Finding the potential of a conducting disk(without laplace)

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Discussion Overview

The discussion centers on finding the electric potential of a conducting disk in space without using the Laplace equation. Participants explore alternative methods, particularly using the potential of a uniformly charged rod, and consider the mathematical challenges involved in this approach.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes using the potential of a uniformly charged rod, noting that its equipotentials are ellipsoids, and suggests that a disk can be viewed as a limiting case of an ellipsoid.
  • Another participant expresses frustration at the lack of responses, implying that the problem should be manageable for university students.
  • Some participants argue that solving Poisson's equation is the simplest method and question the feasibility of the proposed alternative method without it.
  • A participant elaborates on the integration method to find the potential of a uniformly charged rod and how it relates to the potential of a conducting ellipsoid, suggesting that a disk is an ellipsoid with a zero length in one dimension.
  • Another participant questions the approach, suggesting that a brute force integration using the free space Green's function might be more straightforward.
  • One participant defends their approach, asserting that it will yield a simpler final answer compared to the brute force method, which they claim is more complex and difficult to execute.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to find the potential of the conducting disk. There are competing views on whether to use the proposed method involving the charged rod or to solve Poisson's equation directly.

Contextual Notes

Participants express uncertainty regarding the mathematical steps involved in the proposed method and the limitations of their approaches, particularly in taking limits and integrating. There is also a lack of clarity on the assumptions made about the charge distribution and the geometry involved.

sentinel
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I want to find the potential of a conducting disk in space whithout laplace equation.(i know how to do that).I want to use the potential of a uniformly charged rod.because its equipotentials are ellipsoids(ellipses rotated around the rod).now a disk is a limiting case of an ellipsoid.
I can't get through the math part.can someone write and simplyfy it?
because a disk is an ellipsoid where semi major axis approaches zero and semi minor axis approaches R then c approes iR !
 
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This site is turning me down.c'mon guys!its not a hard problem!I mean I may not be able to do it but guys who go to the university must be able to!
 
It's easiest to just solve the poissons equation.
Not sure what alternate method you are suggesting here.
I thought you could only get the field on the symmetry axis without poissons?
 
Let me clear it up.by integration we can simply get the potential of a rod with uniform charge density in space.putting its potential CONST and then doing some algebra we find out its equipotentials are ellipses whose centers are at the edge of the rod(in 2D.in 3D its just ellipses rotated about the rod,i think they call'em ellipsoids).
so if we have a conducting ellipsoid,it means that we we have an ellipsoid at a CONSTANT potential>>so we simply forget about the ellipsoid and put a rod instead.in analogy with what we prooved we can simply find out a rod of what length and charge per length will do the job.
so we have the potential of an ellipsoid in space!(we use the potential of rod except that we right its landa and lenght(L) in terms of V of the conductor and its sami major axis).
now a disk is an ellipsoid really.(in the limit that its z length is zero.)
now we just have to take a limit of what we find for a rod to get the potential of disk in space.but it gets TRICKY!mathematically!
 
Sounds like a steange way to go at it, why not just brute force integrate it with the free space green function?
 
its simple!the final answer is going to come up simple.and the brute force way is HARD!
but we have the potential.we just need to plug in the according prameters!which is much simpler then doing the BRUTE FORCE integral which can't be done!
 

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