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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