Coulomb's Law

Coulomb's law is equivalent to the statement that the electric field ${\bf E}({\bf r})$ generated by a point charge $q'$ located at ${\bf r}={\bf r}'$ is

$${\bf E}({\bf r}) = \frac{q'}{4\pi\,\epsilon_0}\,\frac{{\bf r}-{\bf r}'}{\vert{\bf r}-{\bf r}'\vert^{\,3}}.$$ (148)

The electric force ${\bf F}$ exerted on a point charge $q$ located at position vector ${\bf r}$ is

$${\bf F} = q\,{\bf E}({\bf r}).$$ (149)

Hence,

$${\bf F} = \frac{q\,q'}{4\pi\,\epsilon_0} \,\frac{{\bf r}-{\bf r}'}{\vert{\bf r}-{\bf r}'\vert^{\,3}}.$$ (150)

It follows that the electrostatic force acting between two point charges is inverse-square, central, proportional to the product of the charges, and repulsive if both charges are of the same sign.

Electric fields are superposable (see Section 1.2), which means that the electric field generated by $N$ point charges, $q_i$ , located at position vectors ${\bf r}_i$ , for $i=1, N$ , is

$${\bf E}({\bf r}) = \sum_{i=1,N}\frac{q_i}{4\pi\,\epsilon_0}\,\frac{{\bf r}-{\bf r}_i}{\vert{\bf r}-{\bf r}_i\vert^{\,3}}.$$ (151)

In the continuum limit, the previous expression becomes

$${\bf E}({\bf r}) = \frac{1}{4\pi\,\epsilon_0}\int \frac{\rho({\bf r'})\,({\bf r}-{\bf r}')}{\vert{\bf r}-{\bf r}'\vert^{\,3}}\,dV',$$ (152)

where $\rho({\bf r})$ is the charge density (i.e., the electric charge per unit volume), and the integral is over all space.