Surface density on isolated conductors in space

[cromata]

Homework Statement

There are n isolated condutoctors in space: Sum of charge of all conductors is positive. Show that surface density is postivie everywhere on at least one of conductors

Homework Equations

Using induction

The Attempt at a Solution

-It's clear that if we have one conductor with postive charge that it will have positive surface densitiy everywhere on its surface. But let's have a look on a problem in which we have one positive conductor (with charge q1) and let's say one negative (with charge -q2) where |q1|>|q2|. Charge on each of conductors will redistribute so that the electric field inside of it is 0 (electrostatic influence), it will redistribute in that way that positive charges of conductor1 will be attracted by negativ conductor1 and vice versa.I don't see a reason why shouldn't there be a negative surface charge on some part of the positive conductor due to this redistribution.

-My idea for solving this was to use induction
1)I aready said that it's clear that for n=1 statement is correct
2) let's say that for some n of conductors at least one have positive surface charge everywhere
3)now let's try to add one more conductor to this system and show that at least one conductor will have positive surface area everywhere
Unfortunately I don't know how to show that...perhaps there is some other way for solving this problem.

 

[TSny]

The Attempt at a Solution

-It's clear that if we have one conductor with postive charge that it will have positive surface densitiy everywhere on its surface. But let's have a look on a problem in which we have one positive conductor (with charge q1) and let's say one negative (with charge -q2) where |q1|>|q2|. Charge on each of conductors will redistribute so that the electric field inside of it is 0 (electrostatic influence), it will redistribute in that way that positive charges of conductor1 will be attracted by negativ conductor1 and vice versa.I don't see a reason why shouldn't there be a negative surface charge on some part of the positive conductor due to this redistribution.

Starting with the case of just 2 conductors seems like a good idea to me. It's a "warm up" exercise that you might be able to extend to more than 2 conductors.

Can you argue that there must exist some E field lines that extend from one of the conductors out to infinity?

 

[cromata]

Lines are going from positive conductor to negative conductor, but number of lines going from positive conductor is greater than number of lines going to negative conductor (|q1|>|q2|), which means that some of the lines are going from positive charge to infinity. But origin of line that is directed from some place to infinity has to be positive charge and that solves problem for 2 conductors. Thx

 

[TSny]

Lines are going from positive conductor to negative conductor, but number of lines going from positive conductor is greater than number of lines going to negative conductor (|q1|>|q2|), which means that some of the lines are going from positive charge to infinity.

OK. If you are familiar with Gauss' law, you can make this argument more rigorous.

Edit: Also, note that the problem does not say that one or more of the conductors must have a net negative charge. Some could be neutral, or all of the conductors could be positive.

But origin of line that is directed from some place to infinity has to be positive charge

Yes.

and that solves problem for 2 conductors.

Here, I'm not following you. Yes, at points of the conductor where field lines leave the conductor, the charge density is positive. But why can't there be other points on the surface of the same conductor where the charge density is negative?

 

[cromata]

I'm familiar with Gauss law and I know if I take some closed area around two conductors that flux throgh that area will be equal to (q1-q2)/εo... But also if I take closed area around part of the positive conductor from which field lines are going to infinity, I wolud obtain same result.
And I made wrong asumption that this solves problem, I will have to think thorugh it a bit more to figure it out.

 

[TSny]

I'm familiar with Gauss law and I know if I take some closed area around two conductors that flux throgh that area will be equal to (q1-q2)/εo...

OK. If you take a spherical Gaussian surface around the entire system of conductors, then there must be a net nonzero outward flux through the surface since the net charge of the system is positive. This is true even if the radius of the Gaussian surface gets as large as you wish. So, there must exist field lines somewhere in space that extend from (at least) one of the conductors "outward" to infinity.

 

[cromata]

I have an idea but I am not sure if it's correct.
Let`s assumes that negative conductor has n negative unit charges and positive has m>n positive unit charges. Let's now say that each unit charge is source/sink of E field lines. m-n lines has to go through closed area around counductors (Gauss law), so we have to have m-n positive unit charges on positive conductor...you may now ask but why can't there be no negative charges on positive conductor which will be neutralized by more positive charges on that conductor. The answer is that thete can't be field line that goes from one part of the same conductor to another, becausw that would mean that there is a potential diference between two points on conductor but we know that conducotrs are equipotential, it would also violate rotE=0 (for static fields)

 

[cromata]

And if we have n conductors, we have to have at least one positive conductor from which E field lines are going to infinite (because we have to satisfy the problems condition and Gauss law), and that conductor has to have positive surface density everywhere because of the reasons I stated

 

[TSny]

I have an idea but I am not sure if it's correct.
Let`s assumes that negative conductor has n negative unit charges and positive has m>n positive unit charges. Let's now say that each unit charge is source/sink of E field lines. m-n lines has to go through closed area around counductors (Gauss law), so we have to have m-n positive unit charges on positive conductor...you may now ask but why can't there be no negative charges on positive conductor which will be neutralized by more positive charges on that conductor.

Did you mean the word no to be there?

The answer is that thete can't be field line that goes from one part of the same conductor to another, becausw that would mean that there is a potential diference between two points on conductor but we know that conducotrs are equipotential, it would also violate rotE=0 (for static fields)

OK. But why can't there be negative charge density at some region of the net positively charged conductor such that the electric field lines that terminate at this region originate on the other conductor or perhaps come in from infinity?

 

[cromata]

That 'no' was a mistake in writing...
Well, if they come from infinity that would mean that potential on that region of conductor is negative (because E field is negative grad of potential and we assume that potential is 0 at infinity), which is imposible because conductor is equipontential and it has positive potenetial.
It also can't come from negative conductor because that would mean that it is on lower potential than negative conductor.

 

[TSny]

That 'no' was a mistake in writing...
Well, if they come from infinity that would mean that potential on that region of conductor is negative (because E field is negative grad of potential and we assume that potential is 0 at infinity), which is imposible because conductor is equipontential and it has positive potenetial.

Yes, good.

It also can't come from negative conductor because that would mean that it is on lower potential than negative conductor.

I don't quite follow this statement. Can you explain a little bit more?

 

[cromata]

Yes, good.
I don't quite follow this statement. Can you explain a little bit more?

Both conductors are eqipotential, so it's impossible that some E field lines go from positive conductor to negative and some from negative to positive (it would mean that positive conductor is at the same time on higher and lower potential than negative conductor, which is impossible)

 

[TSny]

Both conductors are eqipotential, so it's impossible that some E field lines go from positive conductor to negative and some from negative to positive (it would mean that positive conductor is at the same time on higher and lower potential than negative conductor, which is impossible)

OK. But, it would be nice if we didn't assume that one of the conductors must have a net negative charge. The only thing that is required is that the net charge of the two conductors together is positive. That does not imply that one of the conductors must be negative.

You know that Gauss' law implies that at least some of the electric field lines must extend from the system of conductors to infinity. Thus, at least one of the conductors must have a positively charged region on its surface where electric field lines originate and go to infinity. Let's call this conductor #1. As you said, this conductor must be at positive potential relative to infinity.

If conductor #1 also has a region of its surface where there is some negative charge, then there would be field lines terminating on #1 in this region. As you said, these lines that terminate on #1 cannot originate at infinity as that would imply that conductor #1 is at a negative potential relative to infinity which we know is not true. You also correctly stated that these field lines cannot originate at some other point of conductor #1 as that would imply that the surface of #1 is not an equipotential surface. So, the field lines must originate at some region of the surface of conductor #2.

Can you continue the argument at this point?

 

[cromata]

It is possible if conductor 2# is on higher potential than conductor #1, but then conductor #2 has positive surface charge everywhere (E field lines can't go from #2 to #1 and from infinity to #2, because potential #2>potential 1#>0), this was for only 2 conductors in space...if we assume that there are more than 2 conductors in space then we can just continue with same arguments on other conductors

 

[TSny]

It is possible if conductor 2# is on higher potential than conductor #1, but then conductor #2 has positive surface charge everywhere (E field lines can't go from #2 to #1 and from infinity to #2, because potential #2>potential 1#>0),

OK. I think that gets it.

 

[TSny]

It is possible if conductor 2# is on higher potential than conductor #1, but then conductor #2 has positive surface charge everywhere (E field lines can't go from #2 to #1 and from infinity to #2, because potential #2>potential 1#>0), this was for only 2 conductors in space...if we assume that there are more than 2 conductors in space then we can just continue with same arguments on other conductors

After rereading your final statement, I'm wondering if you meant the "from #2 to #1" in blue to actually be "from #1 to #2".

 

[TSny]

Sorry, Let me try again. You are right that field lines cannot go from infinity to #2 if there are lines from #2 to #1. Why can't there be a patch of negative charge on #2 where field lines terminate and such that these lines originated on #1? Maybe the reason is obvious, but I would feel better if it was stated explicitly.

 

[cromata]

I already explained that, it is impossible to have E lines that go from one conductor to other in oposite directions, because it would mean that one of the conductors is at higher and lower potential than the other at same time (which is impossible because conductors are equipotential surfaces)...
-To conclude, solution to this problem is: if we have n isolated conductors in space with positive sum of charges, then there is a conductot that has higher potential than all the others, and E lines can only originate from its surface, they can't terminate on that conductor, (I already explained reasons why), and that means that conductor with the highest potential has positive surface area everywhere.

And thank you, it was really helpful to think through this problem step by step.

 

[TSny]

I already explained that, it is impossible to have E lines that go from one conductor to other in oposite directions, because it would mean that one of the conductors is at higher and lower potential than the other at same time (which is impossible because conductors are equipotential surfaces)...

Yes. Good.

-To conclude, solution to this problem is: if we have n isolated conductors in space with positive sum of charges, then there is a conductot that has higher potential than all the others,

Could there be two or more conductors at the highest potential? Then there would not be a conductor that has higher potential than all of the others.

and E lines can only originate from its surface, they can't terminate on that conductor, (I already explained reasons why), and that means that conductor with the highest potential has positive surface area everywhere.

Yes. But, strictly speaking, the argument so far only implies that there must be a conductor with non-negative charge density everywhere (rather than positive charge density everywhere). Could there be a point on the conductor where the charge density is zero?

And thank you, it was really helpful to think through this problem step by step.

This is a fun problem. I'm hoping someone else will chime in with a different approach.

 

[cromata]

There can be more than one conductor on the highest potential, but then all of them would have non-negative surface density everywhere becuase there is no E field between them if they are on same potential (E field is negativ gradient of potential as I said before). But I don't see a reason why couldn't density be 0 at some points on conductor surface.

 

[cromata]

Yes. But, strictly speaking, the argument so far only implies that there must be a conductor with non-negative charge density everywhere (rather than positive charge density everywhere). Could there be a point on the conductor where the charge density is zero?

Hmm, if σ=0 at some part of surface that would mean that ∇E=0(I know that gauss law is ∇E=ρ/εo, but I don't see a reason if σ=0 why wouldn't then also be ρ=0) at that point. And that would mean that E=0 at that point (because it isn't the source/sink of electric field (∇E=0), and there can't be any E field lines passing trough that area because E=0 inside of conductor and component that is tangential to the conductors surface is 0). And this would imply that potential at some area in space(close to this region of surface) is equal to potential on the conductor. And there can't be any any E field lines passing through this region of space because it would imply that that region is not at max potential of that space, and empty space can't be source/sink of electric field because ∇E=0 in empty space.

 

[TSny]

Hmm, if σ=0 at some part of surface that would mean that ∇E=0(I know that gauss law is ∇E=ρ/εo, but I don't see a reason if σ=0 why wouldn't then also be ρ=0) at that point. And that would mean that E=0 at that point (because it isn't the source/sink of electric field (∇E=0), and there can't be any E field lines passing trough that area because E=0 inside of conductor and component that is tangential to the conductors surface is 0).

Yes, if σ = 0 at some point of the surface, then E =0 at that point.

And this would imply that potential at some area in space(close to this region of surface) is equal to potential on the conductor.

I don't follow this. If E = 0 at a point of the surface, I don't see why this implies that the potential V must be constant throughout some region close to the surface.

I haven't been able to show that there can't be a point of the surface where σ = 0. But I also haven't been able to construct an example where σ = 0 at a point of the surface. [I tend to believe the statement of the problem; i.e., there are no points where σ = 0.]

 

[cromata]

I don't follow this. If E = 0 at a point of the surface, I don't see why this implies that the potential V must be constant throughout some region close to the surface

Electric field is gradient of potential (as I said few times already), if we want visualise this, we have to think that E field lines are showing us potentials fastest change.
So if there are some points in space that are not on same potential there must be E field between those two points.
So in our case, if potential around neutral surface of conductor is diferent from conductors potential that would imply that there are field lines between this space and neutral region of conductors surface. But we know that there can't be any field lines going to/out of this neutral surface because E=0.

 

[TSny]

If your argument is valid, then it seems to me that it should apply to any conductor in electrostatic equilibrium.

But it is not hard to come up with an example where you have a conductor with non-negative charge density at every point of the surface of the conductor and such that some points of the surface have zero charge density.

Here is an example. It consists of a large conducting sphere (blue) with a net positive charge Q and two very small conducting spheres (red) which each have a net negative charge -q. The large sphere is symmetrically placed between the two small spheres. The figure shows the direction of the field lines and also shows some equipotential lines.

It is possible to choose the the ratio q/Q such that the charge density is positive at all points of the surface of the large sphere except for the "equator" where the charge density is zero. The two orange dots are points on the equator (the charge density is zero there). Although the field is zero at the equator, the field becomes nonzero as soon as you move perpendicularly off the surface at the equator. For all points in the plane of the equator that are outside the larger sphere, the field points directly toward the center of the large sphere. But as you get closer to the sphere in the equatorial plane, the field gets weaker and goes to zero at the equator.

So, the large sphere is a conductor with non-negative charge density everywhere on its surface and also has points on its surface where the charge density is zero. It seems to me that your argument doesn't work here.

However, for this example, the net charge of the total system is negative. So, it does not correspond to the statement of the original problem! So, we still have an open question as to whether or not the statement of the problem is correct. That is, if the net charge of a system of conductors is positive then is it really true that there must be a conductor in the system with surface charge density that is strictly positive (never zero at any point of the surface)? I don't know.

 

[cromata]

Is the positive conductor in this situation part of the space with the highest potential? If it is then my argument doesn't work.
However, if the positive sphere is not on the highest potential then I think my argument is valid.
Let's get back to situation with net positive charge in space. Part of my argument was that if the surface density was 0 at some point then space close to it should be on same potential. If the potential is the same in some part of the space as on the conductor then that part of space is at max potential. If that part is on max potential that means that field lines can only go from that space (not to it). But since empty space can't be origin of E field lines, it's impossible that any lines are passing through this region of space (if some lines were passing through it, it would imply that isn't on the highest potential in space). And if we think a bit more we could conclude that in this situation all of empty space should be on same (highest potential) which is impossible (unless all of space is empty).

 

[cromata]

Is in your example potential highest in infinity?
Because if it is then my argument must be valid, because in the situation of this example E field lines can be passing through the space near conductor (that is on the same potential as the conductor), because that region of space is not at highest potential.

 

[TSny]

Is the positive conductor in this situation part of the space with the highest potential? If it is then my argument doesn't work.

In my example, the positive conductor is not the place of highest potential. The highest potential is zero (at infinity).

However, if the positive sphere is not on the highest potential then I think my argument is valid.
Let's get back to situation with net positive charge in space. Part of my argument was that if the surface density was 0 at some point then space close to it should be on same potential.

I don't feel comfortable with this part of your argument. If σ = 0 is zero at a point of the conductor, then E = 0 at that point. But I don't see why E must be zero for other points in the neighborhood of this point and exterior to the surface. So, I don't see why it's necessary that the potential be constant in this neighborhood. But I do think you have a key point that the conductor of interest is the region at highest potential. Maybe your argument is fine and I'm just being dense. I just wish it could be "tightened up".

[EDIT: I think the argument we need is essentially the same argument that would prove the following proposition: If you have a single isolated conductor of arbitrary shape and it has a net positive charge, then every point on the surface must have a positive charge density. No point on the surface could have a negative or zero charge density.]