Electric Field of a Point Charge

Take Coulomb's law, $F = k_e q_1 q_2 / r^2$, and divide out the charge you are testing with. What remains, $E = k_e Q/r^2$, belongs to the source charge alone: the field it maintains at every point in the space around it, whether or not anything is there to feel it. The field points radially outward from a positive $Q$ and radially inward toward a negative one, and it dies off with the square of distance because the same total flux is spread over a sphere whose area grows as $r^2$. That is Gauss's law in one sentence, and it is why the exponent is exactly two rather than approximately two.

Numbers give the scale. A 1 nC charge — a modest static charge, the sort a plastic comb picks up — maintains $E = (8.988 \times 10^9 \times 10^{-9})/(0.1)^2 = 899\ \text{V/m}$ at 10 cm. Halve the distance to 5 cm and the field quadruples to 3595 V/m. Take the same charge out to a metre and it is down to 9 V/m. The inverse square is unforgiving in both directions, which is why static problems are almost always about the last centimetre.

Superposition is what makes the formula useful beyond a single charge: the field of any arrangement is the vector sum of the fields of its parts. That is how the dipole, the charged ring, the infinite sheet and the line charge are all derived, and the answers are worth carrying because they behave so differently — a point charge falls as $1/r^2$, a long line as $1/r$, and an infinite sheet not at all.

The first caution is that a point charge is an idealisation and this formula inherits its worst feature: at $r = 0$ it predicts an infinite field. Nature does not do that; what it tells you is that the model has been pushed inside the object it was standing in for. For a charged sphere the formula is exact outside the sphere — with $r$ measured from its centre, by the shell theorem — and completely wrong inside, where the field of a conducting sphere is zero. Second, $k_e$ as used here is the vacuum value; immerse the charge in a dielectric and the field is reduced by the relative permittivity, roughly 80-fold in water, which is exactly why salt dissolves. Third, this is a vector and the sign of $Q$ sets its direction, so solving backwards for $r$ requires $E$ and $Q$ to agree in sign — a field pointing toward a positive charge does not exist at any distance. Fourth, a real conductor near the charge will polarise and distort the field entirely; the formula assumes empty space and nothing else in it.