E = kQ/r² — Electric Field of a Point Charge

Electric field strength around a point charge — with a worked microcoulomb example.

E = \frac{kQ}{r^2}

Electric field equals Coulomb's constant times charge divided by distance squared.

What it means

A charge $Q$ fills the space around it with an electric field that falls off as 1/r² — the same inverse-square pattern as gravity. The field is the force a unit positive test charge would feel; field lines point away from positive charges and toward negative ones.

Variables

Symbol	Name	Unit	Meaning
$E$	Electric field	N/C or V/m	Force per unit charge at the point
$k$	Coulomb constant	N·m²/C²	Coulomb's constant
$Q$	Source charge	C	Charge creating the field
$r$	Distance	m	Distance from the charge

Assumptions

Point charge Q (or r large vs. charge size)
Vacuum or air (k = 8.99×10⁹ N·m²/C²)
Field measured at distance r from the charge center

Worked example

Problem

What is the electric field 0.50 m from a 2 µC point charge?

Given:

Q = 2 µC = 2×10⁻⁶ C
r = 0.50 m
k = 8.99×10⁹ N·m²/C²

Write the formula: $E = kQ/r^2$
Substitute: $E = 8.99\times10^9 \times 2\times10^{-6} / 0.50^2$
Calculate: $E = 17{,}980 / 0.25 = 71{,}920\ \mathrm{N/C} \approx 7.19\times10^4\ \mathrm{N/C}$

≈ 7.19×10⁴ N/C

A 2-microcoulomb charge creates a field of about 72,000 N/C half a metre away — a test charge of 1 C would feel 72,000 N. The 1/r² falloff means doubling the distance quarters the field.