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Calculate the potential inside a uniformly charged solid sphere of radius $R$ and total charge $q$.

My attempt:

There are several ways to solve this problem but I'm curious as to whether this particular method is applicable.

enter image description here

WLOG let the point $P$ lie on the $z$-axis a distance $z$ from the center of the sphere (origin). $z<R$

Consider an infinitesimal volume element whose position vector $\mathbf{r}$ makes an angle $\theta$ with the $z$-axis.

Let $r'$ be the distance between the volume element and $P$. By the cosine rule (as shown in the figure), $$r'=\sqrt{z^2+r^2-2zr\cos\theta}$$

$$\displaystyle dV=\frac{1}{4\pi\epsilon_0}\frac{dQ}{r'}=\frac{1}{4\pi\epsilon_0}\frac{\rho\ d\tau}{r'}=\frac{1}{4\pi\epsilon_0}\frac{\rho r^2\sin\theta\ dr\ d\theta\ d\phi}{\sqrt{z^2+r^2-2zr\cos\theta}}$$

$$\displaystyle V=\int_0^{2\pi}\int_0^\pi\int_0^R\frac{1}{4\pi\epsilon_0}\frac{\rho r^2\sin\theta\ dr\ d\theta\ d\phi}{\sqrt{z^2+r^2-2zr\cos\theta}}$$

The only problem is that the denominator of the integrand goes to $0$ when $\theta=0$ and $r=z$. How do I circumvent this problem?

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closed as off-topic by John Rennie, Jon Custer, Kyle Kanos, GiorgioP, ZeroTheHero May 5 at 23:34

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  • $\begingroup$ First integrate, then worry about limits $\endgroup$ – Tojrah May 2 at 8:07
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    $\begingroup$ Notice that the numerator also goes to zero. $0/0$ doesn't necessarily mean infinity. $\endgroup$ – probably_someone May 2 at 15:29
  • $\begingroup$ Inside the sphere, you probably don't want to integrate all the way to R. $\endgroup$ – R. Romero May 2 at 16:22
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You have a singularity in $(0,0)$. Could you let it out from the problem leading with on integration limits?

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