# Discontinuity in integrand while calculating the electric field of a uniformly charged sphere

Suppose you wanted to calculate the electric potential of a uniformly charged sphere with radius $$R$$ at a point $$r$$ inside the sphere. In Griffith's EM this is done by integrating $$\frac{1}{4 \pi \epsilon_0} \int_V \frac{\rho}{(r-r')}d \tau$$ over the region of the sphere where $$\rho$$ is the uniform charge density and $$d \tau$$ is an infinitesimal volume element.

It seems to me that if $$r$$ was a point in the sphere there would be a discontinuity in the integrand, since $$r-r'$$ would equal 0 at some point we are integrating over. My question is how come this method is valid when $$r$$ is inside the sphere?

The volume factor around the problematic point is is $$4\pi |r-r'|^2 d(r-r')$$ and the $$|r-r'|^2$$ cancels against the $$1/|r-r'|$$ to give a finite integrand.
• When integrating using spherical coordinates you would have $r'^2 sin \theta dr d \theta d \phi$. Where do you get $|r-r'|$ from? Feb 11 at 18:41
• The integrand is rotationally invariant about the problematic point $r'$, so you may as well use $r'$ as the origin when investigating the convergence. The $4\pi$ coes from the integrals over the angles. Feb 11 at 18:47