# Why is it possible to write the electrical potential as a sum of $\int dV \,\rho/R$ and $\int da \,\sigma/R$?

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

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?

• 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).
– user137661
May 10, 2019 at 19:06

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

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

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