My name is Alexander FufaeV and here I will explain the following topic:

Coulomb's Law using 1st Maxwell Equation

In the following, we will derive Coulomb's law from Maxwell's 1st equation and the Divergence theorem.

Divergence theorem illustrated.

The following divergence theorem works for both moving and unmoving charges:

$$ \begin{align} \int_{A} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a} ~=~ \int_{V} \nabla \cdot \boldsymbol{E} ~ \text{d}v \end{align} $$

The following equation for Coulomb's law (with a point charge $ Q $ placed at the coordinate origin) works only for non-moving (stationary) charges, or for charges moving very slowly:

$$ \begin{align} \boldsymbol{F}_{\text e} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{Q \, q}{r^2} \, \boldsymbol{\hat{e}}_{\text r} \end{align} $$

Strictly speaking, Coulomb's law 2 can be derived from divergence theorem 1 only if in addition a spherical symmetry of the electric field $ \boldsymbol{E} $ of a point charge is assumed.

The divergence of the E-field on the right-hand side of the divergence theorem 1, can be rewritten with the first Maxwell equation in differential form $\nabla ~\cdot~ \boldsymbol{E} ~=~ \frac{\rho}{\varepsilon_0}$:

$$ \begin{align} \int_{A} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a} ~=~ \frac{1}{\varepsilon_0} \int_{V} \rho ~ \text{d}v \end{align} $$

The integral of charge density $ \rho $ over volume $ V $ in 3 gives the total charge $ Q $ enclosed by that volume, because charge density is defined as charge per volume $ \rho = Q / V $. Equation 3 thus becomes:

$$ \begin{align} \int_{A} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a} ~=~ \frac{Q}{\varepsilon_0} \end{align} $$

To calculate the surface integral on the left side of 4, the scalar product $ \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a} $ must be calculated first. For this purpose, it is assumed that the electric field $ \boldsymbol{E} $ is constant at any point on the surface $ A $. This is guaranteed if a radial symmetry of the electric field is assumed. Then the electric field always points radially outward.

A point charge enclosed by an imaginary sphere. The E-field vector on the surface of the sphere is parallel to the surface element.

The direction of the E-field is represented by the radial unit vector $ \boldsymbol{\hat{e}}_{\text r} $ in spherical coordinates: $ \boldsymbol{E} ~=~ E \, \boldsymbol{\hat{e}}_{\text r} $. Here $ E := |\boldsymbol{E}| $ is the magnitude of the field vector $ \boldsymbol{E} $.

Next, we calculate the scalar product of the E-field vector with the infinitesimal surface element $ \text{d}\boldsymbol{a} $:

$$ \begin{align} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a} ~&=~ E \, \boldsymbol{\hat{e}}_{\text r} ~\cdot~ \boldsymbol{\hat{e}}_{\text r} \, \text{d}a \\\\
~&=~ E \, \text{d}a \end{align} $$

In the second step, we exploited the fact that the surface element $ \text{d}\boldsymbol{a} = \text{d}a \, \hat{\boldsymbol{a}} $ is also perpendicular to a radially symmetric surface and thus the surface orthonormal vector $ \hat{\boldsymbol{a}} = \boldsymbol{\hat{e}}_{\text r} $ is parallel to the electric field.

Let's insert 5 in 4 and pull the constant magnitude $ E $ of the electric field in front of the integral:

$$ \begin{align} E \, \int_{A} \text{d}a ~=~ \frac{Q}{\varepsilon_0} \end{align} $$

The surface integral in 6 corresponds - because of the spherical symmetry - to the area of a spherical surface and is thus $ A ~=~ 4\pi \, r^2 $:

$$ \begin{align} E \, 4 \pi \, r^2 ~=~ \frac{Q}{\varepsilon_0} \end{align} $$

The magnitude $E$ of the electric field is defined as (electric) force $F_{\text e}$ per charge $q$: $E ~=~ \frac{}{q}$. In other words: If a small charge $q$ (small, so that it does not disturb too much the field $ E $ with its own E-field) is placed somewhere at a place where the electric field $E$ is present. Then this charge experiences an electric force $ F_{\text e} $.

Insert the E-field magnitude into Eq. 7:

$$ \begin{align} \frac{F_{\text e}}{q} \, 4 \pi \, r^2 ~=~ \frac{Q}{\varepsilon_0} \end{align} $$

Rearrange eq. 8 for the electric force. Then you get the Coulomb's law you are looking for, i.e. the magnitude of the electric force for a point charge $ q $ in an electric field caused by another charge $ Q $:

$$ \begin{align} F_{\text e} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{Q \, q}{r^2} \end{align} $$

Together with the radial unit vector $\boldsymbol{\hat{e}}_{\text r}$ you can write Coulomb's law as a vector equation:

$$ \begin{align} \boldsymbol{F}_{\text e} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{Q \, q}{r^2} \, \boldsymbol{\hat{e}}_{\text r} \end{align} $$