V = kQ/r — Electric Potential of a Point Charge

Electric potential (voltage) around a point charge — with a worked example.

V = \frac{kQ}{r}

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

What it means

Electric potential is the potential energy per unit charge at a point — voltage is just a potential difference. Unlike the field (1/r²), potential falls off as 1/r, and because it is a scalar, potentials from many charges simply add up.

Variables

Symbol	Name	Unit	Meaning
$V$	Electric potential	V (volts)	Potential energy per unit charge
$k$	Coulomb constant	N·m²/C²	Coulomb's constant
$Q$	Source charge	C	Charge creating the potential
$r$	Distance	m	Distance from the charge

Assumptions

Point charge Q
Potential defined as zero at infinity
Vacuum or air

Worked example

Problem

What is the electric potential 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: $V = kQ/r$
Substitute: $V = 8.99\times10^9 \times 2\times10^{-6} / 0.50$
Calculate: $V = 17{,}980 / 0.50 = 35{,}960\ \mathrm{V} \approx 3.60\times10^4\ \mathrm{V}$

≈ 3.60×10⁴ V

The potential half a metre from the charge is about 36,000 V — far above household voltage. Potential is a scalar (no direction), which makes it easier to work with than the field vector.