What is the tension in a charged ring?

[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.

 

[etotheipi]

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 ).

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?

 

[Vanadium 50]

Make the ring as small as you can. It is now indistinguishable from a point. Now use the field from a point.

 

[etotheipi]

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

where terms are defined in the paper http://www.mare.ee/indrek/ephi/efield_ring_of_charge.pdf

 

[Vanadium 50]

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.

 

[etotheipi]

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.

 

[Delta2]

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 don't 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?

 

[Vanadium 50]

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.

 

[etotheipi]

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.

 

[Vanadium 50]

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.

 

[etotheipi]

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 . 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!

 

[TSny]

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.

 

[haruspex]

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.

 

[Delta2]

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.

 

[Adesh]

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.

 

[Delta2]

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 !

 

[etotheipi]

In 3D, the field lines from the ring diverge into the third dimension, effectively becoming sparser.

Yes, I think you pinned the tail on the donkey there; when we perform the "shrink" operation on the ring, not all of the field lines in the plane will stay in the plane, but they'll migrate upward and downward to make the field approach a radial one.

And congratulations @Delta2 on 1000 upvotes! I think it's time for a physics party

 

[archaic]

https://www.quora.com/A-ring-has-un...of-tension-in-the-ring/answer/Raushan-Raj-225

 

[etotheipi]

@archaic it is a similar problem, but as far as I can tell they ignore the electric forces on that charge element from the rest of the ring, and only consider the electric force due to the central added charge. That seems a bit problematic, because the field from the rest of the ring becomes infinite at the charge element

 

[archaic]

@archaic it is a similar problem, but as far as I can tell they ignore the electric forces on that charge element from the rest of the ring, and only consider the electric force due to the central added charge. That seems a bit problematic, because the field from the rest of the ring becomes infinite at the charge element

hm i think that the way it is being thought about is that since at the beginning, before having the charge in the middle, the ring is at rest, or, in a way, the "internal" tension at one ring element is equal to the electric force on this element by the others. now we imagine a charge popping into the middle of the ring, this will cause it to expand isotropically until it comes to rest again. at this point, i guess that we could think of the same concept of internal tension (though with lower value this time, since, in some sense, the distance between the elements is larger) cancelling the electric force (which would also be lower in value) of the ring elements on the one we're looking at. though, this time, we have another force; that of the test charge. i guess, then, that in the new "internal tension" there's a component going against the charge's force, hence, in essence, we're calculating this tension.

 

[etotheipi]

I don't know if there's any expansion (because then we'd also have to re-evaluate the force from the rest of the ring), but what they could be doing is looking at the equilibrium case, putting $q$ in the centre and looking at the increment of tension (so they're defining $T = T_{real} - T_0$). That might work, but it still seems a bit strange because you go from one infinite tension to another infinite tension

 

[Keith_McClary]

To compress a continuous charge distribution into a point or a line (segment) requires infinite energy. (A 2D surface can be done with finite energy.) The tension is proportional to the difference in energy between the ring and an infinitesimally stretched ring. I think it must be infinite.

 

[Adesh]

A 2D surface can be done with finite energy.)

Can you please explain that? I’m having trouble with in imaging that.

 

[etotheipi]

Can you please explain that? I’m having trouble with in imaging that.

I might well be wrong, but I think the self energy of a configuration of charges is the integral of the energy density $\frac{1}{2} \epsilon_0 E^2$ through all of space. If the electric field diverges at some points, e.g. at a line of charge, then this will become infinite. But AFAIK the electric field is still finite at a plane of charge, so the energy density doesn't diverge anywhere and we can calculate a finite value for the self energy.

 

[Adesh]

But AFAIK the electric field is still finite at a plane of charge

Is electric field finite at some point on a continuous sheet of charge?

 

[haruspex]

Is electric field finite at some point on a continuous sheet of charge?

The infinities do not arise with charged surfaces, which is just as well as these occur in reality.

 

[Adesh]

The infinities do not arise with charged surfaces, which is just as well as these occur in reality.

Okay for charged surfaces infinities (doing a sin to mathematics) cancel each other out?

 

[haruspex]

Okay for charges surfaces infinities (doing a sin to mathematics) cancel each other out?

Sorry, can't parse that.

 

[Adesh]

Sorry, can't parse that.

If I have a continuous sheet of a charge in $xy$ plane, will electric field at a point on the sheet be finite? If yes then it means that infinities cancel each other out, infinities are arising because charges are so close to each other therefore the $r$ in the denominator is becoming very close to zero.

 

[haruspex]

If I have a continuous sheet of a charge in $xy$ plane, will electric field at a point on the sheet be finite? If yes then it means that infinities cancel each other out, infinities are arising because charges are so close to each other therefore the $r$ in the denominator is becoming very close to zero.

But these are not point charges. You have a finite charge per unit area. If you follow field lines back to the sheet from some point outside they they maintain their spacing, whereas doing that for a line or point charge they become crammed together.
The field from an infinite flat sheet of charge is constant throughout each half space. There is a discontinuity in passing through the sheet, but it remains bounded.

 

[Paul Colby]

The following may be fun to consider / discuss. I'll give it a try.

For what it is worth, the potential of a point charge is an analytic function of the charge location. For example one may write the potential of a charge, $2q$, located at $(x_c,y_c,z_c)$ as

$\Phi(x,y,z) = \frac{-2q}{\sqrt{(x-x_c)^2+(y-y_c)^2+(z-z_c)^2}}$.​

Now, $x_c$, $y_c$ and $z_c$ are simply real parameters. Nothing in principle prevents us from locating our charge at complex locations. Well, except the potential becomes complex as does the resulting electric field. We can fix this by splitting $2q$ into two and placing $q$ at, $z_c = iR$, and the other half at, $z_c = -iR$, where $R$ is a real parameter equal to the charge ring radius discussed previously. The new potential is,

$\Phi(x,y,z) = \frac{q}{\sqrt{x^2+y^2+(z-iR)^2}}+\frac{q}{\sqrt{x^2+y^2+(z+iR)^2}}$.​

Okay, so our new potential is everywhere real for real $(x,y,z)$ real as are the components of the electric field. It also obeys the required equation,

$\nabla^2\Phi = 0$,​

at every point it's defined at which I believe is everywhere except on the ring, $x^2+y^2=R^2$ in the $z=0$ plane. I also believe (but can't prove) that the potential is zero everywhere within the disk of which the ring forms the edge if one treats branch cuts appropriately.

 

[Keith_McClary]

There is a solution by Martin Gales based on the change in electrostatic energy when the radius is changed.

 

[haruspex]

There is a solution by Martin Gales based on the change in electrostatic energy when the radius is changed.

That is a solution to the valid case of a wire with non infinitesimal thickness. The original question in this thread took it to be infinitesimal.

 

[etotheipi]

Yeah, if the ring is toroidal and we have some way of finding the capacitance $C=\frac{\pi R}{ln (\frac{8R}{r})}$ by messy calculation, looking it up, or some other way, then it's not too bad I suppose.

However for a ring of infinitesimal thickness, I think it was concluded that the tension has to diverge, not least because itself energy also diverges. @TSny also gave a very nice analysis a little earlier on, considering the force between two charged arcs, and showing the logarithmic divergence as the gap between them shrinks.

 

[Paul Colby]

I also believe (but can't prove) that the potential is zero everywhere within the disk of which the ring forms the edge if one treats branch cuts appropriately.

Yeah, I can now prove this. We define a complex function on real three space, $\omega(x,y,z)$, given by,

$\omega(x,y,z) = (x^2+y^2+(z + i R)^2)^{-1}$.​

With this our potential (second equation in #90) becomes,

$\Phi(x,y,z) = q(\sqrt{\omega(x,y,z)} +\sqrt{\omega^\ast(x,y,z)})$. (*)​

The important thing to be gleaned from this expression is that given any path in real three space, $(x(t),y(t),z(t))$, the $\omega$ function and its conjugate, $\omega^\ast$, provides two paths in the $\omega$ plane that are conjugates of each other. Any path which terminates on a branch of the first term of (*) implies that the conjugate path taken in the second term will terminate on the opposite branch thus having the opposite sign. Therefore, the potential is zero at every point on the branch or inside the disk bounded by the ring. This implies $E_x$ and $E_y$ are zero also. The third component, $E_z$, is zero because $\Phi$ is an even function of $z$.

So, what is provided by (*) is in fact a closed form expression for the field produced by a ring-source. No integrals required.

 

[Paul Colby]

Yeah, I can now prove this.

Just to close out my suggestion it seems there is a problem with what I'm suggesting. The "solution" posted in #90 etc is indeed a ring-source of some kind however it is definitely not a simple ring of charge. The potential at the ring center is easily shown to be zero. For an actual charge ring, the work bringing a test charge in from infinity can't be zero as claimed in #94. This is corroborated by the exact expression in terms of complete elliptic integrals. To add to the confusion in the far field using the divergence theorem the net charge is indeed what one wants. To add even more obscure issues, following a path in real space around the ring flips the sign of the total charge. One might expect this to be due to a branch discontinuity in either the potential or it's derivatives which, since everything vanishes on the disk bounded by the ring, has no apparent jumps. I'm completely baffled by this.

 

[pablochocobar]

Homework Statement: The ring I'll assume to have negligible thickness, and linear charge density $\lambda$, as well as a radius $R$.
Relevant Equations: N/A

I tried considering a little piece of the ring (shaded black below) subtending angle $d\theta$, and attempted to find the electric field in the vicinity of that piece by a summation of contributions from the rest of the ring:

View attachment 265375

$$dE_x = \frac{dq}{4\pi \epsilon_0 d^2} \cos{\phi} = \frac{\lambda R d\theta}{4\pi \epsilon_0 \cdot 2R^2(1-\cos{\theta})}\cos{\phi}$$ $$E_x = \int_0^{2\pi} \frac{\lambda \sin{\frac{\theta}{2}}}{8\pi \epsilon_0 R(1-\cos{\theta})} d\theta = \frac{\lambda}{8\pi \epsilon_0 R} \int_0^{2\pi} \frac{\sin{\frac{\theta}{2}}}{1-\cos{\theta}} d\theta $$ Problem is, that thing diverges because the denominator goes to zero at the boundaries (the indefinite is easy enough to solve by changing $1-\cos{\theta} = 2\sin^2{\frac{\theta}{2}}$). If we call $$I = \int_{0}^{2\pi} \frac{\sin{\frac{\theta}{2}}}{1-\cos{\theta}} d\theta$$ then the electrostatic force on the piece in terms of this is $$F_e = (\lambda R d\theta) \frac{\lambda I}{8\pi \epsilon_0 R} = 2T\sin{\frac{d\theta}{2}} \approx T d\theta$$ and the tension would be, under the usual equilibrium constraint, $$T = \frac{\lambda^2 I}{8\pi \epsilon_0} = \frac{q^2 I}{32 \pi^3 R^2 \epsilon_0}$$but that doesn't make much sense if $I$ is infinite. I wondered if anyone could help out? Perhaps it is the case that the electric field diverges at the ring, but in that case, how could I modify my calculations to get a sensible answer? Because I am fairly sure that in reality the tension will not be infinite . Thanks!

First I am trying to calculate the field due to any circular arc which subtends angle $\alpha$.

For any sector which subtends same angle the Electric Field will be same. Ie.$| \vec E_1 | = | \vec E_2 |$ Because charge is distributed symmetrically.​

DERIVATION :

$GC\perp{AO}$
$OG = R$, $OC = R\cos\theta$
$OA = 2OC = 2R\cos\theta$The charge on the small arc subtend by angle ${d\theta}$ is, $$dq = \lambda2R\cos\theta d\theta$$
where $\lambda$ is charge per unit length = $\frac{Q}{2\pi{R}}$.
Electric Field due to ${dq}$ is, $$dE = \frac{kdq}{(2R\cos\theta)^2}$$

For a symmetric part OB which subtends same angle $d{\theta}$ the Y-component of electric field $dE_Y$ will cancel.

∴ Net Electric Field due to a arc of angle $\alpha$ at O, $$E_{net} = \int_{-\frac{\alpha}{2}}^\frac{\alpha}{2} dE\cos\theta$$ $$= \int_{-\frac{\alpha}{2}}^\frac{\alpha}{2} \frac{kdq\cos\theta}{(2R\cos\theta)^2}$$ $$= \int_{-\frac{\alpha}{2}}^\frac{\alpha}{2} \frac{k\lambda2R\cos\theta{d\theta}\cos\theta}{(2R\cos\theta)^2}$$ $$= \int_{-\frac{\alpha}{2}}^\frac{\alpha}{2} \frac{k\lambda}{2R}d\theta$$ $$= \frac{k\lambda}{2R}\left[ \theta \right]_{-\frac{\alpha}{2}}^{\frac{\alpha}{2}}$$ $$ = \frac{k\lambda\alpha}{2R}$$

Now for $\alpha=2\pi$ the Electric field will be due to all the charges at any point on the ring.
$$E_{at~any~point~on~ring} = \frac{k\lambda\pi}{R}$$

charge on the marked section = $\lambda2R2d\theta$ and E-field on marked section = $\frac{k\lambda\pi}{R}$

$$2T\sin{d\theta} = \frac{k\lambda\pi}{R}\lambda2R2d\theta$$

∵$d\theta$ is too small, ∴$\sin{d\theta}\approx{d\theta}$

$$T=\frac{Q^2}{8\pi^2\varepsilon{R}^2}$$

 

[Paul Colby]

I’d suggest fattening the ring up ever so slightly, view it as a wire of cross sectional radius $d$ where $d<<R$. Then take the limit $d\rightarrow 0$.

first rate solution. Well done.

 

[pablochocobar]

I’d suggest fattening the ring up ever so slightly, view it as a wire of cross sectional radius $d$ where $d<<R$. Then take the limit $d\rightarrow 0$.

first rate solution. Well done.

You talking about my solution?

 

[Paul Colby]

Well, the first comment was directed at the quoted text in #96. Then I looked through your attachment where a very reasonable solution is arrived at.

 

[TSny]

The charge on the small arc subtend by angle ${d\theta}$ is, $$dq = \lambda2R\cos\theta d\theta$$

This is not the correct expression for $dq$. If you integrate your expression for $dq$ over the ring, you will see that you don't get the correct total charge.

$2R\cos\theta d\theta$ is not the infinitesimal arc length of the ring corresponding to $d \theta$ in your figure. Swinging $2R\cos \theta$ through $d\theta$ gives an infinitesimal arc length that is "perpendicular" to the line segment $2R\cos \theta$. But this arc length is not along the ring. It is not hard to show that the infinitesimal arc length along the ring corresponding to $d\theta$ is $2R\cos \theta d \theta \cdot \large \frac 1 {\cos \theta}$ $= 2R d \theta$.

If you correct this and then go on to find $E_{net}$, you'll find that $E_{net}$ diverges. This was shown in the first post of the thread.