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?


1 Answer 1


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.

  • 1
    $\begingroup$ When integrating using spherical coordinates you would have $r'^2 sin \theta dr d \theta d \phi$. Where do you get $|r-r'|$ from? $\endgroup$
    – User13114
    Feb 11 at 18:41
  • $\begingroup$ 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. $\endgroup$
    – mike stone
    Feb 11 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.