A size from the formula, a direction from the picture

Three facts about charge carry the whole unit, so state them plainly.

1. Two signs. Like charges repel, unlike charges attract. Neutral is not a third kind: it is equal amounts of both.
2. Conserved. Charging moves charge, it never makes charge. A rod that ends at $-4$ nC took those electrons from something that is now at $+4$ nC.
3. Quantized. Every free charge is an integer multiple of $e = 1.6\times10^{-19}$ C. A reported 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 one distinction decides the outcome of nearly every charging or grounding scenario in the next topic, so read the material before predicting anything.

For two point charges separated by $r$,

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

Feed signed charges in and you get a signed number out, and that sign is not an axis. It does not know where you drew $+x$. Slide one charge to the other side of the origin and the arithmetic is unchanged while the true direction reverses, which is the cleanest proof that the sign was never carrying direction.

The reliable procedure has two separate steps:

1. Compute the magnitude from $|q_1||q_2|$. It is always positive.
2. Draw the arrow from the words attract or repel, along the line joining the two charges.

With three or more charges this discipline is what keeps superposition from scrambling: each pairwise arrow gets resolved into components and added as a vector.

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

The guard worth installing now is the ladder, because later in this same unit you meet quantities that really do go as $1/r$:

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

A ratio question answered with the wrong power of two is the visible symptom, and it runs in both directions: force scaled as $1/r$, or potential scaled as $1/r^2$. Write the ratio $F_2/F_1$ explicitly and the exponent has nowhere to hide.

Put a $6\ \mu$C sphere beside a $1\ \mu$C sphere. The two forces are equal in magnitude and opposite in direction. The formula says so directly: $F = k|q_1||q_2|/r^2$ 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 the car again, relocated.

A second scale question is worth settling in the same breath. Between a proton and an electron the electric force beats the gravitational force by about $10^{39}$, and that comparison is correct. It does not extend to planets. Charge comes in two signs and cancels almost perfectly in bulk matter, so nothing electric survives at astronomical distances, while mass comes in one sign and simply accumulates.

Strength per pair and dominance in bulk are separate questions. Electricity wins the first and gravity wins the second.