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
- 1990: I.S. Grant and W.R. Phillips: Electromagnetism (2nd ed.) ... (previous) ... (next): Chapter $1$: Force and energy in electrostatics: $1.1$ Electric Charge