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Voltage at a point.

  1. Nov 12, 2008 #1

    Eo = permittivity of free space

    integral dV = (1/4(pi)Eo) integral (dQ)/a

    Hopefully I have that formula correct.


    dQ = 2(pi)r(dr)(sigma)

    Horizontal components cancel. So there will be a sin(theeta) term at the end of the integral.

    sin(theeta) = z/a

    and a = sqrt(z^2 + r^2)

    and the integral will be between 0 and R

    so the RHS becomes...

    (1/4(pi)Eo) integral(0 -> R) (2(pi)r(dr)(sigma)/sqrt(z^2 + r^2)) * z/sqrt(z^2 + r^2)

    Is this right? I have a feeling I have done something stupid.

    Any help greatly appreciated.
    Last edited: Nov 12, 2008
  2. jcsd
  3. Nov 12, 2008 #2


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    I think that's correct. Since they give you the final answer, you can see if you get it when you do the integral.
  4. Nov 12, 2008 #3
    once you pull all the constants out the front you get....

    (z(sigma)/2Eo) integral(0 -> R) r/(z^2 + r^2) (dr)

    I'm pretty sure that doesn't give that answer.

    Any suggestions?
  5. Nov 12, 2008 #4


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    Why are you talking about components?...The potential is a scalar not a vector(perhaps you are confusing it with the electric field :wink:)....Do you really have a [itex]\sin \theta [/itex] term in your integrand?:wink:
    Last edited: Nov 12, 2008
  6. Nov 12, 2008 #5
    Yerp, confusing sh*t like usual >.<

    But that all works out now! Thanks alot for pointing that out for me :)
  7. Nov 13, 2008 #6


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    Oops, I missed that little detail :blushing: Good save, ggh.
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