B How is surface charge accumulated?

[Dale]

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?

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

 

[gionole]

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

 

[hutchphd]

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.

 

[anuttarasammyak]

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.

 

[gionole]

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.

 

[anuttarasammyak]

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.

Thanks for your reply. To understand your idea please find attatched your idea and mine on the sketch.

 

[gionole]

@Dale

hey Dale, have you had the chance to look at my previous reply ? I think I will convey my final assumptions now and we can close this topic !

ASSUMPTION 1(open circuit): as I said previously, in region 1 and 2, there're different magnitudes of E produced by the battery due to the distance as region 2 is more far than region 1. Due to this, in order not to be constrained to say E=0, in region 1 and 2, we need the opposite magnitude of E so E_net = 0. If we assume surface charges are the same in both regions, we'll be doomed as it can't cancel out both E1 in region 1 and E2 in region 2 as E1 > E2, so in order to cancel out E1, there must be more surface charges in region 1. I think you agreed with me here and I will go ahead now and say that by following this logic, on the positive side(focusing only on positive side for simplicity), we should get surface charge distribution in a way that it decreases each time we get more far away from positive terminal. So region 1 has more surface charge density than region 2 and region 3 even has less surface charge density than region 1. So it's like r1 > r2 > r3. This is only true for sure as long as E produced by the battery in region 3 is less than in region 2. By looking at my open circuit, it might not be true, but you get the point that the more far away the region is from the terminal, the less surface charges will accumulate there than in the region that's closer to the battery. Also, on the very end(Region 7), we will still get surface charges depending on the E by the battery in there which I repeat for sure will be bigger than E in region 3, so region 3 should have less surface charges than region 7. I think this all sums it up roughly. So sum up is more strong E field is by the battery in a region, that region will have more surface charges then other region where E by the battery is lower. This all translates to the distance from the terminals to where we're checking to see surface charges.

The reason I want to get green light about the my assumption above is, on the paper, I looked at fig 7 and i can see that in region 1's some parts, there're less surface charges, then in region 2's some parts. I understand that fig 7 depicts the closed circuit image, but I'm trying not to make a mistake here. hopefully such phenomena wouldn't be produced in open circuit right ? if so, I get the idea and should be enough.