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

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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
    Be careful with your prime and unprimed coordinates and with what $r$ really means – Aaron Stevens Nov 8 at 5:57
  • $r=\sqrt {(x-x')^2+(y-y')^2+(z-z')^2}$ everywhere – Alec Nov 8 at 6:21
  • @AaronStevens: Please elaborate in an answer. – Alec Nov 8 at 10:03
  • I will when I have time. But it seems like you have a good idea already. That value for $r$ is not the same $r$ that appears in the spherical volume element in general. They are only the same if you are trying to find the field at the origin, since then $x=y=z=0$ – Aaron Stevens Nov 8 at 11:06
  • I have yet to understand this. If we are computing electric field at origin $(0,0,0)$ due to a charge distribution, wouldn't both $r$ be same? – Alec Nov 8 at 11:09

Let's start with the equation you give

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

We to realize that $r$ is the distance between the point we are calculating the field at and the coordinate we are integrating over. So then in Cartesian coordinates we have $$r^2=(x-x')^2+(y-y')^2+(z-z')^2$$ However, our volume element only refers to primed coordinates. Therefore $$dx^\prime dy^\prime dz^\prime=r'^2\sin\phi' dr' d\theta'$$ where $r'$ is the spherical coordinate for the primed coordinates: $r'^2=x'^2+y'^2+z'^2$

So as you can see, the $r'^2$ in the volume element is only cancelled by the $r^2$ in the denominator if we are looking at the field at the origin where $x=y=z=0$.

Then our integral for our field at the origin is $$\vec{E}(x,y,z)=\dfrac{1}{4 \pi \epsilon_0} \int ρ\ (r', \phi', \theta')\ \hat{r}\ \sin \phi'\ d \phi'\ d \theta'\ dr'$$

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 $r=0$.

You have to keep in mind that we are integrating over all primed coordinates. The unit $\hat r$ vector is $0$ at the origin (primed coordinates) now, but the value of the integrand at a point in the region we are integrating over does not determine the entire integral.

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

Changing coordinate systems doesn't change what the integral evaluates to. The most likely reason the book uses this example is because it is easy to work through. This also holds true for other $(x,y,z)\neq(0,0,0)$ coordinates that are within our integration region, since we are always free to put our origin anywhere we want.

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

Notice how none of this depends on what $\rho$ actually is. Therefore, this discussion is just as valid for surface or line charges.

  • Thanks... I understand up to here. Please continue in your free time. – Alec Nov 8 at 16:27
  • @Alec I have added more – Aaron Stevens Nov 8 at 22:18
  • Understood all except last part. In case of surface or line charges, the element area (in polar coordinates) and element length will not be having $r^2$ coefficient. Hence the $r^2$ will not cancel out. Am I wrong? – Alec Nov 9 at 4:01
  • @Alec You mean for example if you are integrating over a disk then you infinitesimal area element would be $rdrd\theta$ without an $r^2$ dependency? – Aaron Stevens Nov 9 at 4:24
  • Yes exactly that. – Alec Nov 9 at 5:21

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