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Potential of a point at the z axis of the circular ring?

  1. May 4, 2015 #1
    So i have a thin circular ring laying at the XoY plane, inner radius of the ring is a, outer is b, density of electricity is given by expression ρ=ρ0*b/r , where ρ0 is a constant and r ∈ (a,b). I have to find a potential of the point P with coordinates (0,0,z).

    I tried to do it two ways, but both of them seem wrong.

    Firstly, i tried to first find E and then V since V=∫E*dl , but i ended up with E=ρ0*b*z/2ξ0∫dr/(z2+r2)3/2 , this seems wrong to me since in every other example i end up with, at most, integral with substitution, which this one isn't.

    The other way i tried to do it is, since potential of the single contour gives the value of the potential dV=dQ/4πξ0r potential of the ring in this example would be V=ρ0b*ln(b/a)/2ε0

    which makes no sense since there's no z in the solution, which would mean that distance between point and ring doesn't matter, which means that both of these are probably wrong.
  2. jcsd
  3. May 4, 2015 #2


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    See if this helps..Split the ring into infinite rings of infinitesimal thickness,say,dx and let the radius of that ring be x. Try finding potential due to each ring at the given point (0,0,z),which is on the axis of the ring. You'll get a differential function in terms of x (z is constant). Integrate it between limits a and b. I don't know the exact calculations but I'll first do it myself and then I could be certain..:smile:
  4. May 4, 2015 #3


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    Hello CD, please use the template (equations are missing altogether) and show your workings -- what you ended up with doesn't help us in finding where - if at all - things went wrong.

    First way: seems rather roundabout to do an integral of something you find with an integral instead of just doing the V integral ? Anyway, I don't see what you are doing to get what you ended up with and I sure miss a factor ##2\pi##. [edit] sorry, I see it now: canceled against ##1/(4\pi\epsilon_0)##

    Other way: z is in r in your expression for dV, but you mix it up with the r you are using for dQ.

    If stuck, Hyperphysics is always a good place to look. (ring of charge, disc of charge)
    Last edited: May 4, 2015
  5. May 4, 2015 #4


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    Yupp..I've got the answer..What I suggested seems to be your second approach where you said there's no z.. But I got z..I don't know what's missing in your 2nd approach,but your idea is right..Go ahead with it..:smile:
  6. May 4, 2015 #5
  7. May 4, 2015 #6
    Well, is my work correct, i mean do i only need to integrate it to get correct solution, or did i made a mistake?
  8. May 4, 2015 #7


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    Seems correct to me..:smile:
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