information:classification, linguistic:paraphrasing
Electrostatics

In the special case of a steady state (stationary charges and currents), the Maxwell-Faraday inductive effect disappears. The resulting two equations (Gauss's law ∇ ⋅ E = ρ ε 0 {\displaystyle \nabla \cdot \mathbf {E} ={\frac {\rho }{\varepsilon _{0}}}} and Faraday's law with no induction term ∇ × E = 0 {\displaystyle \nabla \times \mathbf {E} =0}), taken together, are equivalent to Coulomb's law, which states that a particle with electric charge q 1 {\displaystyle q_{1}} at position r 1 {\displaystyle \mathbf {r} _{1}} exerts a force on a particle with charge q 0 {\displaystyle q_{0}} at position r 0 {\displaystyle \mathbf {r} _{0}} of: F 01 = q 1 q 0 4 π ε 0 r ^ 01 | r 01 | 2 = q 1 q 0 4 π ε 0 r 01 | r 01 | 3 {\displaystyle \mathbf {F} _{01}={\frac {q_{1}q_{0}}{4\pi \varepsilon _{0}}}{{\hat {\mathbf {r} }}_{01} \over {|\mathbf {r} _{01}|}^{2}}={\frac {q_{1}q_{0}}{4\pi \varepsilon _{0}}}{\mathbf {r} _{01} \over {|\mathbf {r} _{01}|}^{3}}} where

Note that ε 0 {\displaystyle \varepsilon _{0}} must be replaced with ε {\displaystyle \varepsilon }, permittivity, when charges are in non-empty media.When the charges q 0 {\displaystyle q_{0}} and q 1 {\displaystyle q_{1}} have the same sign this force is positive, directed away from the other charge, indicating the particles repel each other. When the charges have unlike signs the force is negative, indicating the particles attract.To make it easy to calculate the Coulomb force on any charge at position r 0 {\displaystyle \mathbf {r} _{0}} this expression can be divided by q 0 {\displaystyle q_{0}} leaving an expression that only depends on the other charge (the source charge) E 1 (r 0) = F 01 q 0 = q 1 4 π ε 0 r ^ 01 | r 01 | 2 = q 1 4 π ε 0 r 01 | r 01 | 3 {\displaystyle \mathbf {E} _{1}(\mathbf {r} _{0})={\frac {\mathbf {F} _{01}}{q_{0}}}={\frac {q_{1}}{4\pi \varepsilon _{0}}}{{\hat {\mathbf {r} }}_{01} \over {|\mathbf {r} _{01}|}^{2}}={\frac {q_{1}}{4\pi \varepsilon _{0}}}{\mathbf {r} _{01} \over {|\mathbf {r} _{01}|}^{3}}} where

This is the electric field at point r 0 {\displaystyle \mathbf {r} _{0}} due to the point charge q 1 {\displaystyle q_{1}}; it is a vector-valued function equal to the Coulomb force per unit charge that a positive point charge would experience at the position r 0 {\displaystyle \mathbf {r} _{0}}.Since this formula gives the electric field magnitude and direction at any point r 0 {\displaystyle \mathbf {r} _{0}} in space (except at the location of the charge itself, r 1 {\displaystyle \mathbf {r} _{1}}, where it becomes infinite) it defines a vector field.From the above formula it can be seen that the electric field due to a point charge is everywhere directed away from the charge if it is positive, and toward the charge if it is negative, and its magnitude decreases with the inverse square of the distance from the charge.

The Coulomb force on a charge of magnitude q {\displaystyle q} at any point in space is equal to the product of the charge and the electric field at that point F = q E. {\displaystyle \mathbf {F} =q\mathbf {E}.} The SI unit of the electric field is the newton per coulomb (N/C), or volt per meter (V/m); in terms of the SI base units it is kg⋅m⋅s ⋅A.