Potential of a cylinder.

  • #1
850
146

Homework Statement


upload_2017-7-11_6-26-12.png


Homework Equations




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.
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,795
3,153
Your work looks good to me. The last integral does turn out to yield a messy answer.
 

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