D.2.2—Coulomb’s law

Use the inverse-square law

For two point charges, $r$ is their centre-to-centre separation. In a medium of permittivity $\varepsilon$, $k=1/(4\pi\varepsilon)$; in vacuum, $k=8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}}$. Calculate magnitude, then use charge signs to state attraction or repulsion.

Worked example — unlike charges in air

For $q_1=4.5\times10^{-8}\,\mathrm{C}$, $q_2=-1.3\times10^{-7}\,\mathrm{C}$ and $r=3.2\times10^{-2}\,\mathrm{m}$, $F=(8.99\times10^9)|q_1q_2|/r^2=5.1\times10^{-2}\,\mathrm{N}$. The force is attractive because the charges have opposite signs.

Read the scaling

Doubling either charge doubles the force. Doubling the separation reduces the force to one quarter. If the medium has permittivity $\varepsilon$ rather than $\varepsilon_0$, use $k=1/(4\pi\varepsilon)$; greater permittivity reduces the force for the same charges and separation.

Choose the point-charge model

Spherical charged bodies can be treated as point charges at their centres when the geometry permits. Use centre-to-centre separation and convert charge units, such as microcoulombs, before substitution. For several charges, calculate each force vector and add them.

Common trap

Do not use diameter or a single radius as r, and do not forget that a change in separation is squared. Keep the force magnitude positive in the calculation, then state attraction or repulsion separately.