# Electric Field

I will post this a second time, I dont think I was clear the first time. I really need some help on this. If we have a a charged object. Then the potential at all points on the obeject MUST be the same. If they are not, then the charged particles at the surface will move until the potential is the same at all points. I think I figured this part out, and I want you to verify if i am correct please. Lets say we are at some point on the surface, in my picture, the blue point, then the potential at that point, due to the potental of all the chrages around the surface should be exactly equal to the potental at any other point. I can choose my other point to be the olive point. Note that this olive point is coincides with the location of one of the charged particles. Then the potental at this point, due to the potentials of all the other charges, should equal to the SAME value as the blue dot. Is this correct? And also, if these two are NOT equal, then things will move until these two values are equal. In fact, the charges will move until this relationship is true between ALL points on the surface. So if we were at the olive point, despite the fact that the chrage at that spot does not contriubute to the potential, since the distance is zero, the sum of the potential due to the other charges will STILL equal the same value as any other spot on the surface, even though we had to neglect one of the charges since its distance is zero.

Now, IF the potental is different, then the potential ENERGY is different. And things will move. But becuase they move, they will gain kinetic energy, and once they reach the new potential state, they will continue moving past that point, and exhibit simple harmonic motion. But because the electrons collide with eachother, and with the atoms, it will dampen out. And it will finally reach a point where everything is equal in terms of potential and potential energy.

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Tide
Homework Helper
The potential of the surface and the potential energy must both change with the addition of a charge to the surface. In the end the surface will be an equipotential. I think you're trying too hard to make this difficult! :-)

hey tide dont go no where, im going to post another question if u have time to read it. :-)

I worked out a drawing of a nice sphere on grid paper where all the electrons were point masses, measured the distances with a ruler, and calculated the electric field at a point on the surface close to one of the charges. what i found was a field that was NOT perpendicular to the surface! Heres a picture of what I did. Notice that at this point, there was a BIG electric field due to the charge closest to it, but the electric field due to the rest of the charges was small. So the net electric field (orange arrow) was nearly in the same direction as the one closest to it! This is NOT perpendicular to the surface! It seems that the only way for this field to be perpendicular to the surface is if we consider the electrons not to be four points equally spaced, but, hmm im making this up as I go along, they have to somehow be distributed the same over the entire surface, if u were able to break the electrons up and spread them out like dust in a sense, so that there was no point on the surface where the was no charge. Every point on the surface would have to have a "piece" of charge from the electron. And the sum of all these pieces that cover every point on the surface would equal 4*e-. I dont know, im just trying to make sense out of that anwser I got.

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Tide
Homework Helper
Cyrus,

Did I count them correctly - 4 electrons plus a single electron test charge?? ;-)

Aside from the fact that you won't have an equipotential spherical surface with those four electrons those four electrons will not remain in place once you've added the test charge. They will redistribute themselves if you are attempting to model a conductor.

Moreover, you are missing the essential ingredient and spirit of calculating electric fields associated with charged conductors - conductors have LOTS of electrons! That means we take the electron distrubution to be a continuum. For some applications in surface physics one may be interested in (extremely) local effects but for the kinds of calculations you are engaged in the continuum limit is the way to go. Even classically the continuum limit is justified based on the huge number of charge carriers but also the random jitter motion of electrons (thermal motion) constrains us to considering spatially averaged charge distributions.

No, that is not a test charge at that point, its just a point i decided to choose. How come those four charges wont be equipotential? Arent they all equally far away from eachother, and so the potential relative to all points is the same. And for the electric field at that point, I dont need a test charge do I? Electric field is independent of a test charge. My text never talked about a contiuum, So I assumed that you treat each one as a point charge, and where there is not a point charge, there is simply no charge present there. To be clear, are you saying that I should assume that there is charge at every single point on the surface. This seems like a paradox, because if the charge is all over the surface, then how can it be spaced as far apart from eachother at the same time?!

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Tide
Homework Helper
Cyrus,

You can explicitly calculate the electric field or potential from the four charges. Try it. You will convince yourself very quickly that the "circle" is not an equipotential!

Oh, then how would one go about making them equipotential? I dont see any apparent place you could put those four charges to make them equipotential other than equally spaced. They would have to be at four points as in the picture if they are to be the furthest distance from eachother, as required by a conductor.

Also,
My text never talked about a contiuum, So I assumed that you treat each one as a point charge, and where there is not a point charge, there is simply no charge present there. To be clear, are you saying that I should assume that there is charge at every single point on the surface. This seems like a paradox, because if the charge is all over the surface, then how can it be spaced as far apart from eachother at the same time?!

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Tide
Homework Helper
cyrusabdollahi said:
Oh, then how would one go about making them equipotential? I dont see any apparent place you could put those four charges to make them equipotential other than equally spaced. They would have to be at four points as in the picture if they are to be the furthest distance from eachother, as required by a conductor.

Also,
My text never talked about a contiuum, So I assumed that you treat each one as a point charge, and where there is not a point charge, there is simply no charge present there. To be clear, are you saying that I should assume that there is charge at every single point on the surface. This seems like a paradox, because if the charge is all over the surface, then how can it be spaced as far apart from eachother at the same time?!

Did you actually calculate the potential at a few places on the surface to see for yourself whether it is an equipotential surface? Do it! It will be quite instructive!

In the end you will find out that there is no way to arrange the four discrete charges on a spherical surface that will make it an equipotential.

Your textbook may not have used the word "continuum" explicitly but it was there implicitly! That is what they mean when they talk about charge being uniformly distributed along a line, a surface or a volume. Also, you have seen Gauss' law written as an INTEGRAL and NOT a SUMMATION. I'm sure you've also seen elements of charge written as dq = <charge density> X dVolume. All of these are ways of saying your're working with a continuum.

AH! Now I see what you mean. Because it is an elemental dQ, that implies that there must be a piece of charge at EVERY infitestmal interval. So there can not be any piece that does NOT have charge. Thats very hard to pick up on when the book does not say so explicitly. I will use a simple example of line charge density. In the pic, this is how I thought they meant linear charge density, but I guess the correct way of viewing it would be the image on the right.

As an example, say the linear charge density was 4 per length dl. I thought that meant there was 4 evenly spaced electrons along that length dl, and there was no charge at the points where there was not an electron. But I think the more correct picture is the one on the right, where there is a little contribution of charge over the entire length dl, and is continuous. Its like I took those four electrons, and made them into a fine dust, so that they could cover EVERY piece of the wire.

To extend this concept, I know that there is more charge at sharp corners. In that case, the charge density would not be constant but would vary. Does that mean that there is a larger charge density per unit length dl where this sharp corners arise, and less charge density per unit dl where there are smooth flatter areas.

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Another side, I was reading a website about a line of charge. It seems that in my picture, the four points would have to be part of an infinte line of charge.

In reality, if the line of charge were finite, it would pile up on the ends, and look more like this picture.

In my physics book they did a calculation for a wire of finite length and said it had uniform charge distribution, but that cannot be true can it? If it is finite in length, then the ends of the wire MUST contain more charge than the center. So it cannot be a uniformly distributed length of wire in real life. I think they TOLD us that it is to do the calculation, but in reality it is NOT uniform. Also the electric field should be zero inside a conductor. Its clear from the picture that this is true, but it seems that it is not zero at the end points. At each end there is charge only to one side, and no charge opposite to cause an electric field to balance it out. Whats up with that?

I got this picture off of a website. I guess they treating the electron as a point particle like i did tide? Their picture really should not be blue dots, with empty void between them, it should be a distribution of that blue dot all over the piece of wire, but "heavier" distribution near the ends. So its kind of a gradient of distribution across the wire.

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krab
cyrusabdollahi said:
If we have a a charged object. Then the potential at all points on the obeject MUST be the same. .
This is false as a general statement. It is only true of conductors.

krab
cyrusabdollahi said:
In reality, if the line of charge were finite, it would pile up on the ends, and look more like this picture. .
Why? You can put charges where you want them. Are you still assuming everything is a conductor?

Hi Krab, sorry, Yes i did mean in a conductor. Can you please look over my post with a picture of four point charges on a line. If it is stated that the charge distribution along that segment dl is 4 -e, is it incorrect to draw it like I have done on the left. Should it be draw like the right, where I break up the four electrons and "spread" them so that every point along dl now has a piece of charge.

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cepheid
Staff Emeritus
Gold Member
cyrusabdollahi said:
Its like I took those four electrons, and made them into a fine dust, so that they could cover EVERY piece of the wire.

Huh? Surely not!

Umm...just to clarify...I don't think Tide was suggesting that four electrons constitute a continuum! Yet that's what you seem to be implying Cyrus. He was saying that in order to have an equipotential surface, one needs a continuous charge distribution over that surface. Am I right?

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Tide
Homework Helper
cepheid said:
Huh? Surely not!

Umm...just to clarify...I don't think Tide was suggesting that four electrons constitute a continuum! Yet that's what you seem to be implying Cyrus. He was saying that in order to have an equipotential surface, one needs a continuous charge distribution over that surface. Am I right?

Correct!

To make the point consider the following: Take a microCoulomb of electrons and distribute them evenly on the circumference of a circle 1 centimeter in radius. What is the spacing between the electrons? :-)

Hmm, I think I see what you mean now. IF we call it equipotential, then we mean by definition that there is no space? So it implies that everywhere on the surface has a charged particle associated with it?

But i dont know the anwser to your question tide. If we have a microcoulomb of electrons on a sphere of 1 cm, then you are saying that there should be no spacing between the electrons?

Tide
Homework Helper
Cyrus,

If you place a microCoulomb of electrons equally spaced around a circle 1 cm in radius then nearest neighbors will be about $10^{-12} cm$ apart. That's a VERY small separation. For almost any practical purpose it makes little sense to attempt resolving electric fields to such a short scale so we treat the electrons as a continuum. You can do a similar calculation distributing a comparable number of electrons over the surface of a sphere and you'll get similar results for the interelectron spacing.

Oh, so even though there is a spacing, just ignore it. I see. Seems weard in a way, because at that spacing, the potential cannot be the same there, and relative to the size of an electron, the spacing is suddenly not so small anymore.

Ok tide, I read over the posts once again and I see your point now. If we have a charge density, we can consider the spacing to be so small on the macroscopic level that we can treat them as if there is no spacing, but as a continuous distribution. That seems to work well for the macroscopic level, because in practicality we cannot go 10-12cm between electrons.

A few questions arise from this asumption then. First, lets say that there is a body with a charge density x, and an identical body with a charge density 10x. Then this means that the spacing between the 10x one should be much much smaller than the one with the density of x. There has to be a limit to how much charge we can put on that body. Would the limit be when the spacing between electrons is EXACTLY zero. Because adding any more electrons, where could they go on the surface? (assume this is done in a vaccum where we dont have to worry about corona discharge into the surrounding air.)

My second question is about the microscopic scale. This asumption works well macroscopically, but when we look locally between atoms, now that small distance is really quite significant. It seems that our definition of "equipotential" breaks down now. A given amount of charge may or may not ENTIRELY cover the surface, so how can we conclude it is equipotential. Wouldent this be like my drawing of four charges that you said was impossible to make equipotential. (only we would have more charges than four, a whole bunch, none the less, there would still have to be some space between them). Just wondering

PS what time zone are you in tide? I always talk with you at like 1-2 a.m. its taking its toll! :zzz: :zzz: :zzz: :tongue2: :rofl:

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Tide
Homework Helper
Hi, Cyrus!

Yes, the more charge you have the closer the individual charges will be to each other! The way it changes depends, of course, on whether you're talking about a line, surface or volume charge.

On the microscopic level there is no such thing as an equipotential material surface partly because charges themselves are discrete and partly because the surface of a material cannot be resolved to anything better than the nearest neighbor spacing between atoms and molecules that make up that surface.

BTW - I'm on Mountain time but I've fallen into a late schedule because I only have night classes this semester. Sorry about that!

Ok, now it is alot clearer. I thought equipotental surfaces meant as small as you want to go it would still work out, but now you have shown me the light lol.

Can I bother you on a capacitance question? I was just wondering, if a capacitor is attached to a a battery so that I only have one plate of the capacitor hooked up to one terminal of the battery, (lets say the + terminal), then will that one plate of capacitor obtain a charge of (+ whatever the batter was at). Or is it necessary for both plates to be attached to the battery before any charge will be distributed to each plate?

I know that you need a voltage difference before charge will move. And I was wondering if you could consider the plate to be at a voltage of zero, and hooking it up to the one terminal would cause a voltage difference of the Vbattery- Vzero = Vbattery.

Kind of like, having a capicator only half way hooked up in a sense. Should I see one plate have a charge and the other plate have no charge.

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Tide