# Derivation Coulomb's Law using 1st Maxwell Equation

## Video - Maxwell Equations. Here You Get the Deepest Intuition!

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

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

**Divergence theorem**

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:

**Coulomb's law as vector equation**

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 \(\nabla ~\cdot~ \boldsymbol{E} ~=~ \frac{\rho}{\varepsilon_0}\):

**Divergence theorem with inserted first Maxwell equation**

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:

**Surface integral of the E-field corresponds to the enclosed charge**

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*.

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 \( \text{d}\boldsymbol{a} \):

**Scalar product between E-field and surface normal**

~&=~ 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:

**Electric field pulled out of the surface integral**

The surface integral in 6

corresponds - because of the spherical symmetry - to the area of a spherical surface and is thus \( A ~=~ 4\pi \, r^2 \):

**E-field times 4Pi times radius squared is equal to charge divided by vacuum permittivity**

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

:

**Force per charge times 4pi times radius squared equals charge divided by vacuum permittivity**

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 \( Q \):

**Coulomb's law for two point charges**

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

**Coulomb's law as a vector equation**