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:

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

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

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:

Electric field pulled out of the surface integral

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

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

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

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

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:

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 \( Q \):

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