# Functions unbounded at a point on the domain: How to show electric field exists for any continuous volume charge density?

\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.

Hence:

\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.

• 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. – Ruslan Dec 7 '18 at 20:34
• Both limits in the sum $(2)$ are infinite....Can you please elaborate? – N.G.Tyson Dec 8 '18 at 1:42
• Well, just evaluate them one by one, you'll get infinities. After that there's no sense in which they could be added. – Ruslan Dec 8 '18 at 6:52
• @Ruslan: Please have a look at my edited question. – N.G.Tyson Dec 8 '18 at 14:51
• 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. – Ruslan Dec 8 '18 at 17:46

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.
• Thanks for the answer... Anyway in what direction should $\hat{r}$ point at the origin? – N.G.Tyson Dec 6 '18 at 6:04
• Won't $\hat{r}$ be ambiguous at the origin? How shall we deal with it? – N.G.Tyson Dec 6 '18 at 6:32
• @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. – mike stone Dec 6 '18 at 15:21