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Charged ring, integrate for electric potential!

  1. Feb 28, 2014 #1
    1. The problem statement, all variables and given/known data

    A flat ring of inner radius R1 and outer radius R2 carries a uniform surface charge density σ. Determine the electric potential at points along the axis (the x axis). [Hint: Try substituting variables.]

    2. Relevant equations

    V = (kQ)/r


    3. The attempt at a solution

    As you can see from my screenshot, I think I've figured it out mostly I'm just stuck on finding the value for r in the above equation. Shouldn't it be something like sqrt(x^2+y^2)? But that can't be the answer because there's no labelled y-axis...
     

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    Last edited: Feb 28, 2014
  2. jcsd
  3. Feb 28, 2014 #2
    Well, you can't calculate the potential directly from the equation since not every point on the disk will be the same distance from a given point along the x axis. Try splitting the disk into thin disks, and then doing an integration.
     
  4. Feb 28, 2014 #3
    What exactly do you mean by "thin disks"? Don't I still need to find the distance between the x point and any given slice of charge on the disk?
     
  5. Feb 28, 2014 #4
    By thin I mean infinitely thin. With the disk you're using, the distance to the x point is going to vary as you go farther out from the center. You can find an expression for the distance in terms of the x point and the point you go radially outward, but it's going to be variable, so the solution is going to involve integration -- if you just look at an infinitesimally thin disk and find an expression for the potential along its axis, then you can consider the original disk as a collection of these disks, and integrate them all together.
     
  6. Mar 3, 2014 #5
    I'm having trouble finding an expression for the distance between any given point on the ring and the x-point, since there doesn't seem to be any variable that defines the distance between the center of the ring and any given piece of charge on the ring.
     
  7. Mar 3, 2014 #6
    Not explicitly given in the diagram, but that doesn't mean you can't create your own. You can indentify a ring of charge with a radius, or if you really want to identify any point, a radius and an angle.
     
  8. Mar 3, 2014 #7
    Thanks, I was finally able to figure it out, integrating from R1 to R2:

    [itex]\frac{σ}{2ε_{0}}[/itex]*[itex]\int[/itex][itex]\frac{rdr}{\sqrt{x^2+r^2}}[/itex]

    :approve:
     
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