My understanding after reading Mike Stone's answer:

\begin{align} \vec{E} &= k \iiint_{V} \dfrac{\rho(x',y',z')[x-x'\hat{(i)}+y-y'\hat{(j)}+z-z'\hat{(k)}]}{[(x-x')^{2}+(y-y')^{2}+(z-z')^{2}]^{3/2}}dx'dy'dz'\\ &= k \iiint_{V} \dfrac{\rho\ (\hat{r})}{r^2}dV\\ &=k \iiint_{V} \dfrac{\rho\ (\hat{r})}{r^2} r^2\ \sin\theta\ dr\ d\theta\ d\phi\\ &=k \iiint_{V} \rho\ (\hat{r}) \sin\theta\ dr\ d\theta\ d\phi\\ \end{align}

Now our function is $\rho\ (\hat{r}) \sin\theta$.

$\hat{r}$ and $\theta$ are undefined only at the origin. Therefore our function $\rho\ (\hat{r}) \sin\theta$ is undefined only at the origin. So we cannot directly integrate it. Hence we have to use the limit approach:

\begin{align} \vec{E} &= \lim_{\epsilon \rightarrow 0}\ k \left( \iiint_{V \setminus\ \text{sphere with radius $\epsilon$ centered at origin}} \rho\ (\hat{r}) \sin\theta\ dr\ d\theta\ d\phi \right)\\ &-\lim_{\epsilon \rightarrow 0}\ k \left(\iiint_{\text{over a sphere with radius $\epsilon$ centered at origin}} \rho\ (\hat{r}) \sin\theta\ dr\ d\theta\ d\phi \right) \end{align}

In the first term $\rho, \hat{r} \text{ and } \theta$ is defined and finite everywhere. Therefore the integral in the first term is finite.

Since the radius of the sphere $(\epsilon)$ is approaching zero, $\rho$ becomes more and more constant and the second term approaches zero.


\begin{align} \vec{E} = \lim_{\epsilon \rightarrow 0}\ k \left( \iiint_{V \setminus\ \text{sphere with radius $\epsilon$ centered at origin}} \rho\ (\hat{r}) \sin\theta\ dr\ d\theta\ d\phi \right)=\text{finite} \end{align}

However I do not know how to proceed to further simplify this term.

  • $\begingroup$ we use the limit method as written, both limits in the sum $(2)$ are still infinite. I think you actually meant the Cauchy principal value, which is a limit of a sum, not sum of limits. $\endgroup$
    – Ruslan
    Dec 7, 2018 at 20:34
  • $\begingroup$ Both limits in the sum $(2)$ are infinite....Can you please elaborate? $\endgroup$
    – N.G.Tyson
    Dec 8, 2018 at 1:42
  • $\begingroup$ Well, just evaluate them one by one, you'll get infinities. After that there's no sense in which they could be added. $\endgroup$
    – Ruslan
    Dec 8, 2018 at 6:52
  • $\begingroup$ @Ruslan: Please have a look at my edited question. $\endgroup$
    – N.G.Tyson
    Dec 8, 2018 at 14:51
  • 1
    $\begingroup$ We can directly integrate it. Integration is insensitive to removable singularities. See e.g. this Math.SE post. And note that if your $r^{-2}$ was canceled by $r^2$, and the integrand is no longer unbounded, then you indeed have a removable singularity at the origin. $\endgroup$
    – Ruslan
    Dec 8, 2018 at 17:46

1 Answer 1


The element of volume in 3d is $dV= r^2dr d\Omega$ where $d\Omega= \sin\theta d\theta d\phi$ is the angular part. Put the origin of your coordinate system at the point where you want to compute the field. Observe then, that the $r^2$ overcomes the $1/r^2$ divergence in the $\hat {\bf r}/|{|{\bf r}|^2}$ integrand (here $\hat {\bf r}$ is the unit vector). The field therefore remains finite provided the charge density remains finite. Of course if there are point charges the field diverges-so not any charge density works.

  • $\begingroup$ Thanks for the answer... Anyway in what direction should $\hat{r}$ point at the origin? $\endgroup$
    – N.G.Tyson
    Dec 6, 2018 at 6:04
  • $\begingroup$ Won't $\hat{r}$ be ambiguous at the origin? How shall we deal with it? $\endgroup$
    – N.G.Tyson
    Dec 6, 2018 at 6:32
  • $\begingroup$ @faheemahmed400. There is no ambiguity because there is no contribution at the origin. Think of a small sphere of charge density $\rho$. The ${\bf E}$ field at a distance $r$ from the center of the sphere is ${\bf E}= \rho {\bf r}/(3\epsilon_0)$. This is zero at the center of the sphere. Your entire integral is a sum of the contributions from each sphere. $\endgroup$
    – mike stone
    Dec 6, 2018 at 15:21
  • $\begingroup$ Your answer and comment were very useful. As a finishing touch, please have a look at my edited question and ensure whether I am understanding things the right way. $\endgroup$
    – N.G.Tyson
    Dec 7, 2018 at 16:30

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