What is the tension in a charged ring?

[Vanadium 50]

Okay so our ring is a spring, if it's natural length is zero (?)

Natural length is irrelevant. I only look at energy differences. Don't make the problem harder than it has to be.

however here the tension forces do not act in the radial direction

Did I talk about directions at all? You have a spring of one length and I change it's length, so I know how much energy it took. Don't make the problem harder than it has to be.

But here there is no sufficient spherical symmetry for the shell theorem to apply

You aren't going to look at the proof, are you? You will see the proof relies on a lemma involving rings.

But then I thought you replaced the ring with a single point charge,

When in 3D you use the shell theorem on a sphere, does its energy become infinite?

You want to use forces to solve this problem. You should energies.

 

[etotheipi]

Natural length is irrelevant. I only look at energy differences. Don't make the problem harder than it has to be.

I think the energy of a spring is $\frac{1}{2}k \delta^2 = \frac{1}{2}k(l-l_0)^2$ so for an infinitesimal change in length $dE = k(l-l_0)dl = k(2\pi r - l_0) \cdot 2\pi dr = 4\pi^2 k r dr - 2 \pi k l_0 dr$. The second term is dependent on the natural length, but you have dropped it.

Did I talk about directions at all? You have a spring of one length and I change it's length, so I know how much energy it took. Don't make the problem harder than it has to be.

Then how did you calculate the tension? For a straight spring being extended, we could say that $T dR = dE$ if $dR$ is the displacement of the end of the spring, however in this scenario the tension is not obtained by dividing the increment in energy by the change in radius, because the tension forces doing the work are not in the radial direction?

You aren't going to look at the proof, are you? You will see the proof relies on a lemma involving rings.

I have proved it myself earlier, and I did so by integrating up the electric field due to successive rings in the axial direction of those rings. What lemma are you referring to?

When in 3D you use the shell theorem on a sphere, does its energy become infinite?

You want to use forces to solve this problem. You should energies.

I don't know what you mean; the self energy of a sphere can be obtained by integrating $\int \frac{1}{2}\epsilon_0 E^2 dV$ and it is not infinite.

I have thought about using energies but I cannot see a good way of doing so .

 

[Vanadium 50]

The second term is dependent on the natural length, but you have dropped it.

True, but what does that term mean? It means that if there were no charge, that the ring was already under tension (or compression). The way I have defined it, the tension is zero if the charge density is zero. Had I got an answer like T = a + bλ² (you see why it needs to be squared, right?) I would associate a with the initial tension. You shouldn't add in an extra "intrinsic" tension (or compression). Don't make this harder than it has to be.

What lemma are you referring to?

The field contribution of each ring. Which is (in the plane of the ring) $[\frac{1}{4\pi\epsilon_0}]\frac{q}{r^2}$ outside, Directed radially. Inside the ring there is a divergence at r=R, so that can't be a solution. Don't make this harder than it has to be.

 

[etotheipi]

The field contribution of each ring. Which is (in the plane of the ring) $[\frac{1}{4\pi\epsilon_0}]\frac{q}{r^2}$ outside, Directed radially. Inside the ring there is a divergence at r=R, so that can't be a solution. Don't make this harder than it has to be.

Please excuse me if I make a mistake, but I constructed the contribution to the axial electric field from anyone ring as $$dE_r = \frac{Q}{16 \pi \epsilon_0 r^2 R} \frac{r^2 + c^2 - R^2}{c} dc$$ where $c$ is the length of the arm from any point on the ring to the place where we are evaluating the electric field. You say your expression is for the plane of the ring, but I don't know why you need this for the shell theorem.

(N.B. The whole result follows much more nicely from the theorem of Gauss, though)

But nonetheless, in the plane of the ring, outside the ring, I don't think that the field strength is $\frac{q}{4\pi \epsilon_0 r^2}$. How did you obtain that result?

 

[Vanadium 50]

How did you obtain that result?

Draw the field lines (which will be radial) outside the ring. Make the ring smaller. Do I have any more field lines? If so, where did they come from? Do I have less? If so, where did they go? Continue until the ring is a point.

 

[etotheipi]

Draw the field lines (which will be radial) outside the ring. Make the ring smaller. Do I have any more field lines? If so, where did they come from? Do I have less? If so, where did they go? Continue until the ring is a point.

It works for a 3D sphere, yes, but not for a 2D ring. There is not the necessary symmetry of field lines, we cannot yet tell whether the field strength is uniform in all directions.

For a 3D sphere, spherical symmetry implies uniform density of field lines emanating from the sphere. Theorem of Gauss tells me that, with a suitable spherical domain of integration $$\int_S \vec{E} \cdot d\vec{A} = 4\pi r^2 E= \frac{Q}{\epsilon_0}$$ and hence $E = \frac{Q}{4\pi \epsilon_0 r^2}$. But now let us consider a 2D ring. What surface are you integrating over? What symmetry do you invoke to ensure uniformity of field lines over this surface?

 

[Vanadium 50]

DRAW THE FIELD LINES.

Don't just tell me it doesn't work so you don't have to draw them. I'm trying to teach you something here, but I can't if you don't do anything I suggest. But I give up. Do it whatever way you want.

 

[etotheipi]

To the best of my knowledge, here is what I imagine the electric field to look like

black denotes the plane of the ring, red are a few extra outside of the plane of the ring. I missed out a few in the axial directions because I thought it would make the diagram unnecessarily hard to read. Now I construct an arbitrary control volume $V$ that encloses the ring. I can use the integral form of Gauss' law again, $$\int_S \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$$but I don't know how you can extract $E$ from that, because there is only a radial symmetry, and not a spherical symmetry like in the previous case.

I apologise if I upset you @Vanadium 50, but I can assure you I ask out of interest and not stubbornness . I appreciate your help. After all, you have the PhD and I do not

 

[TSny]

If the tension in the ring is finite, then the tension is equal to 1/2 of the force $F$ that one half of the ring exerts on the other half. For the case where the thickness of the wire (or thread) is zero, the tension will not be finite. But, I thought it would be interesting to put a couple of gaps in the ring as shown below and consider how the force $F$ varies with the size of the gap, $S$. $S$ is the arc length of each missing piece of the ring.

The force $F$ can be found by integration and I got the following result. For convenience, let $z$ denote the size of the gap in relation to the circumference of the ring: $z = \frac{S}{2 \pi R}$.

$$F = 2k\lambda^2 \left[ \cos(\pi z) \log \left[ \cot \left(\frac{\pi z}{2} \right) \right] + \sin (\pi z) - 1 \right ]$$
$k$ is Coulomb's constant. I think this is valid for any nonzero gap size; that is, for any $z$ in the range $0<z<\frac{1}{2}$. (As $z$ approaches $\frac{1}{2}$, the gaps consume the whole ring.)

Here are some numbers for a ring of circumference 1 meter and $k \lambda^2=1$N. (So, $\lambda$ is about 10 μC/m.) The gap is varied from 10 cm to 1 × 10^-13 m, which is much smaller than the size of an atom!

The force increases surprisingly slowly as the gap size decreases. For $z << 1$, the formula for $F$ can be approximated as $$F = 2 k \lambda^2 \left[ \log \left(\frac{2}{\pi z} \right) - 1 \right]$$

This shows how the force diverges logarithmically as the gaps shrink to zero.

 

[etotheipi]

The force increases surprisingly slowly as the gap size decreases. For $z << 1$, the formula for $F$ can be approximated as $$F = 2 k \lambda^2 \left[ \log \left(\frac{2}{\pi z} \right) - 1 \right]$$This shows how the force diverges logarithmically as the gaps shrink to zero.

This is very cool, thanks a bunch for the analysis! It's quite fun to play around with the formulae you provided.

I think I am probably quite (very) out of my depth to replicate the result, but I will have a go .

Thanks!

 

[haruspex]

If you don't want to follow the proof, here's the heuristic. In the plane of the ring, field lines are radial by symmetry. I can always replace the ring by one a factor of 2 (or 10, or a million) smaller without changing the field strength anywhere it wasn't zero. (Obviously points that were inside and are now outside see different fields). Do that as many times as you want and you get your ring arbitrarily close to a point.

That only says it is radial, so yes of course it is equivalent to some charge at the central point. I did not dispute that, and I would hope it was obvious to us all. The question is, what charge at the central point? Certainly not the total charge on the ring, as you seemed to imply by referring to the shell theorem without qualification.

Edit: indeed, looking back, I see you specifically stated as much:

The total charge is 2πRλ, Because of the shell theorem, I can treat this as coming from a single point at the center of the loop.

In particular, is it even finite? Everyone else in the thread was coming to the view that it probably was not, and I see @TSny has now confirmed that.

So to get a useful answer to the original question, the ring must be modeled as a torus, at least locally to the "test charge" point.

 

[etotheipi]

That only says it is radial, so yes of course it is equivalent to some charge at the central point.

But it is only equivalent in the plane, right? As far as I can tell there's no telling whether the field above the ring is also like that from a point charge, but my spidey-senses tell me that it's probably not. And that is why we can't deduce even the electric field outside the ring in the plane of the ring by heuristic arguments alone, because the problem lacks spherical symmetry.

I haven't made much progress with the torus today, but I will try harder tomorrow

 

[haruspex]

But it is only equivalent in the plane, right? As far as I can tell there's no telling whether the field above the ring is also like that from a point charge, but my spidey-senses tell me that it's probably not. And that is why we can't deduce even the electric field outside the ring in the plane of the ring by heuristic arguments alone, because the problem lacks spherical symmetry.

Agreed. It will not even seem like a point source when sensed across some small region. What I mean by that is that since the equipotential surfaces are not spheres they will locally have two different radii.
There's an interesting question... at a point on the ring, what is the curvature of the equipotential in a plane normal to the ring and through its centre?

 

[etotheipi]

I downloaded this wolfram project and apparently this is what the electric field looks like around such a charged ring:

It looks like we can vaguely imagine various equipotential surfaces (easier on the left diagram) by tracing through perpendicular to the field lines, as a first "approximation" .

 

[haruspex]

I downloaded this wolfram project and apparently this is what the electric field looks like around such a charged ring:

View attachment 265446View attachment 265447

It looks like we can vaguely imagine various equipotential surfaces (easier on the left diagram) by tracing through perpendicular to the field lines, as a first "approximation" .

Yes, it dawned on me that in the vicinity of the ring the equipotentials will be toroidal shells around the ring.

 

[Vanadium 50]

@etotheipi asked me to come back. Let me make a few comments:

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." That reads more into the problem than is there. Further, they are discovering infinities when they do this. There are. The same infinities come up with any "line of charge" problem. Indeed, these infinities even come up when imagining a test charge as a tiny sphere of charge Q and letting the sphere radius go to zero. So to solve this, we take a "line of charge of charge density λ" at its word and...

It is a mistake to set up a problem so that the very elements of that problem are infinite.

So, I'm am taking the problem at its word - "negligible thickness" means treat this as a 2D problem, not as a 3D problem where I do not neglect the thickness. If you want to complain that this is not physically realizable, you can get in line with people complaining about frictionless surfaces, stretchless ropes, perfect spheres, etc.

Next, let's take a look at the picture in #38. The red lines are out of the plane and therefore irrelevant. I count six field lines in the plane. At 2R, 3R, 100R, I always have six lines. Everywhere outside the ring, the field configuration configuration looks like all of the charge is concentrated at a tiny point in the center of the ring.

Why take the outside field? Because it is finite, and presumably the author of the problem is looking for a finite answer.

OK, so what is the potential energy of this configuration? Let's look at a tiny slice of wire - say 1/1000th of it. It sees the field as if all the charge were at the center, and it's R units away. So it's [1/1000][1/4πε₀][Q²/R]. There are 1000 such pieces, so the total is [1/4πε₀][Q²/R]. A million pieces? Same answer. A trillion? Same answer.

So the energy change of moving the circle from R to R + ΔR is [1/4πε₀][(2πλ)²ΔR/R²]

The energy it takes to stretch a spring from R to ΔR is 4π²kRΔR.

Equating,

4\pi^2kR\Delta R = \frac{1}{4\pi\epsilon_0}\frac{(2\pi R \lambda)^2}{R^2}\Delta R

or

k = \frac{1}{4\pi\epsilon_0}\frac{\lambda^2}{R}

and so (if I didn't make an algebra error)

T = 4\pi^2 k R = \frac{\pi}{\epsilon_0}\lambda^2

 

[haruspex]

It sees the field as if all the charge were at the center

Please explain how you arrive at that. We all accept that by symmetry the net field is radial, but do not understand how you get that it is equivalent to the actual charge on the ring relocated to the centre (as it would be for a sphere).
In particular, you seem by this to magic away an infinity (or, equivalently, a critical dependence on the wire thickness) which provably exists.

 

[Delta2]

It sees the field as if all the charge were at the center

First of all only the field at the plane of the ring looks like as all the charge was at the center of the ring, but then again this is not accurate because we know that the field at the infinitesimally thin ring becomes infinite, while if we take the charge to be concentrated at the center, then the field at the thin ring is not infinite but has a finite value, it is right at the center that the field becomes infinite. So at first glance you move the infinity from the place of the thin ring to the place of center of the ring, so the two configurations do not seem to be equivalent (as they are equivalent in the case of a charged sphere with surface charge density, because we have spherical symmetry there).

 

[Vanadium 50]

but do not understand how you get that it is equivalent to the actual charge on the ring relocated to the centre (as it would be for a sphere).

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.

while if we take the charge to be concentrated at the center,

There is no "center" to a ring of zero thickness (and/or height).

 

[Vanadium 50]

equivalently, a critical dependence on the wire thickness) which provably exists.

Yup. And why am I doing this? Because the problem said to.

 

[haruspex]

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?

 

[etotheipi]

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.

 

[Adesh]

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.

 

[etotheipi]

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.

 

[Adesh]

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.

 

[etotheipi]

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

 

[Adesh]

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.

 

[etotheipi]

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.

 

[Vanadium 50]

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.

 

[Adesh]

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 that's where field is not finite.