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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?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?Go back to counting field lines.
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.Am I right in thinking you have interpreted this as a problem in a 2D universe, effectively turning the ring into an infinite cylinder?
Gauss' law is just an application of divergence theorem in the case of electromagnetism, that isThis 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.
The divergence theorem holds in any number of dimensions. In 2D the closed curve becomes the surface.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.
I would really like to know if there exists some references for that claim.The divergence theorem holds in any number of dimensions. In 2D the closed curve becomes the surface.
From Wikipedia:I would really like to know if there exists some references for that claim.
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.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.
I think ##\oint \mathbf E \cdot dl ## will not be a finite quantity.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.
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.I think ##\oint \mathbf E \cdot dl ## will not be a finite quantity.
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.Am I right in thinking you have interpreted this as a problem in a 2D universe, effectively turning the ring into an infinite cylinder?
Well, I meant our contour is the loop itself and thats where field is not finite.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.
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 ).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.
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 integralsMake the ring as small as you can. It is now indistinguishable from a point. Now use the field from a point.
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.What about this; I found a reference
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.If you read the paper, you would see that he is discussing the field in the XZ plane, not the XY plane.
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.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.
How do we call the center of the circle that corresponds to the ring?There is no "center" to a ring of zero thickness (and/or height).
I am working only in the XY plane. I say it over and over and over again.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.
Are you talking about at each side of the ring? I want an answer so I pick the side that's finite.The field is infinite at the ring in one case and finite at the other case
I am working only in the XY plane. I say it over and over and over again.
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.However, I am working strictly in a plane.
Which is what I explicitly said I was doing in #59.As such, we must use Maxwell's equations in 3D.
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.So it is impossible to "strictly work in a plane"
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.
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.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!
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.The problem says to neglect the thickness, so I am working with a zero thickness line
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.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.
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 !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.