# Calculating electric potential — denominator going to zero [closed]

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

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?

• First integrate, then worry about limits – Tojrah May 2 '19 at 8:07
• Notice that the numerator also goes to zero. $0/0$ doesn't necessarily mean infinity. – probably_someone May 2 '19 at 15:29
• Inside the sphere, you probably don't want to integrate all the way to R. – R. Romero May 2 '19 at 16:22

You have a singularity in $$(0,0)$$. Could you let it out from the problem leading with on integration limits?