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I calculated the electric potential at any point (assuming zero potential at infinity) on the circumference of a uniformly charged ring (non conducting) and after evaluating that integral it turns out to be infinite!

But in case of a uniformly charged non conducting disc the potential (at the edge of the disc) is a finite value.

I want to learn that why this difference occurs (of course that integral tells me the difference, but can someone give me a qualitative reason why potential turns out to be infinite in case of a ring but finite in case of disc?)

EDIT:I can show how I evaluated the potential at circumference of ringenter image description here

I also calculated the electric field at the edge of disc and ring, in both cases the field turns out to be infinity. I'm surprised that in case of disc the potential at edge is finite value but the electric field is INFINITY! How is this possible ? If I have infinite electric field in a region then its potential should be infinite because if I'd take a test charge and come very close to that region (adding up the work performed by external force to get the value of potential) then I would have to apply large external force to place the charge at that point! How can a point have infinite field and finite potential it seems highly counterintuitive to me!

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    $\begingroup$ How do you arrive at your first equation? (Please type equations in your question instead of inserting a picture) $\endgroup$ – Crimson Apr 26 at 22:21
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    $\begingroup$ Possible duplicate of Is the electric field at the edge of a uniformly charged disk infinite? $\endgroup$ – Crimson Apr 26 at 22:38
  • $\begingroup$ @Crimson Since the arc subtends the angle 2(theta) at centre so the differential charge is easily expressed in terms of integration variable (as seen in numerator of the fraction of the picture) also note that if we join the extremeties of diameter to the point which has the differential charge then we would get a right angled trangle (diameter of circle subtends right angle at circumference) giving the distance 2R cos(theta) $\endgroup$ – Shivansh J Apr 27 at 8:01
  • $\begingroup$ @Crimson I'm not able to understand the answer of duplicate question. Sorry but seems I haven't been told anything about the meaning of terms like 'log divergence' 'antiholomoephic functions' and integration of complex numbers in my school. $\endgroup$ – Shivansh J Apr 27 at 8:11
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Here is a non-rigorous qualitative argument: both point charges and one-dimensional charges can be enclosed by a surface that is arbitrarily small, while for a two- or three-dimensional charge there is a minimum size for such a surface. This means that Gauss' law requires the electric field strength to be infinite in the limit when the surface area $\rightarrow$ 0 for point charges and one-dimensional charges.

Specifically: For a ring-shaped charge you can enclose it in a torus, which can be made arbitrarily thin, and thus have a surface area arbitrarily close to $0$. For a disc you can enclose it in a cylinder, but the cylinder must have a surface area that is at least $2\pi R^2$ in order to cover the top and bottom of the disc.

I know the question is about the potential and not the field strength, and I realize that the field strength can be infinite without having an infinite potential, but the point of this argument is that there is a qualitative difference between charges that are distributed over 0 or 1 dimensions versus 2 or 3.

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