# How is surface charge accumulated?

• B
• gionole
In summary: in the case when wire is not connected to battery, but we have circuit open, what happens to the surface charges when we connect wire to battery ? does it cause the surface charges to move or does it just add more surface charges to the wire ?
gionole
TL;DR Summary
How is surface charge accumulated with battery connected, but circuit still open
let's discuss copper wire all along so we only focus on the specific wire. Let's discuss 2 cases(case 1 is hopefully correct, so if it is, would be good to mention it).

I'm sorry that this text got so bigger(didn't expect it), but wanted to mention my thought process. Somehow, none of the videos/articles explain the below. They all explain why surface charges must exist because if they don't, E won't be constant. Maybe they do, but then I'm not understanding it. My questions and theorem are about how they're produced in the first place. Hope you read it and let me know where I'm making mistakes.

Case 1:
We only have copper wire which is not connected to battery or anything else. It's just wire and thats it. In this case, we know there's no surface charges presented on it and the situation is it has got copper atoms inside it in which 1 valence electron from each atom kind of jumps around back and forth within some small area due to electric field of its neighbouring atoms' protons' electric fields. Due to this, electrons do actually move, but within small area and overally, their drift velocity is 0.

Case 2: We connect the wire to battery. Note that we don't close the circuit. Well, in this case, what we know is atoms have been arranged to become surface charges and in this case, E inside wire is 0 everywhere. I wonder about this case and will mention what I'm curious.

So in the wire, there were copper atoms that for a sec, become positive ions, then another sec, become neutral atoms due to electron "seas" jumping into and from it all the time. but when we connected it to the battery(circuit is still open), it's said that on the positive terminal end, wire has positive charges on the surface accumulated, while on the negative terminal end, wire has negative charges on the surface. Could you explain how this happened ? what exactly caused this and why are more surface charges near the terminals of the battery and why they decrease over distance of the wire ?

My Theorem: Maybe what happens is battery's positive terminal's positive charges attract free electrons that are very close to them which causes the wire part that's very close to the terminal to have more positive ions(before battery, it didn't, but now it has). Since positive ions appeared(in the wire, there're more positive ions now than free electrons while before - their count was the same) in the wire near the part of terminal, 2 things take effect now:
• a. these positive ions try to repel from each other.
• b. these positive ions try to attract free electrons that are in the later part of the wire.
due to a), these positive ions end up going from each other as far as possible, so some of it end up on one part of the surface, other ones on another part of the surface. This process happens so quickly that we shouldn't forget the effect of b). While these positive ions were going to the surface, free electrons were attracted by them so some free electrons also go to them. Due to b), in the later part of the wire, we got positive ions accumulated(they lost their free electrons). So now, they also try to get away from each other as far away as possible and end up on the surface.

The good questions now(if ofc, my theorem is right) would be the following:

Please note that I'm only discussing the case during my theorem and in questions(before wire is connected to the battery and when wire is connected to battery but here circuit is still open - the sides of the wires are not connected to each other).

Question 1:
In my explanation, what would you add so that it explains why surface charges are much less the more you go through the wire from positive terminal ?
Question 2: I mentioned that positive terminal of the battery since it had positive charges, when it got connected to wire, it caused some of the nearby free electrons of the wire to be attracted to it, so I guess these free electrons move to battery and stay there ? Which in the end means some of the free electrons completely disappeared from the wire. This seems more logical than to say that free electron and battery's positive charge both go to each other to meet (I think this is wrong as in the positive terminal, there's Cu2+ ion so definitely it's attracting free electrons). So these free electrons move to battery and Cu2+ become Cu. (hopefully I'm explaining this well enough - all these happens before circuit is closed). Thoughts whether this is correct ?
Question 3: Well, I was thinking now how the negative charges end up on the surface on the negative terminal side. Well, on the battery's negative end, we got negative charges or maybe Zn metal that's so eager to lose its electrons. Nearby atoms of the wire definitely can cause these electrons of the battery to get into the wire which logically seems that in the negative terminal's side, the wire has more negative charge(before connecting it to battery, there was the same number of free electrons and positive ions in the wire, but after connecting the wire to battery, as I said, some electrons from battery hopped into the wire, and there're more negative charges overally. I believe this causes the nearby wire's charges to start repelling each other and they end up on the surface. But I have the same question here as in the Question 1: why exactly negative surface charges are less the more you go through the wire ?

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gionole said:
TL;DR Summary: How is surface charge accumulated with battery connected, but circuit still open

Case 2: We connect the wire to battery. Note that we don't close the circuit. Well, in this case, what we know is atoms have been arranged to become surface charges and in this case, E inside wire is 0 everywhere. I wonder about this case and will mention what I'm curious.
Recalling that a capacitor or condensor is a pair of facing metal plates, we may restate your case as why condensor or capacitor is charged when connected to battery ?

vanhees71
anuttarasammyak said:
Recalling that capacitors or condensors are a pair of facing metal plates, we may restate your case as why condensor or capacitor is charged when connected to battery ?
Well, To not take things far and more complicated, I am only asking what I am asking which is you only got battery and wire. I did not mention anything else.

gionole said:
TL;DR Summary: How is surface charge accumulated with battery connected, but circuit still open

let's discuss copper wire all along so we only focus on the specific wire. Let's discuss 2 cases(case 1 is hopefully correct, so if it is, would be good to mention it).

I'm sorry that this text got so bigger(didn't expect it), but wanted to mention my thought process. Somehow, none of the videos/articles explain the below. They all explain why surface charges must exist because if they don't, E won't be constant. Maybe they do, but then I'm not understanding it. My questions and theorem are about how they're produced in the first place. Hope you read it and let me know where I'm making mistakes.

Case 1:
We only have copper wire which is not connected to battery or anything else. It's just wire and thats it. In this case, we know there's no surface charges presented on it and the situation is it has got copper atoms inside it in which 1 valence electron from each atom kind of jumps around back and forth within some small area due to electric field of its neighbouring atoms' protons' electric fields. Due to this, electrons do actually move, but within small area and overally, their drift velocity is 0.

Case 2: We connect the wire to battery. Note that we don't close the circuit. Well, in this case, what we know is atoms have been arranged to become surface charges and in this case, E inside wire is 0 everywhere. I wonder about this case and will mention what I'm curious.

So in the wire, there were copper atoms that for a sec, become positive ions, then another sec, become neutral atoms due to electron "seas" jumping into and from it all the time. but when we connected it to the battery(circuit is still open), it's said that on the positive terminal end, wire has positive charges on the surface accumulated, while on the negative terminal end, wire has negative charges on the surface. Could you explain how this happened ? what exactly caused this and why are more surface charges near the terminals of the battery and why they decrease over distance of the wire ?

My Theorem: Maybe what happens is battery's positive terminal's positive charges attract free electrons that are very close to them which causes the wire part that's very close to the terminal to have more positive ions(before battery, it didn't, but now it has). Since positive ions appeared(in the wire, there're more positive ions now than free electrons while before - their count was the same) in the wire near the part of terminal, 2 things take effect now:
• a. these positive ions try to repel from each other.
• b. these positive ions try to attract free electrons that are in the later part of the wire.
due to a), these positive ions end up going from each other as far as possible, so some of it end up on one part of the surface, other ones on another part of the surface. This process happens so quickly that we shouldn't forget the effect of b). While these positive ions were going to the surface, free electrons were attracted by them so some free electrons also go to them. Due to b), in the later part of the wire, we got positive ions accumulated(they lost their free electrons). So now, they also try to get away from each other as far away as possible and end up on the surface.

The good questions now(if ofc, my theorem is right) would be the following:

Please note that I'm only discussing the case during my theorem and in questions(before wire is connected to the battery and when wire is connected to battery but here circuit is still open - the sides of the wires are not connected to each other).

Question 1:
In my explanation, what would you add so that it explains why surface charges are much less the more you go through the wire from positive terminal ?
Question 2: I mentioned that positive terminal of the battery since it had positive charges, when it got connected to wire, it caused some of the nearby free electrons of the wire to be attracted to it, so I guess these free electrons move to battery and stay there ? Which in the end means some of the free electrons completely disappeared from the wire. This seems more logical than to say that free electron and battery's positive charge both go to each other to meet (I think this is wrong as in the positive terminal, there's Cu2+ ion so definitely it's attracting free electrons). So these free electrons move to battery and Cu2+ become Cu. (hopefully I'm explaining this well enough - all these happens before circuit is closed). Thoughts whether this is correct ?
Question 3: Well, I was thinking now how the negative charges end up on the surface on the negative terminal side. Well, on the battery's negative end, we got negative charges or maybe Zn metal that's so eager to lose its electrons. Nearby atoms of the wire definitely can cause these electrons of the battery to get into the wire which logically seems that in the negative terminal's side, the wire has more negative charge(before connecting it to battery, there was the same number of free electrons and positive ions in the wire, but after connecting the wire to battery, as I said, some electrons from battery hopped into the wire, and there're more negative charges overally. I believe this causes the nearby wire's charges to start repelling each other and they end up on the surface. But I have the same question here as in the Question 1: why exactly negative surface charges are less the more you go through the wire ?
@Dale I asked this(completely different one) on stackexchange but yesterday you enlightened me that surface charges are what makes E constant. So I went ahead and thought the above. If you got time and nerves to bear with my horrible knowledge, I would appreciate it. It is a big text but should be easy to read. Scroll up to the first post here(there should be one theorem - my assumptions and then 3 questions).

Big thank you

The article http://amasci.com/elect/poynt/poynt.html might be of your interest. However there is resistor for the closed surcuit, the discussion is same for open end and plus wire and minus wire form a condensor to store energy between. Free electron excess or scarce front goes with light speed starting from the connection to battery.

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vanhees71
gionole said:
TL;DR Summary: How is surface charge accumulated with battery connected, but circuit still open

We only have copper wire which is not connected to battery or anything else. It's just wire and thats it. In this case, we know there's no surface charges presented on it
Well, if you have a wire, disconnected from everything, you can still have surface charge on the wire. This is called electrostatic induction:

https://en.m.wikipedia.org/wiki/Electrostatic_induction

Some external E field is necessary. This could be provided by the terminals of a battery, even if the battery is not connected to the wire. It also could be provided by an electret, a vandegraff generator, or any number of other sources of an E field.

In the electrostatic case the surface charge will accumulate in such a way as to make the E field inside the wire be zero.

gionole said:
there were copper atoms that for a sec, become positive ions, then another sec, become neutral atoms due to electron "seas" jumping into and from it all the time
I don’t think this is a useful way to view things. Even on a quantum mechanical level the conduction electrons don’t belong to a single atom and jump from one to another. So such a model adds a lot of complexity that is neither necessary nor correct.

It is better to just consider a conductor to have a continuous free charge density that can move in response to an E field according to Ohm’s law. Ohm’s law is so simple that any attempt to explain it at a microscopic level will inevitably add unnecessary complication.

Ohm’s law says: $$\vec J = \sigma \vec E$$ where ##\sigma## is the conductivity. It is hard to beat for simplicity.

gionole said:
Could you explain how this happened ? what exactly caused this and why are more surface charges near the terminals of the battery and why they decrease over distance of the wire ?
So, consider a straight wire in a uniform E field. This wire is not connected to a battery, just in an external E field.

Now, if the wire is initially uncharged then by Ohm’s law we will have current all along the wire. The current cannot jump off the ends of the wire, so it will just stop at the ends.

By conservation of charge that means that the ends will get an increasing charge. The end that the current is flowing towards will gain positive charge and the end that the current is flowing away from will gain negative charge. These are the surface charges, they form on the surface because that is where the current suddenly stops.

This surface charge has its own E field, which points in the opposite direction of the current. By Ohm’s law, this reduces the current caused by the external field. Charge continues to accumulate until the surface charge field is locally as strong as the external field and cancels the external field. Then there is zero E field in the conductor and by Ohm’s law the current stops.

vanhees71
Huge thanks Dale. Whenever you have time, would be lovely if you could reply. I think this is not easy topic and might take a little bit more than 1-2 replies. I feel shy for asking to answer such many things though.

Well, if you have a wire, disconnected from everything, you can still have surface charge on the wire. This is called electrostatic induction:

https://en.m.wikipedia.org/wiki/Electrostatic_induction

Some external E field is necessary. This could be provided by the terminals of a battery
I hope you mean a case when wire is just put beside the battery. None of the wire’s end is connected to battery, but they are just put beside each other. Is this the case you are referring that surface charges will occur in a wire ? If so, then if we got only wire and nothing else, then there is no external E field so no surface charges at all. Then, my reasoning that to produce surface charges in a wire, we need it connected to battery is not ultimately neccessary and just putting them beside each other is enough. I guess distance(how close) we put them together should make a difference whether surface charges will occur or not. All correct ?

I don’t think this is a useful way to view things. Even on a quantum mechanical level the conduction electrons don’t belong to a single atom and jump from one to another. So such a model adds a lot of complexity that is neither necessary nor correct.
I think this is how it is explained everywhere. Even in some well respected courses(MIT). For sure, electron does not belong to one atom, thats what I meant - it jumps around. This somehow seems easy for me as atoms(ions) can attract electron by their E field.

Ohm’s law says: where σ is the conductivity. It is hard to beat for simplicity. J=σE.

J is said to be electric current density. I guess this is electrons density ? As in when current moves, in unit per area, how much electrons are there. Correct ?

Now, if the wire is initially uncharged then by Ohm’s law we will have current all along the wire. The current cannot jump off the ends of the wire, so it will just stop at the ends.

By conservation of charge that means that the ends will get an increasing charge. The end that the current is flowing towards will gain positive charge and the end that the current is flowing away from will gain negative charge. These are the surface charges, they form on the surface because that is where the current suddenly stops.
Well, as it turns out, the same effect(producing surface charges) happens even when battery and wire is put besides each other and not connected(as in battery's terminals not connected to wire at all) and exactly the same effect would happen if terminals of battery is connected to the wire's ends(even when circuit is open). Correct ? Then if so, to discuss surface charges, we don't have to focus battery connected to wire.

So the great question would be how current would be produced in a wire when it's put in external E field. Since you mention uniform E field, then it should have the parallel direction of the wire. if not(perpendicular let's say), electrons wouldn't move in parallel direction of the wire. Ok, if correct, then I've come along to this point. Then what happens is electrons of the wire("sea/free" electrons) start moving in one direction due to this E field, then to the end of the wire(where electrons move to), there, they will accumulate - in this time, other electrons as well go there, but electrons that already got there start repelling new electrons going on there, so at some point, I kind of understand that to the last end, there will be more electrons there than at a previous part of the wire. Similarly, because of the ones that left, positive ions without electrons appeared(well, they were there, but they never were completely ions as I said). but now they're, they start repelling each other so they end up again on the surface, but on the other end of the wire. Would you say I'm right ?

This surface charge has its own E field, which points in the opposite direction of the current.
I think, since charges appear on both surfaces(i mean surfaces of the one end only, the end of the wire has surfaces from each side), then I think if wire is put straight(in horizontal condition), E field of these surface charges's perpendicular components should be canceling each other and their E field only contributes to horizontal component(that is parallel to the wire). Thoughts ?

I don't want to ask too many questions even though I did, but this seems better to be productive then trying to get everything at the same time ?

gionole said:
I hope you mean a case when wire is just put beside the battery. None of the wire’s end is connected to battery, but they are just put beside each other. Is this the case you are referring that surface charges will occur in a wire ?
Yes. Of course any other E field source will have a similar effect, that is just one example.

gionole said:
J is said to be electric current density. I guess this is electrons density ? As in when current moves, in unit per area, how much electrons are there. Correct ?
Yes, ##\vec J## is the current density. It is ##\rho \vec v## where ##\rho## is the charge density of the charge carriers and ##\vec v## is their velocity.

The charge carriers can be electrons in metals, ions in electrolytes or plasmas, etc. The sign of the charge carriers rarely matters.

gionole said:
Then if so, to discuss surface charges, we don't have to focus battery connected to wire.
That is correct. The difference with the battery is that the current doesn’t have to stop at the surface of the wire that is touching the terminal. Otherwise the physics is similar.

gionole said:
now they're, they start repelling each other so they end up again on the surface, but on the other end of the wire. Would you say I'm right ?
Yes. I believe you are correct.

gionole said:
and their E field only contributes to horizontal component(that is parallel to the wire). Thoughts ?
That is correct inside the wire. Outside the wire they produce a standard dipole field.

gionole said:
I don't want to ask too many questions even though I did, but this seems better to be productive then trying to get everything at the same time ?
No worries. You seem to be making good progress in your understanding. So I am content with your questions

vanhees71 and hutchphd
@Dale Thanks so much. This is exciting.

Physics is so interesting. The more you try to dive deeper, the more questions pop up and it's a never ending cycle. I included most of my remaining questions. After these ones, I only got left 1-2 (hopefully the below ones wouldn't take me to the rabbit hole). They're easy to read/follow I believe for you. It seems big, but I mention everything even twice so you know exactly what I mean and don't try to think to know what i meant.

Q1: I have been thinking about your example of uniform E field near the wire. If we imagine that wire is put horizontally, and E is uniform horizontally, then I understand the surface charges situation as I explained in my previous reply. but now imagine E is not only horizontally, it also has y component as well. The funny thing is when electrons start to move(left - direction of field's force), they also move bottom(let's say E field is above the wire only) then, when electrons move, they move left/bottom and they continue like this - when they get to the end, they will be on the bottom surface only. when new electrons come in there, they repel from electrons already present on bottom surface, they then try to move backwards, but they can't also do that as new electrons come in so they repel from them as well(in my opinion - they would repel more from these new coming ones), I don't think they try to move up because they're already bottom and the repelling force is not upwards - even if it was(which is not), E wouldn't let them go up. So we kind of got a situation where electrons would only accumulate on the below surface(not the upper surface) - how nice is this ? haha

Q2: I think I have to confirm something so I understand it very well(I believe this is quite important). So if we imagine only one stream of electrons move to the end(let's say they're 20 electrons), when they move to the end, they will still end up on the surface. In the last example, I said this is due to the fact that new stream of electrons coming as well and their repelling effect doesn't let them move to the opposite direction to the wire, so the try to go to surface. But even if there's no other stream of electrons and this 20 electrons got to the end, now on the end, there's excess of electrons(these 20 and the original ones as there were already atom's original ones) - So these 20 try to move in the opposite direction due to repulsion from original ones, but E field doesn't let them, so since they can't move to the right, from those original electrons, they experience perpendicular component of E(this is E field of original electrons) and still go to the surface. I wonder if we follow this example, how strong E(our uniform field) must be so that it doesn't let these 20 electrons move back a little bit in the opposite direction. Note that if it's not strong enough, repulsion effect from original electrons will be bigger than E field and it will win hence we won't necessarily have surface charges (well we might still have as repulsion effect won't be only horizontal, but in both directions and might end up with surface again but most likely, we won't have surface charges on the very last end of the wire, but some distance away from it). Do you follow this example ? agree with me ?

Q3: So on the surface, from the last end to the wire to the below part, we won't have continous charge distribution as when first electrons got to the last end's wire and new electrons coming in due to E field again, E wants to take them to the very end, but these electrons also feel repelling force from the wire's last end's surface electrons. So they will kind of maintain some distance from them I believe, but now these new electrons will accumulate again in a different region(not on the very end, but a little bit distanced away from it) and then they will experience the same effect and go to surface. That's why I believe we don't have continuous distribution of surface charges. So I believe that unless E field(our uniform one) is strong enough, repulsion effect might override it and we might get a very different case(who knows what). Agreed?

--- Now time to connect to battery ----- The Q4 has a) and b) part.

Q4: When we connect wire to the battery(circuit still open), now E field of positive charges of terminal do the same thing as our uniform E field was doing. I attached the image. (let's focus on the positive terminal's wire side - when I mention wire, I mean that part of the wire). In there, if we look at positive terminal and its side of wire, what exactly is going on ? I believe battery's positive charges attract nearby free electrons, and the region where free electron were, now contains positive ions which try to repel from each other. and they also start repelling from battery's positive charges.

- a) What I wonder is if these positive ions start moving to the end wire or they immediatelly get to the surface that is close to the battery ? but it's hard to imagine they would get to the surface there as even though each ion tries to repel from each other, they also get repulsion from positive charges of battery. you might say that E of battery there is horizontal, but at that region, wire starts to bend where E also is bent. So these ions might be moving below before getting to the surface. Hard to imagine here the flow. This is what I don't get, but I understand how we would get surface charges near the end of the wire. When electrons got attracted to terminal, ions appeared and they start attracting electrons from the next part of the wire and so on and forth. This happened even for the last end of the wire - electrons from it also start moving to the positive terminal's direction and we got left with ions there. So they start repelling each other, but funny part is they couldn't go to the positive terminal's direction as E direction on the below is to the right, so they're stuck and go to the surface.

- b) when electrons got attracted to the battery terminal's charges, what happens to these electrons ? do they get combined with positive charge of battery, making them neutral ? but if this were the case, battery would soon end up having all the neutral charge on its terminal. but if this is not the case, where do electrons go then ? do they actually get attracted to the battery's positive terminal in a way that positive terminal's charges just stay where they're and they just suck in electrons or they meet halfway to the electrons ? but I don't believe they would meet half way as by nearby electrons, we mean in an infinetisemal region close to terminal, electrons would be there(original free electrons), so battery must be sucking them. Not sure.

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gionole said:
let's say E field is above the wire only
What do you mean by this?

gionole said:
So we kind of got a situation where electrons would only accumulate on the below surface(not the upper surface)
You are forgetting that where the mobile charges leave there is a surface charge of the opposite sign left behind. This is automatic. You automatically get charge accumulation on the opposite surface too.

gionole said:
might end up with surface again but most likely, we won't have surface charges on the very last end of the wire, but some distance away from it)
Yes, you will have surface charges. And yes, it is not necessarily the exact end. In electrostatic induction you tend to get the most charge accumulating on portions of the surface with a small radius of curvature. This is why lightning rods are sharp. But surface charges do spread out as much as they can.

gionole said:
That's why I believe we don't have continuous distribution of surface charges.
It is still continuous, but it is not uniform.

gionole said:
what happens to these electrons ? do they get combined with positive charge of battery, making them neutral ?
Yes.

gionole said:
but if this were the case, battery would soon end up having all the neutral charge on its terminal
Yes. While the circuit is open the wire acts as an extension of the terminal. The charge will accumulate on regions of high curvature.

@Dale
You are forgetting that where the mobile charges leave there is a surface charge of the opposite sign left behind. This is automatic. You automatically get charge accumulation on the opposite surface too.
Well, if you imagine E is uniform but with an angle near the wire, (E is tilted so it has x and y component both). So the force is to the left and to the bottom both. If so, electrons not only move to the left, but to the bottom as force is taking them below as well. This is why I said we won't have surface charges on both surface, because even if electrons want to get on on the top surface, they're being taken by the force to the bottom as well. So If E is strong enough, it won't let them get on the top(up) surface. It's true that lots of electrons will start repelling from new electrons coming in to the left and they will start repelling from each other and they would want to go to the up/top surface as well, but if E in the y(below) direction is strong enough, it will override the repelling force and take them to the bottom surface. The first image attached shows what I mean by E field as tilted. Even though, you're right that leaving mobile charges means opposite sign charges are produced, they also go in direction of the force which is to the left and below. Thats why I said we will have surface charges on the below surface.

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Yes, you will have surface charges. And yes, it is not necessarily the exact end.
With Q2, What I meant was if external E is not strong enough, electrons that accumulate on the very end of the wire mightn't necessarily go to the surface right away. This is due to the fact that repulsion effect between excess electrons can cause them to move backwards(in the opposite direction of their movement). The reason they might be repelled backwards is E is not strong, so repelling force > E, so electrons as they repel from each other(some go top, some go down, they also repel horizontally, as E is not strong enough, so some also get repelled backwards). What this results in is not necessarily we could have surface charges on the very end, and the density of surface charges could be less, but if E is strong enough, density of surface charges will be higher(more charges would go to the surface). I think that should be correct assumptions ?
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With your words, what's the reason surface charges on the last end of the wire have bigger density and as you get far away from the end of the wire, density gets less and less ? in copper atom, would all of the valence electrons(each atom has 1) would end up on the surface or some of them might not ?

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With battery connected to the wire(circuit open though), I think on the positive terminal, surface charges accumulate right away on the surface closest to the terminal. On my second attached image, we should have option (A) occuring. I guess its electrons (since they're closest) to the terminal, they right away lose valence electrons, so we got ions produced at the quickest time and at the closest to terminal, hence their repelling from each other make them on the surface, since positive ions accumulate this fast, now, it could be them that's attracting valence/free electrons from the later part of the wire. The place free electrons leave from the later part of the wire, that place contains again positive ions and they again form surface charges. What I don't get though, the electrons that come from the later part of the wire, do they get closer to the ions(the ions that got on the surface - they were closest ones to terminal) ? If so, these ions that were on the surface near the terminal must become again neutral ? but if so, then their E field mightn't help with E = 0. or do these electrons get to the positive terminal's charges ? but note that, when these electrons move from later part of the wire to the terminal, they first have to meet the positive ions that are already on the surface(on 2nd image, option A shows these ions) because these ions also attract them and they're closer. So that got me thinking if electrons jump into them, then they become neutral and this would not cause E =0 in the wire, but if you say they still get to positive terminal's charges, then how ? why they're not attracted more by ions(surface charges) ?

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gionole said:
This is why I said we won't have surface charges on both surface
Well, you can say it all you want, but it isn’t so. You will get most charge on the ends because the curvature of the surface is highest. But some will be on the top and bottom too.

Consider that the wire is pretty long and straight and that we are looking at a part of the wire in the middle (meaning the ends are far away). Now suppose that there is no charge on the top or bottom, only the ends, a positive charge on the left and a negative charge on the right. This produces a field which cancels the left component of the external field, but does not cancel the vertical component. So by Ohm’s law current will flow from the top surface to the bottom. You will get positive charge on the bottom and negative charge on the top, canceling out the vertical component of the external field.

gionole said:
if external E is not strong enough
The strength of the E field doesn’t determine where charges accumulate. It just determines how much charge accumulates. If you double E then you double the surface charge density. You do not change its location/distribution.

gionole said:
With your words, what's the reason surface charges on the last end of the wire have bigger density and as you get far away from the end of the wire, density gets less and less ?
That is because of the curvature of the surface. Remember Ohm’s law. If there is an E field tangent to the surface then current will flow along the surface. So the E field can only be perpendicular to the surface and the charge can only be on the surface. Place a test charge at a place where the curvature changes, say low curvature on the left and high on the right. The E field from surface charges to the left will have a large tangential component but the E field from surface charges to the right will have a smaller tangential component. Therefore, you will need more charges to the right, on the high curvature side, to balance the tangential forces.

gionole said:
With battery connected to the wire(circuit open though), I think on the positive terminal, surface charges accumulate right away on the surface closest to the terminal.
This paper describes how to determine this:

https://www.tu-braunschweig.de/inde...oken=2cc8a71e4fdbf159121c6b8ef8348952a2e0c197

gionole said:
What I don't get though, the electrons that come from the later part of the wire, do they get closer to the ions(the ions that got on the surface - they were closest ones to terminal) ? … if electrons jump into them, then they become neutral
Again, I don’t believe that thinking of electrons and ions and jumping is either useful or correct. Instead, focus on Ohm’s law.

The paper I posted goes into a great amount of detail on this topic. Please read it completely, and then if you have questions remaining we can discuss with specific reference to the paper.

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Lord Jestocost, sophiecentaur and vanhees71
Dale said:
Again, I don’t believe that thinking of electrons and ions and jumping is either useful or correct. Instead, focus on Ohm’s law.
Absolutely. In a metal, each metal atom 'shares' its outer electron with all the others round it; no 'jumping ' is involved as it's more like drifting. The atoms are sometimes described as 'positive ion cores' and there is the equivalent of a dissociated 'electron gas' occupying the spaces in between. Electrons here require (almost) zero energy to move from one location to another and that idea agrees with experimental evidence that wires have very low resistance (almost zero in an ideal case). So there is 'no' internal electric field. All the Field Lines outside a perfect conductor will be perpendicular to the surface.

A picture of charges at the would be a slight difference in charge density right at the surface. The Coulomb force between electrons and ion cores is very high (metals are strong because of this force). This is metallic bonding and the model explains why metals stretch without breaking apart, for instance.

Also, the actual numbers count here. There are more than 1021 electrons in a small piece of metal and each has 1.6e-19 C of charge so even a lot of electrons don't represent a very large charge. The Capacitance of an isolated conductor will be very tiny (a few pF) but two plates, held very close together can have a much higher capacitance because many more electrons can become displaced. (polarisation of charges)

Dale and vanhees71
@Dale

I've read the link you shared, but there's something that still bugs me and maybe due to article complexity, maybe I didn't understand it but it was still written in it.

As far as I understand now, when battery is connected to wire(where wire ends are not connected - circuit is open), there is electrostatic equilibrium which means E(net) = 0. At this situation, the surface charge density throughout the whole wire is uniform(maybe a little bit more at the bend place, but it's not much). "Uniform" means it's the same charge density everywhere, but once we close the gap, surface charge density at different places adjust(not uniform again) and we got E > 0.

on my attached image, I draw some regions so easier for us. Note that drawn lines are not E, but force directions for electrons caused by battery dipole

So instantenously, somehow, we should get elecrostatic equillibrium here where surface charge densities is uniform everywhere. Electrons from region 9 move to region 8. So region 9 now has excess positive charges which start repelling from each other and move to the surface of region 9. Same happens for region 5, as its electrons go to battery and we're left with excess charge of positive. But here is my question. When electrons of region 8 moved to region 7, you might think region 8 got left with net positive, so it should have positive charge on the surface, but at this very time, electrons of region 9 would come to region 8 and region 8's net positive would again become neutral. So until electrons of region 9 also leave region 8, region 8 will not have net positive on the surface. But we know that electron speed is very slow. So since this equilibrium is reached instantenously, I don't think saying that region 9's electrons have to also leave region 8 to have net positive on region 8's surface. but then are we saying region 8's electrons leave the region 8, which causes region 8 to have net positive on the surface, but then when region 9's electrons get to region 8, wouldn't these electrons make region 8's surface neutral again ? I hope now you understand the backbone of my question and also, electrons kind of must be going into the surface of region 8, due to direction of force in there, but they don't.

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gionole said:
there is electrostatic equilibrium which means E(net) = 0.
Note that this is true inside the wire. At the surface there may be a non-zero E field perpendicular to the surface.

gionole said:
At this situation, the surface charge density throughout the whole wire is uniform
This is incorrect. I am not sure what leads you to believe it. Could you explain your thought or what in the paper made you believe it?

gionole said:
So instantenously, somehow, we should get elecrostatic equillibrium
It is not instantaneous, but it is quite rapid. The paper describes the DC situation, but not the dynamic transition to it.

gionole said:
Note that drawn lines are not E, but force directions for electrons caused by battery dipole
Please don’t do this. Don’t focus on electrons. Focus instead on current ##\vec J## and charge ##\rho##. Note that both of those show up in Maxwell’s equations, while electrons do not. Focusing on electrons is not helpful and usually not correct.

So, instead of looking at electrons, let’s look at current. We have in the wire current flowing the opposite direction of your arrows.

Where current starts we get negative charge accumulating, and where current stops we get positive charge accumulating. Where current just passed through there is no accumulation of charge.

So here you have current passing through 8, 7, and 6, so no accumulation. Current starts at 5 and ends at 9. So you get negative accumulation at 5 and positive accumulation at 9.

Note that although there is no net accumulation of charge in 8, you do get a little bit of redistribution of charge to direct the current around the corner. This leads to the charge distribution shown in Figure 8 of the paper.

Where current starts we get negative charge accumulating, and where current stops we get positive charge accumulating. Where current just passed through there is no accumulation of charge.

So here you have current passing through 8, 7, and 6, so no accumulation. Current starts at 5 and ends at 9. So you get negative accumulation at 5 and positive accumulation at 9.
@Dale
You're referring still to the case when circuit is open, right ? I'm currently only interested when circuit is open. So I agree, current starts at 5 and ends at 9. but then, you say: "you get negative accumulation at 5". But we know there should end up positive charges on the surface of region 5.

What I'm trying to determine is how we end up with positive charges on the surface of all regions(5,6,7,8,9). If you're saying we don't, then at

(The link is with timestamp and you can clearly see the open circuit there and all regions on the left side has positive charges on the surface). I couldn't find the exact figure of this in our paper.

I'm thinking in terms of electrons because if I think in terms of current, it even gets confusing more. Even if we say current moves from 5 to 9, (forget about right end - negative terminal for now), then i can't explain the positive charge on the surfaces on 5,6,7,8,9 surfaces.

gionole said:
you say: "you get negative accumulation at 5". But we know there should end up positive charges on the surface of region 5.
I said that based on the current you drew because I thought that was the part you wanted to focus on (since that is what you drew).

gionole said:
But we know there should end up positive charges on the surface of region 5.
You cannot get those with the current you drew, even reversing it. Your drawing neglects the small currents that briefly flow across the wire instead of along the wire.

To see those, use the method in the paper to determine the final charge configuration. Also draw the initial charge configuration. Anywhere that has increased charge was a place where current ended, and anywhere that has decreased charge was a place where current started. Current just passed through everywhere else.

Again, look at Figure 8 in the paper. To get this charge distribution there was a small brief current from left to right across the vertical section and from top to bottom across the horizontal section.

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Dale said:
Please don’t do this. Don’t focus on electrons. Focus instead on current J→ and charge ρ. Note that both of those show up in Maxwell’s equations, while electrons do not. Focusing on electrons is not helpful and usually not correct.
Totally, absolutely. Does @gionole realise that all this basic stuff was sorted out long before electrons had ever been invented (/ discovered). Electrons are like photons in that you should try to ignore them whenever you can.People think they aid understanding but actually they tend to add confusion because the pictures people have in their heads are nearly always flawed.

Dale
sophiecentaur said:
People think they aid understanding but actually they tend to add confusion because the pictures people have in their heads are nearly always flawed.
This!!! When you actually learn about photons and electrons you have to unlearn the flawed pictures. And in the meantime the flawed pictures just add extra complication that is not needed. Especially with respect to Ohm's law which is as simple as a law can be.

gionole said:
I'm thinking in terms of electrons because if I think in terms of current, it even gets confusing more. Even if we say current moves from 5 to 9, (forget about right end - negative terminal for now), then i can't explain the positive charge on the surfaces on 5,6,7,8,9 surfaces.
Like here. The problem was not thinking in terms of current. The problem was that the current of interest was being neglected, both when @gionole was thinking of electrons and when thinking of current.

sophiecentaur
@Dale @sophiecentaur

I actually understand what you both mean but you're saying this because at least you already have a picture of how it works in your mind and you think it's easy and it would be easy for me too, while for me it's quite hard.

By my profession, I'm a developer and in the past, I was learning c language. It was my first year and working with memory management was quite hard, but my professor was still explaining things in terms of some flawed arguments in order to accelerate the process of teaching and not focusing million years on the same thing.

Several years have gone by, now I look at c and even though I've not looked into the memory of c, even now, it's super easy to imagine how it works behind the hood, but if you did explain it back then the same way I know now, well, I'd have not understood the same way.

I know and don't get me wrong, appreciate your inputs but sometimes what you think is correct, mightn't end up the right way for me. Sometimes all I need is a little push on the explanation the same way I'm looking for and then it would be easy to jump to "current" imagination and not "electrons".

Even in Ruth W. Chabay , Bruce A. Sherwood book, they talk about sea electrons and mobile electrons and explain it mostly the way I'm looking for. This one
(
at 14:31 show the same explanation but tricky, it doesn't explain what I'm asking - how unlucky).

Though, one thing is still not clear as how before circuit is closed, positive charges appear on the surface of the positive terminal battery. They only show the example on the bends(image attached). In this example, I understand it. At every point in wire, electrons move to the left, when electrons move from region 2 to region 3, my question was why don't we have excess positive charge in region 2 and the answer is simple, in that time, electrons from region 3 move to region 2. So electrons of region 1 occupied the space of region 2's electrons. This goes on for region 1 as well as region 0's electrons occupy the places for region 1's electrons when they move.

On the attached image, it's all clear. but I'm trying to now see the case of open circuit and why positive charge ends up. On my second attached image, on region 1, why we get positive charges on the surface. I get that electrons start to move to the positive terminal and you might think when they move, they leave excess positive charge, but my argument is when first stream of electrons moved, their place was occupied by the next stream of electrons as this process is so fast.

I think the explanation I come up with is in the positive side wire, E is different magnitude everywhere, stronger closer to terminal and more you go away, weaker. So when the closest electrons get sucked in by the battery's positive charges, the next stream of electrons - they move - but they actually don't reach the space (due to feeling weaker force) that first electrons were occupying. So because of this, the first electrons that left the positive charges behind, those positive charges are still alone(next stream of electrons didn't reach it even though that next stream of electrons moved closer to it), so since they're alone, they have enough time to start repelling each other and distribute on the surface. Not sure if this is absurd.

I also completely understand if you don't want to continue discussing it with my way as you think it's a problematic. maybe I will research more or ask somewhere else and once I catch up to feel my gap, I will come back to this thread so I'm ready. Just let me know your thoughts. Appreciate your help Dale so much FYI.

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gionole said:
I know and don't get me wrong, appreciate your inputs but sometimes what you think is correct, mightn't end up the right way for me. Sometimes all I need is a little push on the explanation the same way I'm looking for
I am sorry. I simply am not willing to push anyone in what I feel is a wrong direction. There are many other sources that use that approach. I won’t.

gionole said:
On the attached image, it's all clear.
I don’t think that either of those images are accurate. I would be a lot more comfortable if they seemed to be rigorously derived.

Do either of them claim to be the result of an actual calculation or mathematical derivation?

gionole said:
I also completely understand if you don't want to continue discussing it with my way as you think it's a problematic.
I absolutely don’t mind you asking. I am just letting you know my opinion on the matter and that my answers will continue to be in the way I believe to be both more correct and easier.

It is true that their images might not be the correct but to me, it feels easier to imagine and I think they are giving intuition by those images. I am sure you think the paper you linked is easy, but to me, it is harder.

I appreciate your opinion on this and I am sure you are more correct than them, but at this point, i need to first understand it a little bit intuitevely and then come to your explanations after that. I will focus on it tomorrow and once I am ready, I will get back here.
Huge thanks !!

gionole said:
It is true that their images might not be the correct but to me, it feels easier to imagine and I think they are giving intuition by those images. I am sure you think the paper you linked is easy, but to me, it is harder.
The problem is that they show the positive side having only positive surface charges and the negative having only negative charges. I don’t think that is right. So you run the risk of building intuition about a false situation. Once intuition is built incorrectly it becomes more difficult to unlearn.

The paper may be difficult, but you can use the pictures confidently even if you don’t use their methods. The surface charges are calculated rigorously.

Dale said:
Please don’t do this. Don’t focus on electrons. Focus instead on current ##\vec J## and charge ##\rho##. Note that both of those show up in Maxwell’s equations, while electrons do not. Focusing on electrons is not helpful and usually not correct.
Given the confusion of the OP, we should be a bit pedantic. What shows up in the local (differential) Maxwell equations are the charge density, ##\rho##, and the current density, ##\vec{J}##. Both are fields, i.e., "local quantities": ##\rho(t,\vec{x})## and ##\vec{J}(t,\vec{x})##.

Their operational meaning is as follows: For the charge density at the
sophiecentaur said:
Totally, absolutely. Does @gionole realise that all this basic stuff was sorted out long before electrons had ever been invented (/ discovered). Electrons are like photons in that you should try to ignore them whenever you can.People think they aid understanding but actually they tend to add confusion because the pictures people have in their heads are nearly always flawed.
I disagree. The discovery of electron theory around 1900 was a big breakthrough in the understanding or electromagnetism and matter, and indeed in a usual metal and the wires in your household currents are due to moving electrons.

Of course, one must be aware of the serious limitations any classical (i.e., non-quantum) theory of matter has, and last but not least that was revealed after struggling with classical electron theory for some decades. The electron very soon after it was discovered was thought of being a (of course then classical point-like) particle with a negative electric charge, and immediately theorists like Heaviside and most prominently Lorentz tried to describe point-particle motion in electromagnetic fields. That's why the corresponding force is known as "the Lorentz force", and it works pretty well for many applications, but as soon as you try to make the theory self-constistent, i.e., to include the interaction of the (accelerated) charge with its own electromagnetic field you run into trouble. Only approximations lead to consistent results, and this is indeed one of the most early realizations of the fact that classical point particles are not a consistent model. Today we know that quantum field theory is needed to solve this very fundamental problem of single elementary charged particles interacting via the electromagnetic interactions.

However, this changes, as soon as you consider many-body systems. Here, semiclassical continuum-mechanical models can at least qualitatively describe many phenomena related to the electromagnetics in media like dielectrics and conductors. The model of a current-conducting wire using the picture of a fluid of valence electrons moving in the positively charged ion-lattice background (Drude model) is very successful. Also the picture of a insulator/dieelectric as consisting of electrons bound elastically to a lattice of positively charged atomic nuclei works pretty well.

Of course all this can finally consolidated by using fully consistent quantum-theoretical many-body theory, and the main reason, why the (semi-)classical continuum-theoretical models work so well is due to the fact that linear-response theory mathematically looks very much alike in both the quantum and the classical models.

vanhees71 said:
The discovery of electron theory around 1900 was a big breakthrough in the understanding or electromagnetism and matter, and indeed in a usual metal and the wires in your household currents are due to moving electrons.
True but until the OP shows competence with Maxwell's level of the topic, is it worth while going beyond? In a 'free for all' clash of ideas then anything goes but I don't see this thread needs one of those. It's all text book stuff which is largely self consistent. The electrons in a metal are bound and behave en-mass so the bullet picture doesn't always constitute a good model.

Motore
vanhees71 said:
The model of a current-conducting wire using the picture of a fluid of valence electrons moving in the positively charged ion-lattice background (Drude model) is very successful.
I do not share your opinion on that. However, if you feel that it is a good didactic approach then you are certainly welcome to answer the OP's questions using your approach. I am simply stating clearly that I will not.

gionole said:
It is true that their images might not be the correct but to me, it feels easier to imagine
So here is an image of a quadrupolar potential like you will get in your suggested arrangement. Notice in the horizontal wire that there is an E field directed vertically. That will lead to positive charges on the top and negative charges on the bottom.

I am not certain this is correct, but it makes me highly suspicious of the drawings.

sophiecentaur said:
True but until the OP shows competence with Maxwell's level of the topic, is it worth while going beyond? In a 'free for all' clash of ideas then anything goes but I don't see this thread needs one of those. It's all text book stuff which is largely self consistent. The electrons in a metal are bound and behave en-mass so the bullet picture doesn't always constitute a good model.
Part of the electrons in a metal are bound, but another part (the conduction electrons) is not, which makes the metal a conductor. I always found the picture of the conduction electrons constituting a gas of charged particles moving in the background of positively charged ions very intuitive. If you want to handle it with a full statistical-mechanics treatment, of course, you have to use Fermi-Dirac statistics, because this "electron gas" is degenerate (under usual "household conditions") to get the specific heat right, but for the purpose of understanding Ohm's Law the naive derivation by Drude is sufficient. It's indeed standard textbook stuff. See, e.g., here:

https://www.physicsforums.com/threa...tance-of-a-series-circuit.994483/post-6401309

Lord Jestocost
vanhees71 said:
Part of the electrons in a metal are bound, but another part (the conduction electrons) is not
This is a moot point I think my statement is valid. The electrons are in a state where they are all close together and subject to mutual effects; I would have said that, as they are not free to go wherever and their separation is very much less than in an achievable electron beam then they can be considered to be 'bound'. Only in a metal, could you get 1C worth of electrons in such a small space, without a lot of trouble.

@Dale

I am willing to only look at the correct pictures as you suggest. One thing I'm worried about is I wanna know the surface charge density in different regions while circuit is open. I don't think the paper contains picture of that and constructing it myself is quite hard. I want to look at surface charge density distribution while battery is connected to wire, but wire's ends are not connected to each other - hence open circuit and no other components in it at all. Q1: Would you be able to find an image for that ?

Here is why I think you might be right. If you look at my attached image, as the book suggests, in region 1 and region 2, we got the same distribution. Now, follow me a little bit ! Let's call E1 to be the electric field in region 1 by ONLY the battery. Also, let's call E2 the electric field in region 2 by ONLY the battery as well. and as image says, in region 1 and 2, surface charge distribution is the same so we call it E(surface) which is electric field produced by only surface charges in that region, but since region 1 and 2 have same surface charge density, E(surface) is the same for both. Now, we know E_net = 0 in both regions. So in order for that to happen, E(surface) = E1 and E(surface) = E2, which says: E1 = E2. The logic now completely shatters, because no chance E1 can be equal to E2. Remember, it's only battery's produced electric field, so E1 will always be bigger than E2 as it's closer to the battery. So there's no chance E(surface) is the same in region 1 and region 2. more importantly, E(surface in region 1) must be more than E(surface in region 2) in order to be able to cancel out their appropriate E1 and E2 by the battery which are not the same. Is this logic 100% correct ? if yes, I definitely agree that the picture is pretty confusing as it suggests uniformity of surface charges while I just proved that it can't. If you agree with me, I'd love to have a correct picture of distribution in the same case as on my attached image. The fig 6 in paper is drawn well, but it's not for this case because circuit looks different + it's closed. The fig 7 is messy, even though the circuit is the same, it's still closed(not open as in my case) and also colors there drive me crazy. Definitely would love to have the image for my case but drawn distribution the same way as in fig 6. Maybe you could quickly find it on google ! I would try, but i'm not sure what to trust anymore. hahaha

Thanks so much.

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gionole said:
I wanna know the surface charge density in different regions while circuit is open. I don't think the paper contains picture of that and constructing it myself is quite hard
You are correct on both counts. I am investigating setting up a finite element calculation myself, but it is not trivial.

Why are you focused on this exactly? What do you hope to learn from it?

gionole said:
Is this logic 100% correct ?
I can’t give you a 100% correct confirmation. All I can do is say it seems good to me. But I can’t even give myself 100% correct.

I don’t have a source for your specific conductor shape. That is why I am looking at doing it myself. But you can find many other shapes. Use “finite element” as one of your search terms, which should narrow it to rigorous plots

Why are you focused on this exactly? What do you hope to learn from it
What I want is to have a very broad logic in my head for when circuit is open and when it is closed.

@Dale
If you agree with my logic, then one interesting question that I have is on the positive side, we should get surface charge distribution in a way that it decreases each time we get more far away from positive terminal. As in region 1 has more density, then region 2 has less, then region 3 even has less than region 2. Otherwise, my logic fails. Note that I am only focusing now on open circuit as in my image. Would it be correct to assume region 2 has less than region 1 and region 3 has even less than region 2 ? If so, figure 7 does not fully explain this as at some points in region 2, there is even more density than in region 1. This is what fig 7 says. Maybe it shows it like this because circuit is closed ? If so, I could understand it as more things going on when its closed so it is a totally different scenario but when it is open, we should not have any place in region 1 where density is less than in region 2 so we should have decreasing density all along. Thoughts ?

Amazing that you are doing it. Kudos

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Dale said:
I do not share your opinion on that. However, if you feel that it is a good didactic approach then you are certainly welcome to answer the OP's questions using your approach. I am simply stating clearly that I will not.
In particular the part that is difficult to convey is that the notion of free particle momentum conservation being replaced by "crystal momentum" (and sometimes dressed mass) makes things quite different. I don't know an easy way to teach the simple theory without the student falling off that cliff at some point with the usual gnashing of teeth.

gionole said:
One thing I'm worried about is I wanna know the surface charge density in different regions while circuit is open. I don't think the paper contains picture of that and constructing it myself is quite hard. I want to look at surface charge density distribution while battery is connected to wire, but wire's ends are not connected to each other - hence open circuit and no other components in it at all. Q1: Would you be able to find an image for that ?

Please find attached the sketches I draw on the picture you quoted. I hope this gives you some insight for your problem.

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anuttarasammyak said:
Please find attached the sketches I draw on the picture you referred. I hope this gives you some idea to solve your problem.

View attachment 328339View attachment 328340
It still does not say much. Can you read my 2 previous replies here ? In open circuit image, I presume that surface charge distribution is decreasing all along the wire in positive side.(same idea to negative side, but focusing on positive only). So the higher is in region 1, less in region 2, even less in region 3. The reason I say this is due to my logic in my previous 2 replies. Your image is still confusing, sorry.

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