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What is the tension in a charged ring?

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  • #51
haruspex
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Go back to counting field lines.
Am I right in thinking you have interpreted this as a problem in a 2D universe, effectively turning the ring into an infinite cylinder?
 
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  • #52
etotheipi
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Am I right in thinking you have interpreted this as a problem in a 2D universe, effectively turning the ring into an infinite cylinder?
This would be interesting. AFAIK there are no constraints that limit Gauss' law to a 3D universe, except the equation would be in a 2D universe,$$\oint \vec{E} \cdot \hat{n} dl = \frac{q}{\epsilon_0}$$where we are integrating around a closed curve, and ##\hat{n}## is orthogonal to the curve. This would mean that the field seen just outside the ring, in the 2D universe, would be (using the radial symmetry)$$E = \frac{q}{2\pi \epsilon_0 r}$$This is like the infinite cylinder in the 3D universe. It would appear to permit a logarithmic electric potential.
 
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  • #53
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This would be interesting. AFAIK there are no constraints that limit Gauss' law to 3D, except the equation would be in 2D,$$\oint \vec{E} \cdot \hat{n} dl = \frac{q}{\epsilon_0}$$where we are integrating around a closed curve, and ##\hat{n}## is orthogonal to the curve. This would mean that the field seen just outside the ring, in the 2D universe, would be (using the radial symmetry)$$E = \frac{q}{2\pi \epsilon_0 r}$$But that would seem to give a logarithmic electric potential.
Gauss' law is just an application of divergence theorem in the case of electromagnetism, that is
$$
\int_V (\nabla \cdot \mathbf E) dV = \oint_S \mathbf E \cdot d\sigma
$$
We have to have a surface.
 
  • #54
etotheipi
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Gauss' law is just an application of divergence theorem in the case of electromagnetism, that is
$$
\int_V (\nabla \cdot \mathbf E) dV = \oint_S \mathbf E \cdot d\sigma
$$
We have to have a surface.
The divergence theorem holds in any number of dimensions. In 2D the closed curve becomes the surface.
 
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The divergence theorem holds in any number of dimensions. In 2D the closed curve becomes the surface.
I would really like to know if there exists some references for that claim.
 
  • #56
etotheipi
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I would really like to know if there exists some references for that claim.
From Wikipedia:
In physics and engineering, the divergence theorem is usually applied in three dimensions. However, it generalizes to any number of dimensions. In one dimension, it is equivalent to integration by parts. In two dimensions, it is equivalent to Green's theorem.
Also have a look here: https://arxiv.org/pdf/1809.07368.pdf. It is noted that in 2 dimensions the electric field of a point charge at ##\vec{r}'## goes as$$\vec{E} = \frac{q}{2\pi \epsilon_0|\vec{r}-\vec{r}'|^2} (\vec{r} - \vec{r}')$$ and as such we can recover the form ##\nabla \cdot \vec{E} = \frac{\sigma}{\epsilon_0}##. The potential is also logarithmic, $$V(r) = -\frac{q}{2 \pi \epsilon_0} \ln{\frac{r}{a}}$$ where ##a## is an arbitrary constant of integration.

Anyway we should probably wait and see if this was what @Vanadium 50 was referring to before speculating any further :wink:
 
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  • #57
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This would be interesting. AFAIK there are no constraints that limit Gauss' law to a 3D universe, except the equation would be in a 2D universe,$$\oint \vec{E} \cdot \hat{n} dl = \frac{q}{\epsilon_0}$$where we are integrating around a closed curve, and ##\hat{n}## is orthogonal to the curve. This would mean that the field seen just outside the ring, in the 2D universe, would be (using the radial symmetry)$$E = \frac{q}{2\pi \epsilon_0 r}$$This is like the infinite cylinder in the 3D universe. It would appear to permit a logarithmic electric potential.
I think ##\oint \mathbf E \cdot dl ## will not be a finite quantity.
 
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  • #58
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I think ##\oint \mathbf E \cdot dl ## will not be a finite quantity.
Note that it is not ##\oint \vec{E} \cdot d\vec{l}##, the unit vector ##\hat{n}## points away from the instantaneous centre of curvature. And in any case, the charge enclosed by a closed curve at a radius ##r## from the centre of the ring (outside the ring) is finite, so the field outside the ring in the 2D universe is surely also finite.
 
  • #59
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Am I right in thinking you have interpreted this as a problem in a 2D universe, effectively turning the ring into an infinite cylinder?
Not exactly - that would change Coulomb's Law as well. I think that's what etotheipi's paper is about. I am not changing Coulomb's Law. However, I am working strictly in a plane. Because that's what the problem said to do. I would not call it an infinite cylinder because I don't know what the role of λ would be in that geometry.

I'm not doing anything but what the problem says. The problem says to neglect the thickness, so I am working with a zero thickness line. But what I am hearing is "No! The line has cross-sectional extent!" and "No, he's really talking about an infinite cylinder!" and now even "He must not be using 3-D Maxwell's equations!"

This feels kind of like an ideal gas problem, and having everyone saying "Any real gas would have liquified!" I have answered the question asked. I have not tried to answer a "better" version of this question.
 
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  • #60
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Note that it is not ##\oint \vec{E} \cdot d\vec{l}##, the unit vector ##\hat{n}## points away from the instantaneous centre of curvature. And in any case, the charge enclosed by a closed curve at a radius ##r## from the centre of the ring (outside the ring) is finite, so the field outside the ring in the 2D universe is surely also finite.
Well, I meant our contour is the loop itself and thats where field is not finite.
 
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  • #61
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Not exactly - that would change Coulomb's Law as well. I think that's what etotheipi's paper is about. I am not changing Coulomb's Law. However, I am working strictly in a plane. Because that's what the problem said to do. I would not call it an infinite cylinder because I don't know what the role of λ would be in that geometry.
Let me explain what I both understand and don't understand about the field line approach you propose (I'll go back to a 3D universe :wink:).

The number of field lines through a surface that encloses the ring is a measure of the flux out of the surface, and is proportional to the enclosed charge. So if we make the ring smaller but maintain its charge, the number of field lines through any surface enclosing the ring will not change. This is okay 😁.

But what I am struggling to understand is why that implies the field just outside of the ring in the radial direction is ##\frac{Q}{4\pi \epsilon_0 r^2}##, at a distance ##r## from the centre. That appears to only be the case if we have spherical symmetry. You did mention that you were only considering field lines in the plane, but I'm not sure this is valid since we are still applying the 3D theorem of Gauss.

Indeed I tried finding an explicit formula without heuristics and ended up with some horrible integrals that I could not solve, which seemed to imply to me that the radial field outside of the ring did not take such a nice form.

I wondered if I had misinterpreted anything you said?
 
  • #62
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Make the ring as small as you can. It is now indistinguishable from a point. Now use the field from a point.
 
  • #63
etotheipi
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Make the ring as small as you can. It is now indistinguishable from a point. Now use the field from a point.
What about this; I found a reference that deals analytically with the electric field from a ring of charge like the one we are describing, and the radial electric field at a point P at ##\vec{p}## does not look like it obeys the relation you suggest, but instead the horrible expression requiring elliptic integrals

1593522462306.png


where terms are defined in the paper http://www.mare.ee/indrek/ephi/efield_ring_of_charge.pdf
 
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  • #64
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What about this; I found a reference
If you read the paper, you would see that he is discussing the field in the XZ plane, not the XY plane. It is possible that if you took the appropriate limit of his calculation you would get an answer different than mine, but I think that's your responsibility to show that. Otherwise, you can deluge me with a bunch of random calculations on random webpages and say "Ha! Prove that this calculation's limiting case matches your derivation" without end.
 
  • #65
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If you read the paper, you would see that he is discussing the field in the XZ plane, not the XY plane.
I thought the point was that because of the symmetry of the ring, we can describe the situation with no loss of generality by positioning the ring in the XY plane and the test charge at an arbitrary point in the XZ plane. For instance, positioning our test charge along the ##y## axis would be equivalent to putting it along the ##x## axis and rotating our piece of paper. The benefit is that we now only require two coordinates.

In any case, we are looking at the specific case in that paper where ##a=0## and ##r>R##, where the terms ##q## and ##\mu## in that formula for ##E_r## are defined as ##q = r^2 + R^2 + a^2 + 2rR## and ##\mu = \frac{4rR}{q}##. At first glance, it doesn't look like that yields a Coulombic field strength for ##E_r##.

I only provided this reference as a counter-example because I am still struggling to understand how the heuristic shrinking of the ring can imply your expression. I have not checked their calculations, however, so I cannot vouch for its accuracy.
 
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  • #66
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Go back to counting field lines. If I have six at 10R and 2R, if I replace the ring with a smaller one - say R/4 - with the same change, I will still have six at 10R and 2R. And at R and R/2. Repeat as necessary.
The field is infinite at the ring in one case and finite at the other case, i dont understand how the field configurations are equivalent. Counting field lines i found them infinite at r=R in one case and finite at the same location in the other case where the charge is in center.

There is no "center" to a ring of zero thickness (and/or height).
How do we call the center of the circle that corresponds to the ring?
 
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  • #67
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I thought the point was that because of the symmetry of the ring, we can describe the situation with no loss of generality by positioning the ring in the XY plane and the test charge at an arbitrary point in the XZ plane.
I am working only in the XY plane. I say it over and over and over again.
If you want to take an XZ calculation and see what happens when you set z = 0, I think it's your responsibility to do that. You can't just toss calculations at me for something else.

The field is infinite at the ring in one case and finite at the other case
Are you talking about at each side of the ring? I want an answer so I pick the side that's finite.
 
  • #68
etotheipi
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I am working only in the XY plane. I say it over and over and over again.
However, I am working strictly in a plane.
I think this is actually the problem. We are considering a ring embedded in a 3D space. As such, we must use Maxwell's equations in 3D. So it is impossible to "strictly work in a plane" in the sense of your heuristic argument with field lines.

As an example, here is my proof of the shell theorem. Take a spherically symmetric charge distribution, and a spherical surface ##S##. Then$$\int_S \vec{E} \cdot d\vec{A} = 4\pi r^2 E = \frac{Q}{\epsilon_0}$$hence the configuration is equivalent to a point charge at the origin. The key stipulation here was the spherical symmetry, since that allows us to remove the electric field from the integral i.e. it is of uniform magnitude across ##S##.

Now for a ring of charge, we can also surround it by a Gaussian surface and write $$\int_S \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$$except now we cannot deduce anything further because there is not sufficient symmetry. Although the flux/number of field lines ##\int_S \vec{E} \cdot d\vec{A}## is indeed constant however you choose your surface, that does not imply shell theorem for a ring, because we can no longer pull ##E## out of the integral: it is not of constant magnitude over ##S##!

In your terms, as you make the ring smaller and smaller, we are actually changing the distribution of field lines. We can thus infer nothing about the magnitude of the field strength outside the ring.
 
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  • #69
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As such, we must use Maxwell's equations in 3D.
Which is what I explicitly said I was doing in #59.

So it is impossible to "strictly work in a plane"
I discussed this when I said 'People are, often implicitly, interpreting "negligible thickness" as "the ring is a torus, and we are taking the limit as the minor axis goes to zero.' As I pointed out then, doing this leads to infinities.

Let me point out that I am the only person who has actually been able to get to an answer to this problem.

The complaints fall into two categories. One is "you need to consider a finite extent of the ring (aka 3D vs 2D)" where my response has been that I am following the question as asked. It says "negligible" so I am neglecting it. (And would argue that not neglecting it when told to is a mistake) I have been very explicit about it. The other objection is that in a case with ambiguity ("inside" or "outside" of the ring) I pick the finite answer. I have been very explicit about that as well.

I've said all that I intend to say. This is the answer to the question posed. It is not an answer to any different question, even if that question is "better" or more realistic.
 
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  • #70
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Alright, thanks for your help. I still do not think that the field outside the ring is ##\frac{q}{4\pi \epsilon_0 r^2}##, although I would agree that the field reduces to this in the limit ##r \gg R##. I don't think either of the shell theorems apply to rings in a 3D universe.

So perhaps we will agree to disagree :wink:. You did manage to get an answer, though, but I think I am happier not having an answer than accepting one that I don't understand. In any case, an interesting discussion!
 
  • #71
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What if you consider a small element on the right, and calculate the tension on that and take the limit as the size of the element goes to zero?

I don't know if that would work out. It's just an idea.
Actually that's a pretty good idea, have some (no longer infinitesimal) control region on the right that subtends an angle ##\alpha## or something, and try to come up for an expression for the electrostatic force on that.

It's slightly more complicated maths-wise and I'm not so sure how best to do it yet, but I think that would give something useful! Let me think about it for a while 😁

Thanks!
This a good approach. For an infinitesimal segment subtending angle ##\alpha##, the force on the segment is due to the net electric field of all of the charge of the ring excluding the charge in the segment itself. (The field due to the segment does not exert a net force on the segment itself.) So you just need to modify the limits of the integral ##I## in the OP. Everything else in that post looks good. When I carry out this procedure, I get an expression for ##T## that agrees with ##T = F/2## in post #39 in the limit of infinitesimally small ##\alpha##. In particular, you get the logarithmic divergence as ##\alpha## approaches zero.
 
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  • #72
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The problem says to neglect the thickness, so I am working with a zero thickness line
But it does not say space has only two dimensions, whch is what is needed for your field line counting. Your argument results in an inverse law, not an inverse square law.
In 3D, the field lines from the ring diverge into the third dimension, effectively becoming sparser.

Indeed, if you consider the infinite cylinder in 3D, axis along Z, you will see that the field lines in the XY plane do exactly what you are saying.
 
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  • #73
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We all are chickens, hehe, no one dared to deal with the case where we take the ring to be a torus. We would have to do a "fearsome" triple integral over the region of the torus to find the electric field. Something tells me that working in cylindrical coordinate system might be the best for working with torus, but i am not sure. The good thing when we take the ring to be a torus , is that the volume charge density becomes a nice finite continuous function over the region of the torus, so we get rid of the infinity of the dirac delta function for the charge density which is apparent when we take the ring to be a circle.
 
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  • #74
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We all are chickens, hehe, no one dared to deal with the case where we take the ring to be a torus. We would have to do a "fearsome" triple integral over the region of the torus to find the electric field. Something tells me that working in cylindrical coordinate system might be the best for working with torus, but i am not sure. The good thing when we take the ring to be a torus , is that the volume charge density becomes a nice finite continuous function over the region of the torus, so we get rid of the infinity of the dirac delta function for the charge density which is apparent when we take the ring to be a circle.
I really appreciate the idea, and yes I worked for the magnetic field of a torus and cylindrical coordinates fits best (the phi component vanishes). I got this honor to give you the 1000th like, sir.
 
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  • #75
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I really appreciate the idea, and yes I worked for the magnetic field of a torus and cylindrical coordinates fits best (the phi component vanishes). I got this honor to give you the 1000th like, sir.
Thanks @Adesh :D, I was wondering the same thing not too long ago, that is who would be the person that will give me the 1000th like !
 
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