Question
Asked 18th Mar, 2022

Will the surface charge density be uniform?If not, where will it be minimum?

A thin, circular disc of radius R is made up of a conducting material. A charge Q is given to it, which spreads on the two surfaces.
Will the surface charge density be uniform? If not, where will it be minimum?

Most recent answer

30th Mar, 2022
George Dishman
Thales Group, UK
Thank you Joerg Fricke, those are excellent resources.

All Answers (10)

18th Mar, 2022
Elias Sadeghi Malvajerdi
Shahid Beheshti University
Certainly the surface charge density is not uniform. The surface charge density is highest at the margins and lowest at the center of the disk.
Best regards
18th Mar, 2022
Elias Sadeghi Malvajerdi
Shahid Beheshti University
Of course, what I said was assuming that the material of the disk is homogeneous and isotropic and the surface of the disk is smooth and polished.
19th Mar, 2022
Stam Nicolis
University of Tours
A homework problem, that's stated in ambiguous terms, which leads to endless discussions. This is material that's part of a course, not part of research.
What is, hopefully, taught is what is needed to make unambiguous statements, not statements that don't mean anything.
23rd Mar, 2022
Mohit Rattanpal
Bhabha Atomic Research Centre
Elias Sadeghi Malvajerdi Shouldn't the charge density at all patches on the flat surface be uniform as the curvature is same there??
23rd Mar, 2022
Mohit Rattanpal
Bhabha Atomic Research Centre
Stam Nicolis Actually I have encountered this problem while I have been studying Electrodynamics from a textbook where the answer to this problem is not given. The author has himself asked to discuss such conceptual questions with friends. So i posted on Research Gate.
23rd Mar, 2022
Elias Sadeghi Malvajerdi
Shahid Beheshti University
The electric charge accumulates at the edges and sharp points. Therefore, the electric charge should have a maximum value at the edges of the disk, and the farther we go from the edges to the center, the lower the electric charge should be.
27th Mar, 2022
Paul Pistea
Try to find it by using derivative on set functions
28th Mar, 2022
George Dishman
Thales Group, UK
I'd be interested in the answer this as well. It's not homework for me, I'm recently retired and this is just a pet project. My aim is to calculate the path of a charged particle fired at an acute angle towards any point on charged disc. What I've found so far are these discussions on the field and the potential which may be helpful:
29th Mar, 2022
Joerg Fricke
FH Münster
when I don't "see" the answer to a problem immediately, I find it sometimes worthwhile to play around with the help of a math program. For example, with integrals such program replaces a heavy book like Gradshteyn, Ryzhik: "Table of Integrals, Series, and Products", and if an integral cannot be solved symbolically, it can be solved at least numerically for a few cases.
Assuming a unit disk, and using polar coordinates (r, φ), for a charge at a certain point (p, 0) on the surface one can split the surface into an inner disk with radius = p - ϵ and an outer ring with inner radius = p + ϵ, for a small ϵ (and outer radius 1, of course). Then one can write an equation for the r component of the force (the φ component is zero due to symmetry) which takes into account the 1/d^2 dependence of the force (d being the distance between p and an arbitrary point on the disk) as well as the angle of the force. 2D integration over the inner disk and over the outer ring results in the forces exerted by both, and addition gives the total force. Please see the attached figure 1. Since this procedure involves no term expressing a dependence of charge density on location, it covers evenly distributed charge, and figure 1 refutes this idea: Except at the center, the total force isn't zero, so the charge density would change (here, positive values stand for centrifugal forces and vice versa).
If the term to be integrated is expanded by a factor modeling a charge density dependent on r, then the results change accordingly. The correct term, 1/sqrt(1 - r^2) in the case of a unit disk, results in figure 2. In this case, the total force is zero everywhere, except at the edge of the disk.
These notes by Ted Bunn might be of interest because they explain how to arrive at the correct term:
1 Recommendation
30th Mar, 2022
George Dishman
Thales Group, UK
Thank you Joerg Fricke, those are excellent resources.

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