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A conducting disc is given a charge Q. As the disc is conducting it must be equipotential and hence the charge distribution will not be uniform. From what I know the charge will have a tendency to spread out more towards the rim and will arrange itself such that the net field is perpendicular everywhere on the surface of the disc. Is it possible to find this charge distribution and the electric field produced by it ? If so then how ?

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Find $E$ field of a ring of charge $Q$ on an axis perpendicular to the center of the ring.

Assume a uniform charge distribution - ignore the edge effects.

Then use the solution of the ring to find the $E$ field for the disk by integrating the ring solution over the area of the disk.

The charge density on the disk will be $\sigma = \frac{Q}{2\pi r^2}$ where $r$ is the radius of the disk. Again, ignore the edge effects.

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  • $\begingroup$ I asked the same for a conducting disc and not for a non-conducting disc. What you said can only be done in case of an insulating disc. $\endgroup$ – Nisarg Upadhyaya Mar 19 at 14:48
  • $\begingroup$ Why do think the solution is for a non-conducting disc? In the problem statement you indicated it was a conductor but you didn't indicate a width - so I assumed the width was small compared to the radius. Since it's a conductor, there is no electrical field inside the disc. $\endgroup$ – Cinaed Simson Mar 20 at 1:44
  • $\begingroup$ "As the disc is conducting it must be equipotential" is true but "hence the charge distribution will not be uniform" is false. It's conductor and it's assumed to be uniformly flat. $\endgroup$ – Cinaed Simson Mar 20 at 2:00

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