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.

$F={\frac {1}{4\pi \epsilon _{0}}}\,{\frac {\left|q_{1}q_{2}\right|}{r^{2}}}$

Constants

• $k={\frac {1}{4\pi \epsilon _{0}}}=(10^{-7}\ \mathrm {N} \cdot \mathrm {s} ^{2}/\mathrm {C} ^{2})c^{2}\approx 8.988\times 10^{9}\ \mathrm {N} \cdot \mathrm {m} ^{2}/\mathrm {C} ^{2}$
• $c=2.99792458\times 10^{8}\ \mathrm {m} /\mathrm {s}$ (the speed of light for above equation)
• $\epsilon _{0}\approx 8.854\times 10^{-12}\ \mathrm {C} ^{2}/(\mathrm {N} \cdot \mathrm {m} ^{2})$
• $e=1.6021765314\times 10^{-19}\ \mathrm {C}$ (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 ($F_{e}/F_{g}$)

${\begin{aligned}F_{e}&={\frac {1}{4\pi \epsilon _{0}}}\,{\frac {q^{2}}{r^{2}}}\\F_{g}&=G\,{\frac {m^{2}}{r^{2}}}\\{\frac {F_{e}}{F_{g}}}&={\frac {1}{4\pi \epsilon _{0}G}}\,{\frac {q^{2}}{m^{2}}}=3.1\times 10^{35}\end{aligned}}$

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