# Physics: Electric field

This question involves maths. I do not think it is inappropriate here.

I am reading Purcell and Morin's Electricity and Magnetism 3rd Edition.

Equation ($$1.22$$): $$\vec{E}(x,y,z)=\dfrac{1}{4 \pi \epsilon_0} \int \dfrac{ρ\ (x^\prime, y^\prime, z^\prime)\ \hat{r}\ dx^\prime, dy^\prime, dz^\prime}{r^2}$$

On page 22, it says:

"A continuous charge distribution $$ρ\ (x^\prime, y^\prime, z^\prime)$$ that is nowhere infinite gives no trouble at all. Equation $$(1.22)$$ can be used to find the field at any point within the distribution. The integrand doesn’t blow up at $$r = 0$$ because the volume element in the numerator equals $$r^2 \sin \phi\ d \phi\ d \theta\ dr$$ in spherical coordinates, and the $$r^2$$ here cancels the $$r^2$$ in the denominator in Eq. $$(1.22)$$. That is to say, so long as $$ρ$$ remains finite, the field will remain finite everywhere, even in the interior or on the boundary of a charge distribution."

According to the above quoted paragraph, equation $$(1.22)$$ becomes:

$$\vec{E}(x,y,z)=\dfrac{1}{4 \pi \epsilon_0} \int ρ\ (x^\prime, y^\prime, z^\prime)\ \hat{r}\ \sin \phi\ d \phi\ d \theta\ dr$$

Here there is no particular direction for $$\hat{r}$$ at $$r=0$$. Then how can we say that in spherical coordinates the integral doesn't blow up at $$r=0$$.

I have more questions on this:

(2) How can we be sure that the integral doesn't blow up at $$r=0$$ in other coordinate systems?

(3) Are there any analogous expressions for electric field (independent of $$r$$) due to surface charge density and line charge density?

## 1 Answer

1. The question wether the integral blows up or not is not a "pointwise" question, since the singleton $$\{r=0\}$$ has $$0$$ volume, we have $$\int_{\overrightarrow r\in \mathbb R^3} f(r) \mathrm d \overrightarrow r = \int_{\overrightarrow r\in \mathbb R^3 \setminus\{0\}} f(r) \mathrm d \overrightarrow r$$ This remark is valid in any coordinate system. In you case, you have a gentle function writtent in spherical coordinates which is well defined everywhere but at $$r=0$$. No problem : you can choose the value of the function to be whatever you want at $$\{r=0\}$$ even infinite, the integral will still be well defined and has the same value.
2. Here his a theorem for yout peace of mind : let $$(X,\mu)$$ be a measures set, let $$f:X\rightarrow \mathbb R$$ be a measurable function and let $$g:Y\rightarrow X$$ be change of variables, then $$\int_{X} |f\circ g|\, \mathrm d \mu< +\infty \quad\Leftrightarrow\quad \int_{Y} |f|\,g\#\mathrm d \mu < +\infty$$ This theorem is valid in a very wide generality : the change of variable $$g$$ do not need to be continuous or bijective and it includes all change of variable you might encounter. I wrote $$\phi \# \mathrm d \mu$$ for the image measure. For instance, let $$X=\mathbb R^3$$ and $$Y= [0,2\pi]\times [0,\pi]\times \mathbb R_ +^*$$, let $$g : \begin{matrix}Y&\rightarrow &X \\ (\theta,\phi,r)&\mapsto& (r\cos(\theta)\cos(\phi),r\sin(\theta)\cos(\phi),r\sin(\phi))\end{matrix}$$ You recognize your spherical change of variable and $$\mathrm d\mu = \mathrm d x\mathrm dy \mathrm d z \quad\quad \quad g\#\mathrm d \mu = r^2\sin(\phi)\mathrm d \theta \mathrm d \phi \mathrm d r$$ here $$f=\rho \hat r$$.
3. If you're interested for more precise mathematical treatement of non-smooth electrival charge distribution, you could have a look at distribution theory. It allows you to write such integrals even for "line-wise" or "surface-wise" or "point-wise" charges. In this case, you only need to understand the most basic examples of distributions which are actually measures as in Lebesgue measure theory. If you go in that direction, you rewrite $$E(\rho)= \int_{\mathbb R^3} \frac{\rho\left(\hat r\right) \hat r}{r^2} \mathrm d \hat r = \int_{\mathbb R^3} \frac{ \hat r}{r^2} \left(\rho(\hat r)\mathrm d \hat r\right) = \int_{\mathbb R^3} \frac{ \hat r}{r^2} \mathrm d \mu = E(\mu)$$ Your $$\mathrm d \mu$$ can now be any charge distribution you want as long as the behaviour around the origin is not too pathological.