# Gauss’s Law from Coulomb’s Law

## Equivalence of Gauss’ Law for Electric Fields to Coulomb’s Law

Coulomb’s law is often one of the first quantitative laws encountered by students of electromagnetism. It describes the force between two point electric charges. It turns out that it is equivalent to Gauss’s law. Coulomb’s law states that the force between two static point electric charges is proportional to the inverse square of the distance between them, acting in the direction of a line connecting them. If the charges are of opposite sign, the force is attractive and if the charges are of the same sign, the force is repulsive. Mathematically, Coulomb’s law is written as

(43)$\mathbf{F} = \frac{qQ}{4\pi \varepsilon_0 |\mathbf{r} - \mathbf{r'}|^2}~\mathbf{\hat{\underline{r}}} \;,$

where $$\mathbf{F}$$ is the force between the two charges $$q$$ and $$Q$$, $$|\mathbf{r} - \mathbf{r'}|$$ is the distance between the charges and $$\mathbf{\hat{\underline{r}}}$$ is a unit vector in the direction of the line separating the two charges.

Having defined Coulomb’s law, one might next naturally ask the question how would a standard reference charge behave in the presence of any distribution of electric charge we might dream up? Answering this question brings us to the concept of the electric field. We follow the presentation of [Gri99]. We can define the electric field of an arbitrary charge $$Q$$ as the force experienced by a unit charge $$q$$ due to $$Q$$

(44)$\mathbf{e} = \frac{\mathbf{F}}{q}.$

Dividing both sides of Coulomb’s law by $$q$$ and substituting the definition of $$\mathbf{e}$$, we get that the electric field of a point charge $$Q$$ is

(45)$\mathbf{e}(\mathbf{r}) = \frac{Q}{4\pi\varepsilon_0 |\mathbf{r} - \mathbf{r'}|^2}~\mathbf{\hat{\underline{r}}}\;.$

It is important to note here that the electric field obeys the principle of superposition, meaning that the electric field of an arbitrary collection of point charges is equal to the sum of the electric fields due to each individual charge.

(46)$\mathbf{e}\left(\sum_{k=1,n} Q_i\right) = \sum_{k=1,n} \mathbf{e}(Q_i)$

If we consider the the electric field due to a spatially extended body with charge density $$\rho$$, the sum becomes an integral over infinitesimal volume elements of the body

(47)$\mathbf{e} = \frac{1}{4\pi\varepsilon_0}\int_V \frac{\rho}{|\mathbf{r} - \mathbf{r'}|^2}\;~\mathbf{\hat{\underline{r}}}\;\mathrm{d}v,$

where $$|\mathbf{r} - \mathbf{r'}|$$ is now the distance from a point in the charged body to the point at which the electric field is to be evaluated. The integral is over the charged body.

We can show that (47) is equivalent to Gauss’s Law directly from the definition of divergence,

$\boldsymbol{\nabla} \cdot \mathbf{e} = \underset{\Delta V \rightarrow 0}{lim} ~\frac{1}{\Delta V} \oint_{S} \mathbf{e}~da,$

where the integral is over $$S$$, the closed surface bounding the volume $$\Delta V$$. Applying this definition to the electric field of a point charge $$q$$ at the origin gives

$\boldsymbol{\nabla} \cdot \mathbf{e} = \underset{\Delta V \rightarrow 0}{lim} \left[ \frac{1}{\Delta V}\frac{q}{4\pi\varepsilon_0 |\mathbf{r} - \mathbf{r'}|^2} \oint_{S} ~da \right].$

Taking $$\Delta V$$ as a closed sphere of radius $$|\mathbf{r} - \mathbf{r'}|$$ centered at the origin, we can easily evaluate the integral, giving

\begin{align}\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{e} &= \underset{\Delta V \rightarrow 0}{lim} \left[ \frac{1}{\Delta V} \frac{4 \pi |\mathbf{r} - \mathbf{r'}|^2\;q }{4\pi\varepsilon_0 |\mathbf{r} - \mathbf{r'}|^2} \right ]\\~ &= \underset{\Delta V \rightarrow 0}{lim} \left[ \frac{1}{\Delta V} \frac{q}{\varepsilon_0} \right ].\end{aligned}\end{align}

In the limit $$\Delta V \rightarrow 0$$, $$\frac{q}{\Delta V}$$ is simply the charge density $$\rho$$. This establishes the desired result

$\boldsymbol{\nabla} \cdot \mathbf{e} = \frac{\rho}{\varepsilon_0}.$

For a more detailed discussion, see page 36 of [Fle08]. For an alternate derivation and discussion, see pages 65-70 of [Gri99].