Formula: Coulomb's Law for Two Point Charges Force Electric charge Distance Relative permittivity
$$F_{\text e} ~=~ \frac{1}{4\pi \varepsilon_0 \, \varepsilon_{\text r}} \, \frac{\class{red}{q_1} \, \class{blue}{q_2}}{r^2}$$
$$F_{\text e} ~=~ \frac{1}{4\pi \varepsilon_0 \, \varepsilon_{\text r}} \, \frac{\class{red}{q_1} \, \class{blue}{q_2}}{r^2}$$
$$\class{red}{q_1} ~=~ 4\pi \varepsilon_0 \, \varepsilon_{\text r} \, \frac{r^2 \, F_{\text e}}{\class{blue}{q_2}}$$
$$\class{blue}{q_2} ~=~ 4\pi \varepsilon_0 \, \varepsilon_{\text r} \, \frac{r^2 \, F_{\text e}}{\class{red}{q_1}}$$
$$r ~=~ \sqrt{ \frac{1}{4\pi \varepsilon_0 \, \varepsilon_{\text r}} \, \frac{\class{red}{q_1} \, \class{blue}{q_2}}{F_{\text e}} }$$
$$\varepsilon_{\text r} ~=~ \frac{1}{4\pi \varepsilon_0} \, \frac{\class{red}{q_1} \, \class{blue}{q_2}}{r^2 \, F_{\text e}}$$
$$\varepsilon_0 ~=~ \frac{1}{4\pi \varepsilon_{\text r}} \, \frac{\class{red}{q_1} \, \class{blue}{q_2}}{r^2 \, F_{\text e}}$$
Force
$$ F_{\text e} $$ Unit $$ \mathrm{N} = \frac{\mathrm{kg} \, \mathrm{m}}{\mathrm{s}^2} $$
Electrostatic force (also called Coulomb force) is the attractive or repulsive electric force between two charges \( q_1 \) and \( q_2 \).
Electric charge
$$ \class{red}{q_1} $$ Unit $$ \mathrm{C} = \mathrm{As} $$
This charge is the property of the first charge carrier participating in the electrical interaction. Depending on the sign of the charge, the charge carrier repels or attracts other charge carriers. A proton (positive sign) attracts an electron (negative sign).
Electric charge
$$ \class{blue}{q_2} $$ Unit $$ \mathrm{C} = \mathrm{As} $$
This charge is the property of the second charge carrier that participates in the electrical interaction.
Distance
$$ r $$ Unit $$ \mathrm{m} $$
The distance between the charges \( q_1 \) and \( q_2 \). The larger this distance is, the smaller is the electrostatic force between the charges.
Relative permittivity
$$ \varepsilon_{\text r} $$ Unit $$ - $$
This dimensionless quantity describes the medium in which the two charges are located. If the two charges are in vacuum then \( \varepsilon_{\text r} = 1 \). And, if they are in water, for example, then \( \varepsilon_{\text r} = 80 \). The greater the relative permittivity of the medium, the more this medium weakens the force between the charges.
Vacuum Permittivity
$$ \varepsilon_0 $$ Unit $$ \frac{\mathrm{As}}{\mathrm{Vm}} $$
The vacuum permittivity is a physical constant that appears in equations involving electromagnetic fields. It has the following experimentally determined value:
$$ \varepsilon_0 ~\approx~ 8.854 \, 187 \, 8128 ~\cdot~ 10^{-12} \, \frac{\mathrm{As}}{\mathrm{Vm}} $$