# Homework Help: Surface density on isolated conductors in space

1. Jul 1, 2017

### cromata

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Using induction

3. 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 statment 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.

2. Jul 1, 2017

### TSny

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?

3. Jul 1, 2017

### 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

4. Jul 1, 2017

### TSny

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.

Yes.

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?

5. Jul 1, 2017

### 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.

6. Jul 1, 2017

### TSny

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.

7. Jul 1, 2017

### cromata

I have an idea but im 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)

8. Jul 1, 2017

### 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

9. Jul 1, 2017

### TSny

Did you mean the word no to be there?
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?

10. Jul 1, 2017

### 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.

11. Jul 1, 2017

### TSny

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

12. Jul 1, 2017

### cromata

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)

13. Jul 1, 2017

### TSny

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?

14. Jul 1, 2017

### 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

15. Jul 1, 2017

### TSny

OK. I think that gets it.

16. Jul 2, 2017

### TSny

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".

17. Jul 2, 2017

### 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.

Last edited: Jul 2, 2017
18. Jul 2, 2017

### 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.

19. Jul 2, 2017

### TSny

Yes. Good.
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.

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?

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

20. Jul 2, 2017

### 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.