

It is not always a (non conservative) electric field $\mathbf{E_m}$ that plays the role of the EMF. I am not sure what happens inside a battery but in the case of a rotating coil its the $\mathbf{v}\times\mathbf{B}$ force that is the real EMF and of course it is obvious it is not an electric field.

 

[rude man]

It is not always a (non conservative) electric field EmEm that plays the role of the EMF. I am not sure what happens inside a battery but in the case of a rotating coil its the v×Bv×B force that is the real EMF and of course it is obvious it is not an electric field.

v x B is equivalently an E field. The Lorenz force plays the part of the E field. In the coil there are thus two E fields, one Em and one Es, and the two add to zero as they must unless the coil carries current. Exactly as in a battery.

It can be shown rigorously using relativity that E and v x B are the same field, basically. cf. Feynman Lectures on Physics.

ALSO: if you're a hardy bug sitting on your rotating coil there is no v, thus no Lorentz force. The bug sees only the emf and electrostatic fields. he can measure the line-integrated Es field with a voltmeter provided its leads are outside the magnetic field.

 

[Delta2]

v x B is equivalently an E field. The Lorenz force plays the part of the E field. In the coil there are thus two E fields, one Em and one Es, and the two add to zero as they must unless the coil carries current. Exactly as in a battery.

It can be shown rigorously using relativity that E and v x B are the same field, basically. cf. Feynman Lectures on Physics.

Ok if you bring in relativity then I guess $\mathbf{v}\times\mathbf{B}$ becomes equivalent to an electric field. But i suck in relativity (so i didn't know that) and in the regime of classical physics alone the magnetic and the electric field are different quantities.

 

[rude man]

Ok if you bring in relativity then I guess $\mathbf{v}\times\mathbf{B}$ becomes equivalent to an electric field. But i suck in relativity (so i didn't know that) and in the regime of classical physics alone the magnetic and the electric field are different quantities.

See the "ALSO" note to my previous post. You see, it's all "relative"!
BTW I suck in relativity too but I do remember Feynman discussing this equivalence in his lectrure notes somewhere

 

[etotheipi]

Yes, there are two sources of E fields inside the battery, one I call $E_m$ which is non-conservative $(\nabla \times \bf E \neq 0)$ and the other electrostatic which of course is conservative which I call $E_s$.

But I thought there is only one $\mathbf{E}$ field; $\mathbf{f}_s$ doesn't appear to be an electric field here, since it's of a chemical origin?

 

[Adesh]

I am not sure how we can show from here that these two quantities are equal

EMF and Voltage difference are two different quantities.

 

[etotheipi]

EMF and Voltage difference are two different quantities.

Yes, but they're numerically equal for a battery in a circuit w/ zero internal resistance. In CA we usually think of the EMF being "across the cell", however here it seems they're only equal if the EMF is evaluated around a closed loop.

 

[Adesh]

Yes, but they're numerically equal for a battery in a circuit w/ zero internal resistance. In CA we usually think of the EMF being "across the cell", however here it seems they're only equal if the EMF is evaluated around a closed loop.

For qualitative explanation, EMF is the energy required to maintain the potential difference between the terminals (case for DC circuits).

In AC circuits, Electric Field doesn’t have a potential, so the qualitative definition of EMF is altered and we say “the total work done in moving a charge around a closed loop”.

 

[etotheipi]

I think EMF should always be considered around closed loops, cause it then shows up which is the "real EMF" that is which force does the real work supplying with energy the circuit.

I think this has to be correct. I think it is still consistent with the electrochemical approach. In any given half cell there is an EMF of $\mathcal{E}_i$ across the double layer. Then around a closed loop through a battery containing two half cells, the total EMF will be $\mathcal{E}_1 - \mathcal{E}_2 = \mathcal{E}$.

I have usually thought of this as an EMF across the terminals (i.e. the sum of two EMFs across two double layers) but I suppose it is perhaps more accurate to think of it as the EMF around the whole loop?

 

[Delta2]

I think this has to be correct. I think it is still consistent with the electrochemical approach. In any given half cell there is an EMF of $\mathcal{E}_i$ across the double layer. Then around a closed loop through a battery containing two half cells, the total EMF will be $\mathcal{E}_1 - \mathcal{E}_2 = \mathcal{E}$.

I have usually thought of this as an EMF across the terminals (i.e. the sum of two EMFs across two double layers) but I suppose it is perhaps more accurate to think of it as the EMF around the whole loop?

I am not sure which example you have in mind here (I can't access the paper linked in the OP), but yes I believe its more accurate to take EMF around closed paths only. That way the contribution from conservative E-fields is being vanished. Or you can consider EMFs in open paths as long as you don't count in the conservative E-fields but only the non conservative E-fields(that come from time varying B-fields or from $v\times B$ type, that is from transformer EMF or motional EMF E-fields).

 

[etotheipi]

v x B is equivalently an E field. The Lorenz force plays the part of the E field. In the coil there are thus two E fields, one Em and one Es, and the two add to zero as they must unless the coil carries current. Exactly as in a battery.

It can be shown rigorously using relativity that E and v x B are the same field, basically. cf. Feynman Lectures on Physics.

ALSO: if you're a hardy bug sitting on your rotating coil there is no v, thus no Lorentz force. The bug sees only the emf and electrostatic fields. he can measure the line-integrated Es field with a voltmeter provided its leads are outside the magnetic field.

I am not too sure if I follow this part. With Stoke's Law + Reynold's transport theorem we can manipulate $$\nabla \times \vec{E} = -\partial_t \vec{B}$$ into $$\mathcal{E} = \oint (\vec{E} + \vec{v} \times \vec{B}) \cdot d\vec{r} = -d_t \Phi$$Now it appears you are saying we can perform a Helmholtz decomposition of $\vec{E} = \vec{E}_s + \vec{E}_m$, where $\vec{E}_s$ is such that $\oint \vec{E}_s \cdot d\vec{r} = 0$. That's certainly possible, however then you're left with $$\mathcal{E} = \oint (\vec{E}_m + \vec{v} \times \vec{B})\cdot d\vec{r}$$And that doesn't appear very helpful? Of course, if $\vec{B} = \vec{0}$ or if $\vec{B} \parallel \vec{v}$ then the EMF would just become $\mathcal{E} = \oint \vec{E}_m \cdot d\vec{r}$.

But I would be inclined to say that the other force inside the battery is not a non-conservative component of the electric field, but merely another chemical force.

 

[Adesh]

If I take A to be the negative terminal and B to be the positive terminal, then the EMF from A to B is the line integral of all forces (per unit charge),

$$\varepsilon=\int_{A}^{B}(\frac{1}{q} \mathbf{F_{chemical}} +\mathbf E) \cdot d\mathbf r$$​

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.

 

[etotheipi]

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.

Right but this is what's causing some of the confusion, in that I'm not sure if it makes sense to speak of an open loop EMF. Certainly as was shown in Griffiths the EMF around the entire loop reduces to the integral of the chemical force between the terminals.

But it would be strange if the definition of EMF differed depending on what type of loop (open/closed) you evaluate around, in terms of neglecting some forces in some cases and not in others? So after reflection I don't think it makes sense to speak of an EMF across an open loop.

 

[Dale]

Thank you @anorlunda for quoting my post. It may have been my earlier one on the subject.

Yes, there are two sources of E fields inside the battery, one I call $E_m$ which is non-conservative $(\nabla \times \bf E \neq 0)$ and the other electrostatic which of course is conservative which I call $E_s$.

I introduced this fact as an effort to show more generally that every emf is associated with a non-conservative E field. I gave several other examples which require understanding of the separate sources of $E_m$ and $E_s$ fields (cf. link below).

In a battery the chemical reaction generates an $E_m$ field which is used to move charge from the cathode to the anode against the $E_s$ field which builds up just as in a capacitor. Equilibrium between the two forces is reached when $E_m = -E_s$. $E_m$ points from cathode to anode; $E_s$ from anode to cathode. So the net E field in the battery is zero.

If anyone wants more of the same they can look at my blog at https://www.physicsforums.com/threads/how-to-recognize-split-electric-fields-comments.984872/

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. It is therefore also conservative.

There is therefore no non-conservative potential in a battery in an open circuit.

 

[etotheipi]

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. It is therefore also conservative.

This is a perfect explanation. There is only the one $\vec{E}$ field; in this case it is conservative and inside the battery it points from the positive to the negative terminal. There is also a chemical force (which as you say is derivable from a chemical potential) that opposes this electric force inside the battery. This is in full accordance with the explanation of Griffiths.

I agree that we can always perform a Helmholtz decomposition of an electric field into a conservative and non-conservative part, as you can do with any vector field, however in this case it is a very boring operation since there is already only a conservative component and you end up right back where you started!

 

[Dale]

There is also a chemical force (which as you say is derivable from a chemical potential) that opposes this electric force inside the battery. This is in full accordance with the explanation of Griffiths.

I agree that we can always perform a Helmholtz decomposition of an electric field into a conservative and non-conservative part, as you can do with any vector field, however in this case it is a very boring operation since there is already only a conservative component and you end up right back where you started!

Yes. If you want to separate the chemical potential from the electric potential then the Helmholtz decomposition is not the way to go.

I have no problem with doing a decomposition of the field, but it isn’t conservative vs non-conservative.

 

[rude man]

For qualitative explanation, EMF is the energy required to maintain the potential difference between the terminals (case for DC circuits).

In AC circuits, Electric Field doesn’t have a potential, so the qualitative definition of EMF is altered and we say “the total work done in moving a charge around a closed loop”.

AC circuits also have potential. If I have an ac generator connected to a resistor the resistor has potential only across it. The generator has emf and potential in its windings. The windings, being conductors, have to have zero E field. This is done by magnitude-equal emf and potential E fields. The circulation of the potential around a loop is zero (Kirchhoff); the circulation of the emf field is the emf (Faraday in this case).

AC is no different than DC in this regard.

 

[rude man]

But I thought there is only one $\mathbf{E}$ field; $\mathbf{f}_s$ doesn't appear to be an electric field here, since it's of a chemical origin?

There is room for semantic difference in the case of a battery but in all electrical effects $f_s$ is an electric field.

Please see attached excerpt from a well-known textbook (H.H. Skilling, Fundamentals of Elecric Waves, 2nd edition, 10th printing, pp. 73-75).

 

[etotheipi]

There is room for semantic difference in the case of a battery but in all electrical effects $f_s$ is an electric field.

I am not sure about this. Griffiths says

The physical agency responsible for $\mathbf{f}_s$ can be many different things: in a battery it’s a chemical force; in a piezoelectric crystal mechanical pressure is converted into an electrical impulse; in a thermocouple it’s a temperature gradient that does the job; in a photoelectric cell it’s light; and in a Van de Graaff generator the electrons are literally loaded onto a conveyer belt and swept along. Whatever the mechanism, its net effect is determined by the line integral of $\mathbf{f}$ around the circuit:

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.