The force between two charges

Three facts do a lot of work later, so state them plainly now.

1. Two signs. Like charges repel, unlike charges attract. There is no third kind, and neutral is not a kind: it is equal amounts of both.
2. Conserved. Charging never creates charge; it moves it. Rub a rod with fur and the rod goes to $-4$ nC exactly when the fur goes to $+4$ nC. The pair still sums to what it started at.
3. Quantized. Every free charge is an integer multiple of $e = 1.6\times10^{-19}$ C. A measured charge of $2.4\times10^{-19}$ C is not a small charge; it is an impossible one.

Conductors let charge move through the material; insulators hold it where it was put. That single distinction decides what happens in nearly every charging or grounding scenario, so read the material before predicting the redistribution.

The magnitude of the force between two point charges separated by $r$ is

$$F = \frac{k|q_1||q_2|}{r^2}, \qquad k = \frac{1}{4\pi\varepsilon_0} = 8.99\times10^9 \ \text{N}\cdot\text{m}^2/\text{C}^2.$$

The absolute value bars are not decoration. If you feed signed charges in, you get a signed number out, and that sign is not an axis: it does not know where you drew $+x$. A student who writes $F = k(-2\,\mu\text{C})(+3\,\mu\text{C})/r^2$, gets a negative number, and concludes "the force points in the $-x$ direction" has invented a direction the algebra never contained.

The reliable procedure is two separate steps:

1. Compute the magnitude with $|q_1||q_2|$. It is always positive.
2. Draw the arrow from the physics: unlike charges attract (arrows point toward each other), like charges repel (arrows point apart). The arrow lies along the line joining the two charges, always.

All the sign information you need is already in the words "attract" and "repel". Once there are three or more charges, this discipline is what keeps the superposition from scrambling, because then each pairwise arrow gets resolved into components and added as a vector.

Coulomb's constant $k$ and the permittivity of free space $\varepsilon_0$ are two ways of writing the same constant:

$$k = \frac{1}{4\pi\varepsilon_0}, \qquad \varepsilon_0 = 8.85\times10^{-12} \ \text{C}^2/(\text{N}\cdot\text{m}^2).$$

So the force law can be written either way:

$$F = \frac{k|q_1||q_2|}{r^2} \qquad \text{or} \qquad F = \frac{|q_1||q_2|}{4\pi\varepsilon_0 r^2}.$$

What is never correct is both at once. Writing $F = \dfrac{k|q_1||q_2|}{4\pi\varepsilon_0 r^2}$ double-counts the constant and lands about eleven orders of magnitude off, which students then hunt for as an arithmetic slip. Pick one costume per equation. Gauss's law is where $\varepsilon_0$ becomes the natural choice, so both forms are worth being fluent in.

The $r^2$ in the denominator is the single most-tested scaling in the unit. Doubling the separation does not halve the force; it quarters it. Tripling it divides by nine.

A useful guard: later you will meet potential energy $U = kq_1q_2/r$ and potential $V = kq/r$, which really do go as $1/r$. Keep the ladder straight from the start.

$$F \ \text{and} \ E \ \sim \frac{1}{r^2}, \qquad U \ \text{and} \ V \ \sim \frac{1}{r}.$$

Finally, the third law holds here with no exceptions. Put a $5\ \mu$C charge next to a $1\ \mu$C charge and the two forces are equal in magnitude and opposite in direction. This surprises people, so look at the formula: $F = k|q_1||q_2|/r^2$ contains both charges symmetrically, so there is no way for it to produce a different number depending on which charge you ask about. What genuinely differs is the acceleration, since that divides by each object's own mass. It is the truck-and-car collision again, relocated to charges.