Finding the potential of a conducting disk(without laplace)

  • Thread starter sentinel
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  • #1
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 !
 

Answers and Replies

  • #2
sentinel
18
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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!
 
  • #3
GarageDweller
104
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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?
 
  • #4
sentinel
18
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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!
 
  • #5
GarageDweller
104
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Sounds like a steange way to go at it, why not just brute force integrate it with the free space green function?
 
  • #6
sentinel
18
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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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