= Coulomb's law
{c}
{title2=$\mathbf F_{2\leftarrow1}=kq_1q_2\mathbf r/|\mathbf r|^3$}
{wiki}
<Coulomb's law> gives the mutual <electric force> between stationary <point charges> in <vacuum>. Put <electric charges> $q_1$ and $q_2$ at $\mathbf r_1$ and $\mathbf r_2$, and define the <displacement> \b[from charge 1 to charge 2] by $\mathbf r=\mathbf r_2-\mathbf r_1$, with separation $r=|\mathbf r|>0$ and <unit vector> $\widehat{\mathbf r}=\mathbf r/r$. In <SI> units, the <force> on charge 2 due to charge 1 is
$$
\boxed{\mathbf F_{2\leftarrow1}=kq_1q_2\frac{\mathbf r}{|\mathbf r|^3}=\frac{kq_1q_2}{r^2}\widehat{\mathbf r},\qquad k=\frac{1}{4\pi\varepsilon_0}.}
$$
Here $\varepsilon_0$ is the <vacuum permittivity>. The magnitude is $k|q_1q_2|/r^2$. If $q_1q_2>0$, the <force> on charge 2 points away from charge 1; if $q_1q_2<0$, it points toward charge 1. Interchanging the charges gives $\mathbf F_{1\leftarrow2}=-\mathbf F_{2\leftarrow1}$, consistent with <Newton's third law> in this electrostatic setting. https://openstax.org/books/university-physics-volume-2/pages/5-3-coulombs-law[OpenStax, University Physics II, §5.3: Coulomb's law]{ref}.
For a fully computed example, stipulate $q_1=+2\,\mathrm{nC}$, $q_2=-3\,\mathrm{nC}$, $\mathbf r_1=\mathbf0$, $\mathbf r_2=0.20\,\mathrm m\,\widehat{\mathbf x}$ and the approximate value $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$. Then
$$
\mathbf F_{2\leftarrow1}=\frac{(8.99\times10^9)(2\times10^{-9})(-3\times10^{-9})}{(0.20)^2}\,\mathrm N\,\widehat{\mathbf x}=-1.3485\times10^{-6}\,\mathrm N\,\widehat{\mathbf x}.
$$
The <force> is attractive, of magnitude $1.3485\times10^{-6}\,\mathrm N$. The source <electric field> at charge 2 is $\mathbf E_1(\mathbf r_2)=449.5\,\mathrm{N/C}\,\widehat{\mathbf x}$, pointing away from the positive source. Multiplication by the negative $q_2$ reverses the direction to give the attractive <force>. This field excludes charge 2's own self-field.
Holding the <electric charges> fixed while changing the separation from $R$ to $2R$ or $3R$ gives the normalized <inverse-square law> values
$$
\frac{|\mathbf F(R)|}{|\mathbf F(R)|}=1,\qquad \frac{|\mathbf F(2R)|}{|\mathbf F(R)|}=\frac14,\qquad \frac{|\mathbf F(3R)|}{|\mathbf F(R)|}=\frac19.
$$
For the stipulated example, separations $0.20$, $0.40$ and $0.60\,\mathrm m$ give magnitudes $1.3485\times10^{-6}$, $3.37125\times10^{-7}$ and $1.498333\ldots\times10^{-7}\,\mathrm N$. The diagram computes these ratios and uses one common arrow-length scale for both sign choices.
\Image[/coulomb-force-distance.png]
{title=Equal and opposite Coulomb force vectors: like charges repel, unlike charges attract, and doubling or tripling separation gives one quarter or one ninth of the force}
The <inverse-square law> has a geometric explanation through <Gauss's law>: a source <point charge> gives fixed total <electric flux> through a concentric <sphere>, whose <surface area> is $4\pi r^2$. Dividing that flux by the area gives a radial <electric field> proportional to $1/r^2$. This connects a pairwise <force> calculation to a global conservation-like accounting of field sources. For several stationary <point charges>, the <superposition principle> adds the individual <electric fields> as <vectors>, and $\mathbf F_i=q_i\mathbf E_{\mathrm{other}}(\mathbf r_i)$ uses all sources except charge $i$ itself. The <Coulomb potential energy> gives an alternative route when the desired quantity is <work> rather than instantaneous <force>.
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