# Is the electric field of a volume charge distribution well defined?

Consider a volume charge distribution $\rho({\bf r'})$. The electric field at ${\bf r}$ is

$${\bf E}({\bf r})=\iiint\frac{1}{4\pi\epsilon_0}\frac{\rho({\bf r'})}{R^2}\hat{\bf R}\, \mathrm d^3{\bf r'}$$

where ${\bf R}={\bf r}-{\bf r'}$.

My question is at positions ${\bf r}$ where $\rho \ne 0$, the integrand blows up when ${\bf R}={\bf 0}$. But I know from EM textbooks that ${\bf E}({\bf r})$ at where $\rho({\bf r})\ne 0$ is well-defined.

Can one explain this mathematically?

Edit: Here is own thought, but I am not sure whether it's correct.

Consider this as a purely mathematics question first.

Then I think the question is similar to whether $$\int_0^1\frac{1}{\sqrt{x}}dx$$ is undefined or equals $2$.

In my opinion, strictly speaking, the above integral is undefined, because the integrand is undefined at $x=0$.

However, since $$\lim_{\delta\rightarrow0^+}\int_\delta^1\frac{1}{\sqrt{x}}dx=2$$ we usually just say $$\int_0^1\frac{1}{\sqrt{x}}dx=2$$

So in my opinion, when we write $${\bf E}({\bf r})=k\iiint\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$ we are actually saying $${\bf E}({\bf r})=\lim_{\delta\rightarrow0^+}k\iiint_{D_{\delta}}\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$ where $D_\delta$ is the space minus a ball with radius $\delta$ centered at ${\bf r}$.

It can be shown that if $\rho({\bf r})$ is finite or does not go to infinity $\textit{too fast}$, then the limit exists.

If the above is true, then the physics question one should ask is whether this gives the correct answer in physics.

In classical EM, the $\textit{macroscopic}$ field ${\bf E}$ is in fact the average of the $\textit{microscopic}$ field ${\bf e}$. (Landau and Lifshitz, $\textit{Electrodynamics of Continuous Media}$, p.1.)

So you can consider ${\bf E}({\bf r})$ as the average of ${\bf e}$ over a ball with radius $\delta$ centered at ${\bf r}$.

Now, it can be shown that

(1) the average field over the ball due to charges outside the ball is the same as the total field due to all charges outside the ball at the center of the ball, and

(2) the average field over the ball due to charges inside the ball is $$-k\frac{{\bf p}}{\delta^3}$$ where ${\bf p}$ is the total dipole moment inside the ball.

(Griffith, $\textit{Introduction to Electrodynamics}$, p. 156-157, problem 3.41).

So when $\delta$ is small macroscopically, ${\bf p}\rightarrow{\bf 0}$, and we have $${\bf E}({\bf r})=k\iiint_{D_{\delta}}\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$

• Maybe this helps. – Dominik Apr 14 '16 at 15:03
• These fields are treated classically and are piecewise continuous only. Further, this representation is already in macroscopic form, which means one has assumed a spatial ensemble average a priori... – honeste_vivere Apr 14 '16 at 15:17
• @Kyson Yes, for surfaces and line charges the expression diverges on the surface/line (but is fine everywhere else). – Dominik Apr 14 '16 at 15:54
• @Dominik, it is true the integral diverges on the surface, but in fact electric field in the physical sense is defined in the surface. For a planar uniformly charged disk, electric field at its center is zero; for uniformly charged sphere, electric field in the surface is half of the field $E = \sigma/\epsilon_0$ outside the sphere just above the surface. So it seems natural that as the former is continuously deformed into the latter, the field in the surface is defined at all stages and changes from 0 to $E/2$ continuously. – Ján Lalinský Apr 14 '16 at 20:18
• Your integral is in fact well defined as a triple improper integral, so should be treated as such. – Ruslan Apr 24 '16 at 8:40

I actually agree with Ruslan's comment. You cannot say that the integral blows up when $\textbf{r} = \textbf{r}'$, with $\textbf{r}'$ spanning the integration domain where $\rho \neq 0$.

The reason is simply that this is a triple integral and that the volume form in the integral may compensate the diverging behaviour of the Green function.

To see this you can make a change of variable $\textbf{r}' \rightarrow \textbf{R}$ and swap to spherical coordinates $(R, \theta, \phi)$ (with $\theta$ and $\phi$ defined with phase shift to account for the minus sign arising with the change of variable), you will see that your formula for the electric field becomes:

\begin{equation} \textbf{E(r)} = \iiint \frac{\rho(\textbf{r}-\textbf{R})}{R^2} \hat{\textbf{R}} \: R^2 \sin \theta \: d\theta \: d\phi \:dR = \iiint \rho(\textbf{r}-\textbf{R})\hat{\textbf{R}} \: \sin \theta \: d\theta \: d\phi \:dR \end{equation}Of course the price to pay to write it like that in general is that the domain of integration becomes very non trivial and all the difficulty is encapsulated in expressing the function $\rho$ as a function of $R, \: \theta$ and $\phi$.

I would say yes, the electric field is well defined in a volume charge distribution. You raise some interesting points, that I think ultimately comes down to:

1. Is the electric field consistent and well defined mathematically?
2. What is the physical interpretation of the mathematics?
3. Is a volume charge distribution physical and realistic?

To answer the first question, you're correct in your interpretation of the integral as a limit excluding the divergent point. This can be viewed as the Cauchy principal value of the improper integral. This kind of integral with singularities is very common when discussing integral solutions to differential equations (in this case, Gauss's law), and doesn't mean there's a problem with the electric field. Also keep in mind that the electric field is defined mathematically by Gauss's law, and that this integral just happens to be the solution to Gauss's law. If the final electric field solution is consistent with Gauss's law, then whether the integral has a singularity doesn't really matter.

Physically, we like to think of this integral as a summation of the Coulomb fields of a number of point charges with a charge of $\rho \mathrm dV$. Since the electric field of a point charge blows up, shouldn't the electric field for a volume charge distribution (which is a summation of point charges) also blow up? No, it shouldn't. Though we think of it as a summation of point charges, a volume charge distribution is continuous and smooth, and doesn't inherit the divergence of a point charge.

Realistically, a volume charge distribution is more likely to be an approximation to a more complex charge distribution, such as with charges attached to points on the molecules making up a material. It's not clear to me whether the electric field is well-defined close to a charge at this size scale. It's at this size scale that some kind of averaging technique would be necessary to define the electric field, which could be different depending on the averaging technique used. This is more due to the breakdown of classical EM at small size scales though, and less to do with the classical electric field being poorly defined.

Edit: there seems to be some debate about whether the Cauchy principal value is necessary to evaluate the integral. For integrals with divergent integrands, the value of the integral depends on how the divergent point is approached. In this case, the order of limits and how the domain is subdivided becomes important. I'm going to show that this is the case for the electric field integral.

First, I'm going to divide the integral into two domains, one centered around the divergent point where we wish to evaluate the electric field, and one including the rest of the domain. The choice of my domain around the divergent point is a cylinder, which I'll denote $C$. I'll also use this as my origin, thus $\mathbf{r} = 0$.

$${\bf E}=\int_C \frac{1}{4\pi\epsilon_0}\frac{\rho(\mathbf{r'})}{r'^2}\hat{\mathbf{r'}}\, \mathrm dV + \int_{D \setminus C}\frac{1}{4\pi\epsilon_0}\frac{\rho(\mathbf{r'})}{r'^2}\hat{\mathbf{r'}}\, \mathrm dV$$

To evaluate the first integral, say that the cylinder has an infinitesimal cross section of area $\mathrm dA$, which we'll later take to approach 0. We'll assign the cylinder's axis to be the $x$ axis, and divide the cylinder into two sections, one for $x>0$, and one for $x<0$. We'll then discretize each of these sections into $N_1$ and $N_2$ sub-sections lengthwise, and convert the integral into a summation. For simplicity, also say that the charge distribution is constant, or approximately constant over this small domain. We'll also assign the cylinder a total length $l$

$$\int_C \frac{1}{4\pi\epsilon_0}\frac{\rho(\mathbf{r'})}{r'^2}\hat{\mathbf{r'}}\, \mathrm dV = \frac{\mathbf{\hat{x}}\rho \mathrm dA}{4\pi\epsilon_0} \lim_{N_1 \rightarrow \infty} \sum_{n=1}^{N_1} \frac{l/N_1}{(\frac{nl}{N_1}-\frac{l}{2N_1})^2}-\frac{\mathbf{\hat{x}}\rho \mathrm dA}{4\pi\epsilon_0} \lim_{N_2 \rightarrow \infty} \sum_{n=1}^{N_2} \frac{l/N_2}{(\frac{nl}{N_2}-\frac{l}{2N_2})^2}$$

Each summation diverges, resulting in $\infty-\infty$. In cases like this, the result can take on different values depending on how quickly each term approaches infinity. When they approach the divergence at the same rate (e.g. $N_1=N_2$), then this is the Cauchy principal value, and the two cancel. Otherwise, you can manufacture the result to produce infinity, or other values.

In response to gatsu's answer, by writing the integral in spherical coordinates around the divergence, you ensure that the divergence is approached equally from all sides, ensuring that the integral respects the Cauchy principal value.

• Cauchy principal value can exist even when improper integral diverges. In the OP's case the integral converges (given good enough $\rho$), so no need in Cauchy principal value. – Ruslan Apr 24 '16 at 11:48
• Nice answer. Note that at the molecular scale, charges (say, electrons) are not points but "spots" (wave packets), so the charge density is indeed continuous. – L. Levrel Apr 24 '16 at 13:34

The electric field at a point outside the volume charge distribution is well defined, certainly. What about the electric field at a point inside the charge distribution? Let $\vec{r'}$ denote a point that lies in the charge distribution and $\vec{r}$ denote a point where the electric field is to be determined. The problem is the determination of the electric field at $\vec{r}=\vec{r'}$. The electric field at such a point is $\displaystyle\vec{E(\vec{r})}=\int\frac{\rho (\vec{r'})(\vec{r}-\vec{r'})}{\lvert (\vec{r}-\vec{r'})^{3}\rvert}d^{3}\vec{r}$+$\displaystyle\int ^{'}\frac{\rho (\vec{r'})(\vec{r}-\vec{r'})}{\lvert (\vec{r}-\vec{r'})^{3}\rvert}d^{3}\vec{r}$, where the first integral is over a tiny sphere centred at $\vec{r}$ and the second integral is over the rest of the the charge distribution. The first integral can be simplified as follows. Since the spherical region is very small, the charge density is a constant equal to $\rho(\vec{r})$.It can be taken out of the integral.The order of magnitude of $\displaystyle\frac{(\vec{r}-\vec{r'})}{\lvert(\vec{r}-\vec{r'})\rvert^{3}}$ is $\displaystyle\frac{1}{\lvert(\vec{r}-\vec{r'})\rvert^{2}}$. The volume of the sphere is of order $\lvert(\vec{r}-\vec{r'})\rvert^{3}$ and so the integrand is of order $\lvert(\vec{r}-\vec{r'})\rvert$ which goes to zero in the limit as $\vec{r}\rightarrow\vec{r'}$. So the contribution from the first integral is zero. As a result, the electric field at a point inside the charge distribution comes entirely from the second integral. So, the electric field at a point inside a charge distribution is given by the same formula that applies to a point outside the charge distribution.

I agree that the volume form in spherical coordinates cancels the $1/r^2$ divergence from Coulomb's law, but I think there's a more physical way to interpret this mathematical fact: remember that strictly speaking, for a continuous charge distribution with no delta functions in the density, $\rho({\bf x})$ is not the amount of charge at point ${\bf x}$. There is zero charge exactly at point ${\bf x}$. Instead, the quantity $\rho({\bf x})\ d^3x$ is the amount of charge contained in the volume $d^3 x$ centered at ${\bf x}$. So the point exactly at ${\bf R} = {\bf 0}$ by itself does not contribute to the integral - just as no single point of any integral ever affects the value of the integral! That is, changing the value of a single point of a function never affects the value of the function's integral, exactly because there is "no charge" exactly at that point (assuming no delta functions). (Moreover, $\hat{\bf R}$ is undefined at ${\bf R} = {\bf 0}$, so it's not even clear what this vector quantity would be even if it weren't infinite.) So you can safely ignore the point exactly at ${\bf R} = {\bf 0}$ and only focus on the nearby points - and as the improper integral converges for smooth charge distributions, everything works out fine.

This is related to the fact that to find the electric field on a point charge, you only add up the fields from the other charges and ignore the point charge's own field, because it's infinite at that point. (This question of "self-energy" - whether a charge "feels" its own electric field - is quite subtle; see Griffiths section 2.4.4. Things get really hairy when the charges move relativistically; see Griffiths section 11.2.3.)

I think the philosophy is like this: Coulomb's law was originally discovered to describe the interaction between finite numbers of charged point particles. As you discuss, there is some subtlety as to how to generalize it to continuous charge distributions. Now the "integral form" of Gauss's law $\oint_S {\bf E} \cdot d{\bf S} = Q_\text{enclosed} / \epsilon_0$ is completely equivalent to Coulomb's law in the case of a finite number of point particles, but it has the nice property that it's also well-defined for continuous charge distributions. So if you like, you can think of it as a generalization of Coulomb's law that also works (straightforwardly) for continuous distributions. In practice, when dealing with continuous charge distributions, it's almost always easier to use Gauss's law (usually in its "differential form" ${\bf \nabla} \cdot {\bf E} = \rho / \epsilon_0$) than Coulomb's law, because the subtleties that you mention don't arise.

We put (four pi times) the permittivity $4\pi\varepsilon=1$ equal to one from now on (in the SI unit system) for simplicity. Let ${\bf r}_0\in \mathbb{R}^3$ be a fixed position.

The electric field in the $i$'th Cartesian direction at ${\bf r}_0$ is

$$\tag{1} E^i({\bf r}_0)~=~- \int_{[0,\infty[\times [0,\pi]\times[0,2\pi]} \mathrm{d}r~\mathrm{d}\theta~\mathrm{d}\phi~ \underbrace{\sin(\theta)~\rho({\bf r}+ {\bf r}_0) \frac{x^i}{r} }_{\text{integrand}},$$

where we have used spherical coordinates $(r,\theta,\phi)$ in a hopefully obvious notation. [For starter it is implicitly implied that the Cartesian coordinate $x^i$ depends on $(r,\theta,\phi)$ in the standard way. Note that the denominator in $\frac{x^i}{r}$ vanishes when expressed in spherical coordinates.]

Notice that $\left|\frac{x^i}{r}\right| \leq 1$. A majorant function for the integrand (1) is

$$\tag{2} [0,\infty[\times [0,\pi]\times[0,2\pi] ~\ni~(r,\theta,\phi) \quad\mapsto \quad\sin(\theta) |\rho({\bf r}+ {\bf r}_0)|~\geq ~0.$$

Thus if the integrand (1) is Lebesgue measurable and the majorant function (2) is Lebesgue integrable, then the integrand (1) is Lebesgue integrable, and the integral (1) is well-defined.

Much of chapter I, section 9 in "Foundations of potential theory, Oliver Dimon Kellogg, Berlin: Verlag von Julius Springer, 1929", parts of which appear in the answers by Mathaholic, Procyon and Qmechanic, is devoted to answer this question.

Let $v$ be a small region of arbitrary shape, containing $P$ (defined by $\vec{r}$) in its interior. We consider the integral: $$\vec{E}(\vec{r}) \equiv \iiint_{V-v} \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|^3} (\vec{r} - \vec{r}')d^3 r',$$ as the maximal chord of $v$ tends to $0$. If this integral converges, we define it as the field, following Newton's or Coulomb's law, of a distribution of mass or charge.

A necessary and sufficient condition that $\vec{E}$ converges to a limit is that for all $\epsilon >0$, there exists $\delta >0$, such that if $v$ and $v'$ are any two regions containing $P$ and contained in the sphere $\sigma$ of radius $\delta$ about $P$, then $$\left|\iiint_{V-v} \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|^3} (\vec{r} - \vec{r}')d^3 r' - \iiint_{V-v'} \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|^3} (\vec{r} - \vec{r}')d^3 r'\right|<\epsilon.$$ If $B$ is an upper bound for $|\rho|$ in $\sigma$, then we have: $$I\equiv \left| \iiint_{\sigma- v} \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|^3} (x - x')d^3 r' \right| \leq B \iiint_{\sigma-v} \frac{|x-x'|}{|\vec{r} - \vec{r}'|^3} d^3 r'.$$ If we take a component of the integral on the rhs, use spherical coordinates with $P$ as pole, then the integral is no longer improper and we have: $$I< B \int_0^{2\pi} \int_0^\pi \int_0^\delta \frac{|\varrho\cos\theta|}{\varrho^3} \varrho^2 \sin\theta d \varrho d \theta d \phi < 2 B \pi \delta.$$ If we let $\delta< \epsilon/(4B\pi)$, then the condition for convergence is satisfied. Thus, we observe that every integral of such type with $\rho$ bounded converges.

I agree with gatso, but I think it might be explained more simply by noting that the volume element can be written as $dV=4\pi r^2dr$ which cancels out the $r^2$ in the denominator.

I also agree that the problem is similar to integrating $\left(\large{\frac1{\sqrt x}}\right)dx$ from $x=0$, because here the ratio $\large{\frac{dx}{\sqrt x}} = 2\sqrt x\frac{d(\sqrt x)}{\sqrt x} = 2d(\sqrt x)$.