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Potential of a cylinder.

  1. Jul 10, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-7-11_6-26-12.png

    2. Relevant equations


    3. The attempt at a solution


    The position of the point (where V is to calculated) on the z-axis would be ##u = z_0 + l/2##.


    So in cylindrical coords,

    $$V(u) = \int_V {k \rho \over (s^2 + (u -z)^2)^{1/2}} dV = k \rho \int_0^L \int_0^{2\pi} \int_0^R {k \rho \over (s^2 + (u -z)^2)^{1/2} } \ ds\ d\phi\ dz \\= 2\pi k \rho \int_0^L (R^2 + (u -z)^2)^{1/2} - (u - z)\ dz =2\pi k \rho \left[\int_0^L (R^2 + (u-z)^2)^{1/2} dz - z_0L \right]$$

    This integral is very complex and cubersome to calculate. I think I made a mistake some where.
     
  2. jcsd
  3. Jul 10, 2017 #2

    TSny

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    Homework Helper
    Gold Member

    Your work looks good to me. The last integral does turn out to yield a messy answer.
     
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