Coulomb's Law of Electrostatics/SI Units

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Physical Law

Coulomb's Law of Electrostatics gives the force between two stationary charged particles $a$ and $b$ as:

$\mathbf F_{a b} \propto \dfrac {q_a q_b {\mathbf r_{a b} } } {r^3}$

where:

$q_a$ and $q_b$ are the charges on $a$ and $b$ respectively
$\mathbf F_{a b}$ is the force exerted on $b$ by the electric charge on $a$
$\mathbf r_{a b}$ is the displacement vector from $a$ to $b$
$r$ is the distance between $a$ and $b$.
the constant of proportion is defined as being positive.


In SI units, the law becomes:

$\mathbf F_{a b} = \dfrac 1 {4 \pi \varepsilon_0} \dfrac {q_a q_b {\mathbf r_{a b} } } {r^3}$

where:

$q_a$ and $q_b$ are measured in coulombs
$r$ is measured in metres
$\mathbf F_{a b}$ is measured in newtons
$\varepsilon_0$ denotes the vacuum permittivity:
$\varepsilon_0 \approx 8 \cdotp 85418 \, 78128 (13) \times 10^{-12} \, \mathrm C^2 \, \mathrm N^{-1} \, \mathrm m^{-2}$


Also presented as

Coulomb's Law of Electrostatics expressed in SI units can also be presented as:

$(1): \quad \mathbf F_{a b} = \dfrac 1 {4 \pi \varepsilon_0} \dfrac {q_a q_b \hat {\mathbf r}_{a b} } {r^2}$
where $\hat {\mathbf r}_{a b}$ is the unit vector in the direction from $a$ to $b$.
$(2): \quad \mathbf F_{a b} = \dfrac 1 {4 \pi \varepsilon_0} \dfrac {q_a q_b} {\size {\mathbf r_a - \mathbf r_b}^3} \paren {\mathbf r_a - \mathbf r_b}$
where $\mathbf r_a$ and $\mathbf r_b$ are the position vectors of $q_a$ and $q_b$ respectively.


Sources