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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?

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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.

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    $\begingroup$ 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

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