What is the charge of each conductor afterwards?

[Rijad Hadzic]

Homework Statement

Two identical conductors are brought into contact. Initially one conductor has a charge of +30 x10^(-6) C, what is the charge of each conductor afterward? Does it matter how the contact is made?

Homework Equations

The Attempt at a Solution

Well since they are conductors, electrons can roam freely throughout the material..

Basically, I think the question is implying that the other conductor is neutral, correct?

Anyways, The electrons in other conductor want free space, but since one has +30 x10^(-6) C charge, it is willing to take half that charge in electrons since they are similar, correct?

So both conductors will end up with +15 x 10^(-6) C ?

And it doesn't matter how contact is made because they are conductors and electrons move around easily with conductors.

 

[NFuller]

Looks good to me.

 

[collinsmark]

Looks good to me.

Actually, I'm not sure if it's quite so simple.

Homework Statement

Two identical conductors are brought into contact. Initially one conductor has a charge of +30 x10^(-6) C, what is the charge of each conductor afterward? Does it matter how the contact is made?

Homework Equations

The Attempt at a Solution

Well since they are conductors, electrons can roam freely throughout the material..

Basically, I think the question is implying that the other conductor is neutral, correct?

Anyways, The electrons in other conductor want free space, but since one has +30 x10^(-6) C charge, it is willing to take half that charge in electrons since they are similar, correct?

So both conductors will end up with +15 x 10^(-6) C ?

And it doesn't matter how contact is made because they are conductors and electrons move around easily with conductors.

I think you are supposed to assume that one conductor initially has zero charge, but the problem statement is a little ambiguous on this point. (If you want to get nit-picky, you could make the argument that each conductor has 30 microcoulombs of charge -- the problem statement does specify that they are initially "identical," and doesn't specifically state what the charge on the other conductor is, leaving the reader to assume it must have a charge identical to the first. I doubt you're supposed to use this line of reasoning though.)

Regarding your claim about the "electrons move around easily [within] conductors," be careful here. Free electrons (or lack thereof: holes) will arrange themselves such that the electric field within the conducting material is always zero. This places some heavy restrictions about where within the material the charges are.

This problem would be pretty easy if the conductors are spherical. But what if they are not? What if they are of some complicated shape? And more importantly, what if the point of contact is asymmetrical, for example the end of one conductor touching the middle of the other?

 

[rude man]

I agree with the OP even though collinsmark makes some good points.

 

[haruspex]

more importantly, what if the point of contact is asymmetrical

I agree. The question ought to state that they are brought into contact in a symmetrical arrangement.

 

[rude man]

I agree. The question ought to state that they are brought into contact in a symmetrical arrangement.

I don't think it matters how contact is made, as long as contact is made somewhere for at least picoseconds. Potentials are equalized and the charges rearrange themselves very rapidly until as collinsmark has pointed out the internal E fields are zero..

 

[haruspex]

I don't think it matters how contact is made, as long as contact is made somewhere for at least picoseconds. Potentials are equalized and the charges rearrange themselves very rapidly until as collinsmark has pointed out the internal E fields are zero..

But if the configuration is asymmetric, will it equalise the charges?

 

[rude man]

But if the configuration is asymmetric, will it equalise the charges?

Charge will essentially instantly arrange over both conductors so as to effect a zero electric field in each; whether one or both are pre-charged makes no difference. The amount of charge on each is determined by this tendency, so if they're asymmetric one will likely wind up with more charge than the other. And the distributions will change again if the two conductors are distanced from each other after contact. Potentials will be equal upon contact unless the conductors are subsequently moved away from each other (as I found out on one of your earlier posts with the two spheres of differing radii if you remeber.)

 

[haruspex]

Charge will essentially instantly arrange over both conductors so as to effect a zero electric field in each; whether one or both are pre-charged makes no difference. The amount of charge on each is determined by this tendency, so if they're asymmetric one will likely wind up with more charge than the other. And the distributions will change again if the two conductors are distanced from each other after contact. Potentials will be equal upon contact unless the conductors are subsequently moved away from each other (as I found out on one of your earlier posts with the two spheres of differing radii if you remeber.)

That's all true, but I am not sure whether you are agreeing or disagreeing with me. We seem to be saying the same now.

I tried to find an example where I could prove the charge would not be distributed equally between the two objects if their contact is asymmetric, but it is surprisingly diffiicult.

 

[rude man]

That's all true, but I am not sure whether you are agreeing or disagreeing with me. We seem to be saying the same now.

I tried to find an example where I could prove the charge would not be distributed equally between the two objects if their contact is asymmetric, but it is surprisingly diffiicult.

If one conductor is larger (larger surface area) than the other, that larger conductor will have more charge.
I interpret your post #7 as having some doubt - somewhere. I have been completely consistent in my posts I believe.

And I disagree with your post #5. It matters not how contact is made, as I said in my post #6.

 

[collinsmark]

If one conductor is larger (larger surface area) than the other, that larger conductor will have more charge.
I interpret your post #7 as having some doubt - somewhere. I have been completely consistent in my posts I believe.

And I disagree with your post #5. It matters not how contact is made, as I said in my post #6.

We can assume that the conductors are identical in surface area (they are identical per the problem statement).

But I'm still not convinced that the end result will be equal charge on each conductor. Suppose the conductors are rod shaped and the end of one rod touches the other rod in the middle. I haven't worked out the math* (as @haruspex mentioned, it's a difficult problem) but I suspect that the charge on each conductor will might be different. [Edit: or perhaps a better statement is that I have yet to convince myself that they must be equal.]

*[Electrostatic modeling software would come in handy; unfortunately, I don't have access presently.]

 

[haruspex]

It matters not how contact is made, as I said in my post #6.

And as I responded in post #7, I see no justification for that conclusion. Equal potentials does not imply equal charges since the potential in each is affected by the distribution of charges in the other, and because of the asymmetric contact these affects are not symmetric.

Edit: the simplest example I could think of which might result in different charges is this...
Two pairs of identical spheres radius r. Each pair is connected by a thin rod, not necessarily straight, but keeping the sphere centres distance s apart.
One sphere of one pair is sited halfway between the spheres of the other pair. The other sphere of the first pair is off to the side at right angles.
Using spheres, the difficulty is that the charges will not be uniformly distributed on them. If we ignore that then I can show the two pairs can have unequal charges, but the discrepancy is so small that it could be explained by the approximation.

Using point charges instead is no good because the potential at a point charge is unbounded.

 

[rude man]

And as I responded in post #7, I see no justification for that conclusion. Equal potentials does not imply equal charges since the potential in each is affected by the distribution of charges in the other, and because of the asymmetric contact these affects are not symmetric.

I have to concede that I jumped the gun here. For two spheres making contact there is one and only one way that contact is made, so the exact contact points do not matter.. For similar but arbitrary shapes that is not the case. I guess the only thing we can positively state is that the charges will distribute so that the E field inside both conductors is zero.

 

[rude man]

We can assume that the conductors are identical in surface area (they are identical per the problem statement).

But I'm still not convinced that the end result will be equal charge on each conductor. Suppose the conductors are rod shaped and the end of one rod touches the other rod in the middle. I haven't worked out the math* (as @haruspex mentioned, it's a difficult problem) but I suspect that the charge on each conductor will might be different. [Edit: or perhaps a better statement is that I have yet to convince myself that they must be equal.]

*[Electrostatic modeling software would come in handy; unfortunately, I don't have access presently.]

Yes, I jumped the gun. Very likely the relative charge distribution on each conductor will depend on where contact is made.

I finally realized that there is only one way to make contact with two dissimilar spheres so the relative charge on each sphere is the same irrespective of where exactly the spheres make contact. And even this simple charge ratio took Maxwell some effort to derive. (Haruspex sent out a link to a paper wherein this charge ratio is derived.) Do you know what a digamma function is? it's involved in even this simple situation! So probably you'd be spinning your wheels trying to derive the charge disributions for e.g two similar rods, the end of one touching the middle of the other, etc.
EDIT: you need the digamma function only if you want the potential of the touching ensemble. But you still need Maxwell's (or othe involved - Legendre polynomials?) equation to determine the charge ratio.

 

[NFuller]

So here's a general scenario. Imagine two identically shaped blobs of conducting material, one is charged and one is not. The two objects are brought into contact such that they have equal potential across their surfaces, as is the nature of conductors. The blobs are now separated at infinity such that they do not interact with each other. They will both produce identical electric fields since they are the same shape and have the same potential at their surfaces. Since the electric potential uniquely determines the electric field by
$$\mathbf{E}=-\nabla\phi$$
and the surface charge density is uniquely determined at the surface by
$$\sigma=-\epsilon\frac{\partial \phi}{\partial n}$$
I imagine that two identically shaped conductors with identical electric fields will have identical surface charge densities. Doesn't this mean they both have equal total charge?

 

[haruspex]

since they are the same shape and have the same potential at their surfaces.

Only if the charges are guaranteed the same, which is what we we would like to prove or disprove.
The potentials while in contact are not the same as the potentials when separated.

 

[NFuller]

The potentials while in contact are not the same as the potentials when separated.

I expect this would require the charges to instantly redistribute themselves at the instant of separation, which doesn't make sense to me since the charges will become stranded on the conductor at separation and unable to move to the other conductor.

 

[haruspex]

I expect this would require the charges to instantly redistribute themselves at the instant of separation, which doesn't make sense to me since the charges will become stranded on the conductor at separation and unable to move to the other conductor.

No, at an infinitesimal separation, after contact, any charge redistribution on each conductor is likewise infinitesimal. The total charge on each is fixed.
As the separation increases, the charges will continue to redistribute on each as the field from the other becomes weaker.

 

[NFuller]

Ok, so doesn't this mean that at infinitesimal separation the potentials on each surface are still equal?

 

[haruspex]

Ok, so doesn't this mean that at infinitesimal separation the potentials on each surface are still equal?

Only if they had the same total charge when in contact. Otherwise they will be infinitesimally different.

 

[NFuller]

But otherwise this would require the potential to change instantaneously at separation, which would require an instantaneous redistribution of charge.

 

[haruspex]

But otherwise this would require the potential to change instantaneously at separation, which would require an instantaneous redistribution of charge.

I don't get you.
While in contact, they will have the same potential but perhaps different charges.
On separation, each will retain the charge it had, but the potential at each will gradually change as the separation increases.
Remember that the potential at each is due to its own charge and the field from the other. Since the field from the other affects its distribution of charge this is quite a complicated problem. See the link rude man posted for the two sphere case.

 

[NFuller]

Maybe I'm not explaining my idea very well. Here's another thought. Since both conductors are the same they both have equal self capacitance. If at the moment of separation the potential measured from infinity is the same and the charge can no longer move between conductors then doesn't ##Q=CV## show the the charge on each conductor must be the same?

 

[haruspex]

Maybe I'm not explaining my idea very well. Here's another thought. Since both conductors are the same they both have equal self capacitance. If at the moment of separation the potential measured from infinity is the same and the charge can no longer move between conductors then doesn't ##Q=CV## show the the charge on each conductor must be the same?

Can't say I follow the reasoning.
Does your argument still apply if it's two identical spheres but with a point charge nearer to one than to the other at the moment of separation?

 

[NFuller]

Can't say I follow the reasoning.

The self capacitance of a conductor is determined by its shape. I think we agree that the potential is the same right after separation so using the definition of capacitance, the charge on each conductor is the same right after separation.

Does your argument still apply if it's two identical spheres but with a point charge nearer to one than to the other at the moment of separation?

I'm not sure in this case since each conductor now has a mutual capacitance between it and the point charge on top of its self capacitance. The conductor closer to the point charge would have a different value of ##C## then the conductor further away.

 

[rude man]

I expect this would require the charges to instantly redistribute themselves at the instant of separation, which doesn't make sense to me since the charges will become stranded on the conductor at separation and unable to move to the other conductor.

Can't say I follow the reasoning.
Does your argument still apply if it's two identical spheres but with a point charge nearer to one than to the other at the moment of separation?

@haruspex, perhaps you should send the OP the link to that new zealand paper with the maxwell derivation for Q and V for two dissimilar spheres. I think it would give the OP an idea of just how difficult this problem is. I would insert the file myself but it seems attachments are no longer facilitated in this forum.

 

[haruspex]

I'm not sure in this case since each conductor now has a mutual capacitance between it and the point charge on top of its self capacitance. The conductor closer to the point charge would have a different value of C then the conductor further away.

In the same way, the two objects would have mutual capacitance. Because of their asymmetric arrangement, this may affect them differently.

 

[NFuller]

I keep trying to mathematically prove this one way or the other but keep hitting snags. I think I might post this as a new question and reference this thread. Maybe some other people here have some ideas. I asked a few people in my department today about this and they were also unsure of how to proceed.

 

[haruspex]

@haruspex, perhaps you should send the OP the link to that new zealand paper with the maxwell derivation for Q and V for two dissimilar spheres. I think it would give the OP an idea of just how difficult this problem is. I would insert the file myself but it seems attachments are no longer facilitated in this forum.

This paper may be more relevant: https://arxiv.org/pdf/0906.1617.pdf.
It tackles the case of multiple spheres, so could be applied to my two pairs of two spheres. But it's hard work and still involves approximations.

 

[rude man]

I was looking into the Uniqueness theorem in electrostatics, which says (I think) that if the potential of the connected conductors were known, the E field would everywhere be unique. And a unique E field implies unique surface charge.
The snag is that we have not shown that the potential of the connected conductors is the same irrespective of how contact is made. I suppose the prima facie belief is that it is not, that V varies with how contact is made.

 

[haruspex]

I was looking into the Uniqueness theorem in electrostatics, which says (I think) that if the potential of the connected conductors were known, the E field would everywhere be unique. And a unique E field implies unique surface charge.
The snag is that we have not shown that the potential of the connected conductors is the same irrespective of how contact is made. I suppose the prima facie belief is that it is not, that V varies with how contact is made.

I agree that the overall potential is likely to depend the configuration, but this is quite a different question, right? If it does depend on the configuration, it could still turn out that the two charges are equal, and even if it is independent of configuration the charge split could differ for the same potential.

Edit: As an indicator that the potential does depend on configuration, consider a large number of identical spheres. Arranged as a tight ball the potential would be more than when arranged as a spherical shell.
The same model says the charges will be different for different spheres in the tight ball.
Showing these results for two objects will be tougher.

 

[rude man]

If connecting two similar charged rods in two different ways always resulted in the same ensemble potential, then you could slide one rod against another without losing contact until you get the same juxtaposition for both connecting ways. Then by the uniqueness theorem the charge distributions would have to be the same since potential and juxtaposition would be the same. Unfortunately, I guess there would be two differing ensemble potentials so the Q distributions would also differ.

 

[haruspex]

If connecting two similar charged rods in two different ways always resulted in the same ensemble potential

Yes, but I think it's easy to see that it won't. Placed side by side would surely create a higher potential than end to end. Likewise a pair of plates.

 

[rude man]

Yes, but I think it's easy to see that it won't. Placed side by side would surely create a higher potential than end to end. Likewise a pair of plates.

Not obvious to me, but no argument either.

 

[haruspex]

Not obvious to me, but no argument either.

This paper, http://www.colorado.edu/physics/phys3320/phys3320_sp12/AJPPapers/AJP_E&MPapers_030612/Griffiths_ConductingNeedle.pdf, gives exact expressions for charge distribution in and equipotentials near an infinite thin ribbon. See section I
V.
It should be possible to compare two such ribbons placed edge to edge (doubling λ and a) with placing them face to face (doubling λ only).