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A standard problem in finding the field is that of a uniformly charged disc, on its axis, but for this problem I'm supposed to find the potential and the field on the edge of the disc, i.e. in the plane of the disc. I'm having trouble doing this. I easily set up the integral for the potential:

$$\phi=k\sigma \int_{r=0}^{r=R} \int_{\theta=0}^{\theta=2\pi} \frac{rdrd\theta}{\sqrt{R^2+r^2-2Rr\cos\theta}}$$

where $R$ is the radius of the disc, and $r$ and $\theta$ are the traversing variables. The idea is that $dV = kdQ/r$, and $dQ=\sigma rdrd\theta$ (the non-sigma part being area), and then the part in the square roots is the distance which we need to use. I used the law of cosines. But evaluating this integral on wolfram alpha is horribly ugly. Did I do this right?

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    $\begingroup$ I think that your expression is correct, but I'm getting an ugly integral involving elliptic functions, too. That's probably the reason that most assignments of this type ask for the field along the axis and not in the plane of the disk. $\endgroup$ – user93237 Dec 20 '15 at 19:23
  • $\begingroup$ Would integrating the other variable first work? $\endgroup$ – Faraz Masroor Dec 20 '15 at 19:37
  • $\begingroup$ I used Mathematica to try integrating the 'theta' and 'r' variables in both of the two different orders and both methods were difficult. $\endgroup$ – user93237 Dec 20 '15 at 20:48
  • $\begingroup$ You can calculate the potential as in this image i.ytimg.com/vi/mY9e7Sbf70g/maxresdefault.jpg from the youtube . The field at the edge is discussed here: physics.stackexchange.com/questions/40919/… . $\endgroup$ – Keith McClary Dec 21 '15 at 4:35
  • $\begingroup$ Please note that Physics.StackExchange is not a homework help site. Please read this Meta post on asking homework-like questions and this Meta post for "check my work" problems. $\endgroup$ – Kyle Kanos Dec 21 '15 at 11:19