- #36
vcsharp2003
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Delta2 said:Nope I don't think there is an analytical solution, cause wolfram can't find it either.
Did you mean that there is no method available to solve the integral?
Delta2 said:Nope I don't think there is an analytical solution, cause wolfram can't find it either.
Let me consider a point P', radially opposed to P. If I applied the same reasoning, I would conclude the field at P' points radially inward. Therefore, we should have a negative charge (sink) at the center but I can't see that charge.vcsharp2003 said:From above argument we can conclude that electric field at P is not zero but pointing radially inward.
Gordianus said:Let me consider a point P', radially opposed to P. If I applied the same reasoning, I would conclude the field at P' points radially inward. Therefore, we should have a negative charge (sink) at the center but I can't see that charge.
By convention, field lines start at positive charges and end at negative charges. If I understood your reasoning correctly, at any point inside the ring (except the center) the field points radially inward. Thus, we should have a negative charge at the center.vcsharp2003 said:I am not sure if your reasoning is correct. If it's correct, then a positive charge has electric lines of force emanating from it which must be directed towards a negative charge. We know this is not true. A positive charge can exist by itself and have its own electric field independent of a negative charge being paired with it.
How about electric field due to an isolated positive charge?Gordianus said:By convention, field lines start at positive charges and end at negative charges
I'd say there are no isolated positive charges. Negative charges must be somewhere, perhaps very far away.vcsharp2003 said:How about electric field due to an isolated positive charge?
Do you have a reference for that?Delta2 said:This is one of the cases where wikipedia (and you) are wrong. The definition I know requires invariance both in ##z## and ##\phi##. Anyway I think you have to agree with me that this cylindrical symmetry you propose is of no use with Gauss's law in integral form.
Gordianus said:I'd say there are no isolated positive charges. Negative charges must be somewhere, perhaps very far away.
vcsharp2003 said:My view on this is that there is a 3 dimensional electric field existing due to the ring and the lines of force from the ring will spread in a 3 dimensional manner.
vcsharp2003 said:They will bend outward towards the ring's axis. Sort of like a funnel pattern on each side of the ring.
Check post #23.hutchphd said:So please explain to me what it is that you "know".
I taught the undergraduate EM course from Griffiths and still have no idea what you are talking about. In particular what is the difference between axial and cylindrical symmetry ? A few sentences should suffice.
Post #23 says that the cylindrical symmetry I know has invariance with ##z## as well with ##\phi##. You know it only with invariance to ##\phi##. If that's objectively nothing and doesn't explain my objection then ok, I would say that your eyes can't see what you don't want to read.hutchphd said:Post #23 still says objectively nothing. If you cannot even explain your objection, I shall value your opinion appropriately.
yours is not the customary definition of cylindrical symmetry. You are welcome to use your own idiosyncratic definitions. You are not welcome to define the rest of the community as incorrect.haruspex said:Apart from one related to GR, all the items I could find online regard cylindrical, circular, azimuthal and axial symmetry as interchangeable terms.
It might not be the customary definition, but in my opinion the rest of the community is incorrect and the fact that they use many different names for it (axial, azimuthal, cylindrical, circular) proves that the community doesn't know what they talk about.hutchphd said:yours is not the customary definition of cylindrical symmetry. You are welcome to use your own idiosyncratic definitions. You are not welcome to define the rest of the community as incorrect.
Steve4Physics said:Here's a decent video (though a bit long at 13mins) deriving the field at a point inside a charged ring...
Not sure how to interpret "yours" in the above. You are addressing Delta2, yes? In the post of mine you quote, I was saying nearly all the online references I could find agree with you.hutchphd said:I did not understand what you meant. Your definition of cylindrical symmetry is extraordinarilly restrictive: only infinitely long right cylinders are included. As mentioned above
yours is not the customary definition of cylindrical symmetry. You are welcome to use your own idiosyncratic definitions. You are not welcome to define the rest of the community as incorrect.
Good to clear this up.
Yes of course he is addressing me, he is just using you as his lawyer :P.haruspex said:Not sure how to interpret "yours" in the above. You are addressing Delta2, yes? In the post of mine you quote, I was saying nearly all the online references I could find agree with you.
Research assistant?Delta2 said:Yes of course he is addressing me, he is just using you as his lawyer :P.
That's a good approach, but it would not be quite enough to show that the same choice of pairing does not cancel; you would need to show that for this or some other pairing the net field is always the same way.bob012345 said:Then show for the 2D case of a charged ring they don't cancel thus avoiding messy integrations.
Thanks, so I take it that's a confirmation that pairings do cancel for the sphere. In either case, the pairings would always be general not based on specific angles.haruspex said:That's a good approach, but it would not be quite enough to show that the same choice of pairing does not cancel; you would need to show that for this or some other pairing the net field is always the same way.
It must be the case that well chosen pairings cancel for the sphere, since we know the net field is zero. I have not checked whether this is true of the pairing you illustrate.bob012345 said:Thanks, so I take it that's a confirmation that pairings do cancel for the sphere.
Not sure what you mean by that. You would need a relationship between the angles, ##\phi=f(\theta)##, such that the band ##(\theta, \theta+\delta\theta)## cancels the band ##(f(\theta), f(\theta+\delta\theta))## .bob012345 said:In either case, the pairings would always be general not based on specific angles.
haruspex said:It must be the case that well chosen pairings cancel for the sphere, since we know the net field is zero. I have not checked whether this is true of the pairing you illustrate.
Not sure what you mean by that. You would need a relationship between the angles, ##\phi=f(\theta)##, such that the band ##(\theta, \theta+\delta\theta)## cancels the band ##(f(\theta), f(\theta+\delta\theta))## .
Call angle BPO ##\alpha##.bob012345 said:View attachment 288165
The element from the right side is proportional to ##\Large \frac {R^2 sin(\gamma) d\gamma}{s^2}## and from the left side ##\Large \frac {R^2 sin(\beta) d\beta}{s'^2}## but in opposite directions along the axis. I know that ## sin(\beta) = {\Large \frac{s'}{s }}sin(\gamma)## and I believe ## d\beta = {\Large \frac{s'}{s }}d\gamma## making the two differential elements cancel at all angles. But I am having difficulty proving the relationship between ##d\beta## and ##d\gamma##. It is easy to see when the angles are very small and when ##s## and ##s'## are vertical but not easy to see in between. I did precision numerical calculations however to verify that the relation ## d\beta = {\Large \frac{s'}{s }}d\gamma## holds for wiggles around any angle. All that just makes
$$ \large \frac {R^2 sin(\beta) d\beta}{s'^2} → \large \frac {R^2 sin(\gamma) d\gamma}{s^2}$$
Thanks. I finally understand your derivation for ##d\beta\ = \Large\frac{s'}{s} \large d\gamma##. The difference between ##B## and ##B'## and ##C## and ##C'## is greatly exaggerated.haruspex said:Call angle BPO ##\alpha##.
Angle PBO= Angle PCO=##\gamma-\alpha##.
##s\delta\alpha=R\delta\gamma\cos(\gamma-\alpha)##
##s'\delta\alpha=R\delta\beta\cos(\gamma-\alpha)##.
I think that would involve elliptical integrals which are best left alone here.haruspex said:You could try writing down the integral for the net field at P and look for extrema as the distance of P from the centre of the hoop varies.
Since only extrema are of interest, it might not be necessary to solve the integral.rude man said:I think that would involve elliptical integrals which are best left alone here.
Since the OP only asked if it is zero or not at one point along the radius, you would not have to actually do any integrals but just set the problem up. I found the formula worked out here*. The answer does involve Elliptical integrals of the first and second kind. If it wasn't enough to just look at the formula and see it is not zero, one can evaluate the elliptical integrals by lookup table at that point and see the value is not zero. I still think none of this is necessary based on the previous strategy.haruspex said:Since only extrema are of interest, it might not be necessary to solve the integral.