Coulomb's Law

The magnitude of electric force ($F$) between two point charges is directly proportional to the product of the charges ($q_1$ and $q_2$) and inversely proportional to the square of the distance between them ($r$)

$F=k\frac{\left|q_1{q}_2\right|}{r^2}$

This equation will often be written in the following form for simplification purposes in the future.

$F=\frac{1}{4\pi {ϵ}_0}\frac{\left|q_1{q}_2\right|}{r^2}$

Constants

• $k=\frac{1}{4\pi {ϵ}_0}=(10^{-7}\mathrm{N}\cdot {\mathrm{s}}^2/{\mathrm{C}}^2)c^2\approx 8.988\times 10^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2$
• $c=2.99792458\times 10^8\mathrm{m}/\mathrm{s}$ (the speed of light for above equation)
• ${ϵ}_0\approx 8.854\times 10^{-12}{\mathrm{C}}^2/(\mathrm{N}\cdot {\mathrm{m}}^2)$
• $e=1.6021765314\times 10^{-19}\mathrm{C}$ (the charge of an electron)

Example

An α particle is the nucleus of a helium atom:

• mass ($m$) = 6.64 × 10⁻²⁷ kg
• charge ($q$) = $+2e$ = 3.2 × 10⁻¹⁹ C

Compare the force of the electric repulsion between two α particles with the force of gravitational attraction between them ($F_e/F_g$)

$\begin{array}{rl}{F}_e& =\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^2}\\ F_g& =G\frac{m^2}{r^2}\\ \frac{F_e}{F_g}& =\frac{1}{4\pi {ϵ}_{0}G}\frac{q^2}{m^2}=3.1\times 10^{35}\end{array}$

As stated in the footnotes, the gravitational force is negligible, hence the huge number from the resulting ratio.