Two charges at rest push or pull on each other, and the size of that push is Coulomb's law: $F = \dfrac{k_e q_1 q_2}{r^2}$ — read aloud F equals k-e q-one q-two over r squared. Name every letter before it works for you. $F$ is the electrostatic force in newtons. $q_1$ and $q_2$ are the two charges in coulombs — the subscripts simply label them, one for each charge, and swapping which is which changes nothing, because they multiply. $r$ is the centre-to-centre separation in metres. And $k_e$ is Coulomb's constant, $8.99\times 10^{9}\ \mathrm{N\cdot m^2/C^2}$, which this chapter prints as $9.00\times 10^{9}$ the way every first-year text does. In this lesson the two charges and the separation are always given, and $F$ is what you are solving for.

The $r^2$ is the headline. Double the separation and the force falls to a QUARTER, not a half. Halve it and the force quadruples. Gravity does exactly the same thing with masses, which is not a coincidence — it is what a force spreading out over the surface of a sphere always looks like.

The real difficulty in this lesson is not the physics, it is the prefixes. A whole coulomb is a monstrous amount of charge; benchwork lives in microcoulombs, $1\ \mathrm{\mu C} = 10^{-6}\ \mathrm{C}$. Two of them multiplied together carry $10^{-12}$, and that tiny factor is the only thing standing between $k_e$'s nine billion and an answer you could hold in your hand. Convert both charges before you touch the calculator, every single time.