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Calculating E-field and potential of a charged ring

  1. Aug 13, 2013 #1
    Hello folks!

    I've been trying to calculate the E-field of a charged ring. It seems well documented for a symetric point(a line from the center etc.) but what I'm interested in is say if I'm slightly of the center of the ring, how can I make a more general equation?

    I've tried calculating the potential and the field from there but I get a dominating 0 somewhere so that must be wrong ( V = ∫(λ(x')*dl'(1/(|x-x'|))) and E = -∇V , substituting dl' and λ(x') I get the product of those vectors to (-ρcos(θ')*ρsin(θ') + ρcos(θ')*ρsin(θ'))λdθ' = 0 ??).

    Anybody knows where I'm wrong and what to do?
     
  2. jcsd
  3. Aug 13, 2013 #2

    Simon Bridge

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    Welcome to PF;
    The reason everyone does the field along the axis is because the symmetry is easy.
    Off that axis, things get trickier.

    Specify cylindrical polar coords, and specify a vector ##\vec{r}## to an arbitrary point.
    The resulting field should look a lot like that in a plane from two like charges - then rotated about the z axis.

    It may be easier to solve Poisson's equation or follow the differential form.

    Anyway - you could also search for "off axis electric field of a ring of charge"
    http://www.mare.ee/indrek/ephi/efield_ring_of_charge.pdf
    http://electron.physics.buffalo.edu/sen/documents/field_by_charged_ring.pdf
     
  4. Aug 14, 2013 #3
    Thank you for the welcome. And thanks, I will check it out, although I found something out that I did miss. Apparently I've treated λ and dl' as vectors when they probably should not have been. Now I went straight to the general integralform of the E-field equation and solved it as λ being a constant over the integration and dl' = ρ*dθ'. For x,y = 0 I get the same equation as for the symetric equations as expected. I'm going to use it in electron path simulation so I solved the integral numericaly, probably going to do it proper later though.
     
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