I'm trying to understand why one can write the electrical potential as follows \begin{equation} 4\pi\varepsilon_0\phi(\mathbf r) =\int d^3 r\,\dfrac{\rho(\mathbf r')}{\|\mathbf r - \mathbf r' \|} + \int d^2 r\,\dfrac{\sigma (\mathbf r')}{\|\mathbf r - \mathbf r' \|} \label{phi} \end{equation}

I know that the surface density charge can be written as $$\dfrac{\sigma}{\varepsilon_0} = \partial_n\phi_1-\partial_n\phi_2 $$ (Where $\partial_n = \nabla\cdot\mathbf n$ is directional derivative in the normal direction.)

But introduce two new potentials that I don't know how to deal with. Any clue about how to get the such expression for $\phi$?

EDIT: Or a better question: Under what kind of conditions can I write $\phi$ as it is written above?

  • $\begingroup$ I think this page by Wolfram addresses your question scienceworld.wolfram.com/physics/SurfaceChargeDensity.html As you can see, that form for the surface density arises from Gauss's law (it is stated in section 1.1 of Schwinger's book). $\endgroup$
    – user137661
    May 10, 2019 at 19:06

1 Answer 1


The reason it is possible to define an electric potential is that $\mathbf{\nabla} \times \mathbf{E} = 0$. We define $\phi\left(\mathbf{r}\right) = - \int_{r_0}^{\mathbf{r}}\mathbf{E} \cdot d\mathcal{\mathbf{l}}$, with $r_0$ as some reference point. This leads to $\mathbf{E}=-\mathbf{\nabla}\phi$. So, for a point charge $q$, we get

$$\phi\left(\mathbf{r}\right) = \frac{1}{4\pi\epsilon_0}\frac{q}{r},$$

where $r$ is the distance from the charge to $\mathbf{r}.$ For a group of $N$ charges, we use the superposition principle and write

$$\phi\left(\mathbf{r}\right) = \frac{1}{4\pi\epsilon_0}\sum_{i}^{N}\frac{q_i}{r_i}.$$

For a continuous distribution of charge, this becomes

$$\phi\left(\mathbf{r}\right) = \frac{1}{4\pi\epsilon_0}\int\frac{dq}{r'}.$$

To work out the integral, we need to trade $dq$ for a function of $\mathbf{r}$. We can do this in any of three ways:

  1. Define a volume charge density $\rho\left(\mathbf{r'}\right) \equiv dq / d^3r'$;
  2. Define a surface charge density $\sigma\left(\mathbf{r'}\right) \equiv dq/d^2r'$;
  3. Define a linear charge density $\lambda\left(\mathbf{r'}\right) \equiv dq/dr'$.

We pick the charge density that matches the geometry of the charge distribution. In a situation where all three are present, we could write

$$4\pi\epsilon_0\phi\left(\mathbf{r}\right) = \int{\frac{\rho\left(\mathbf{r'}\right)}{r}}d^3r' + \int{\frac{\sigma\left(\mathbf{r'}\right)}{r}}d^2r' + \int{\frac{\lambda\left(\mathbf{r'}\right)}{r}}dr'.$$

  • $\begingroup$ Good answer, +1 for including linear density. Points too, but it's all the same. $\endgroup$
    – user196418
    May 10, 2019 at 20:35
  • $\begingroup$ Apparently, it is straightforward using the superposition principle. I was confused because of an exercise in Jackson's electrodynamics. Thanks!! $\endgroup$ May 10, 2019 at 23:44

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