Coulomb's Law — Physica

Equivalent forms

F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1 q_2|}{r^2}

\vec{F} = k_e \frac{q_1 q_2}{r^2} \hat{r}

The electric cousin of Newton's gravitational law — same inverse-square form, but 10³⁶ times stronger and comes in two polarities.

Symbol	Name	SI unit	Dimension	Range
$F$	Electrostatic forceoutput Magnitude of the force between two point charges	N	M·L·T⁻²	0 – 100
$q₁$	Charge 1 First point charge	C	I·T	-0.00001 – 0.00001
$q₂$	Charge 2 Second point charge	C	I·T	-0.00001 – 0.00001
$r$	Separation distance Distance between the two point charges	m	L	0.01 – 1

Unit systems

Symbol	SI	CGS	Imperial
$F$	N	dyn	lbf
$q$	C	statC	C
$r$	m	cm	ft

Where it holds

Valid for point charges or spherically symmetric charge distributions in vacuum. Breaks down at subatomic distances where quantum electrodynamics applies. In media, replace

\varepsilon _{0}

with

\varepsilon _{0}\varepsilon

ᵣ.

Dimensional analysis

$[F] = [k_e]\cdot [q]^{2}\cdot [r]^{-2} \to (N\cdot m^{2}\cdot C^{-2})(C^{2})(m^{-2}) = N \checkmark$

Discovery

Charles-Augustin de Coulomb · 1785

Coulomb used a torsion balance to precisely measure the force between charged spheres, proving it follows an inverse-square law analogous to gravity.

Try this

Why does a balloon stick to the wall after you rub it on your hair?

Two point charges of +3 μC and −5 μC are separated by 0.2 m. Find the magnitude and direction of the electrostatic force.

Research status: stable

Real-world applications

Electrostatic precipitators in smokestacks
Xerographic printing and photocopiers
Van de Graaff generators in physics demonstrations
Modeling molecular bonds in computational chemistry

Common misconceptions

Coulomb's law applies only to point charges or spherical distributions, not arbitrary shapes
The force is along the line joining the charges — it has no transverse component
The law assumes static charges; moving charges require the full Lorentz force

Experimental verification

Coulomb's torsion balance (1785) established the inverse-square dependence. Modern Cavendish-type experiments confirm the exponent is 2 to within

\pm 10^{-16}

. Equivalently tested via Gauss's law and absence of charge inside a hollow conductor.

Derivation

Experimentally determined using a torsion balance.

The constant

k_e = 1/(4\pi \varepsilon _{0}) = 8.9875 \times 10^{9} N\cdot m^{2}/C^{2}

.

No derivation from first principles in classical physics — it is a fundamental empirical law.

Limiting cases

r \to 0

⟶

F \to \infty

Force diverges at zero separation — in reality, quantum effects dominate at atomic scales.

r \to \infty

⟶

F \to 0

Force vanishes at large distances due to inverse-square falloff.

q_{1} or q_{2} \to 0

⟶

F \to 0

No charge means no electrostatic interaction.

What if…

What if the distance doubles?

Force drops to 1/4 of the original value. At $r = 0.4\,\mathrm{m}$ , $F = 3.37/4 = 0.843\,\mathrm{N}$ — inverse-square dependence.

What if one charge triples?

Force triples. $F = 3 \times 3.37 = 10.1\,\mathrm{N}$ . Force scales linearly with each charge.

What if a dielectric

(\kappa = 4)

fills the space?

Force reduces by factor $\kappa$ : $F = 3.37/4 = 0.843\,\mathrm{N}$ . The medium's polarization partially screens the charges.

1

Force between two charged spheres

Given ·

q1:: 0.000003
q2:: -0.000005
r:: 0.2

Find · F

Steps

Identify: $q_{1} = +3 \mu C = 3\times 10^{-6} C$ , $q_{2} = -5 \mu C = -5\times 10^{-6} C$ , $r = 0.2\,\mathrm{m}$
Apply Coulomb's law: $F = k_e \times |q_{1}q_{2}| / r^{2}$
Compute numerator: $8.9875\times 10^{9} \times 15\times 10^{-12} = 0.1348 N\cdot m^{2}$
Divide by $r^{2}$ : $0.1348 / 0.04 = 3.37\,\mathrm{N}$
Opposite signs → force is attractive, along the line joining the charges

Result ·

F = k_e \times |q_{1}q_{2}| / r^{2} = 8.9875\times 10^{9} \times |3\times 10^{-6} \times 5\times 10^{-6}| / 0.2^{2} = 3.37\,\mathrm{N}

(attractive)

2

Equilibrium position for a third charge

Given ·

q1:: 0.000004
q2:: 0.000016
d:: 0.3

Find · x (position where net force on a test charge is zero)

Steps

Place $q_{1} at$ origin and $q_{2} at x = 0.3\,\mathrm{m}$ . Test charge at distance x from $q_{1}$
Set forces equal: $k_e \times q_{1}/x^{2} = k_e \times q_{2}/(0.3 - x)^{2}$
Cancel k_e: $4\times 10^{-6}/x^{2} = 16\times 10^{-6}/(0.3 - x)^{2}$
Simplify: $(0.3 - x)^{2}/x^{2} = 4 \to (0.3 - x)/x = 2$
Solve: $0.3 - x = 2x \to x = 0.1\,\mathrm{m}$

Result ·

x = 0.1\,\mathrm{m}

from

q_{1}

(one-third of the way from the smaller charge)

Electric Field of a Point Charge→Gauss's Law→Electric Potential (Point Charge)→