Coulomb's Law

Two point charges exert forces on each other proportional to the product of the charges and inversely proportional to the square of their separation: $F=\dfrac{Q_1Q_2}{4\pi\varepsilon_0r^2}$ , with $\varepsilon_0=8.85\times10^{-12}$ F m⁻¹. Like charges repel; unlike attract. A point charge (or charged sphere, from outside) creates a radial field $E=\dfrac Q{4\pi\varepsilon_0r^2}$ . Fields from several charges add as vectors.

Cambridge 9702 Paper 4Edexcel YPH11 WPH14AQA 7408 Paper 2≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use Coulomb's law for point charges and charged spheres.
Use $E=\dfrac Q{4\pi\varepsilon_0r^2}$ for a radial field.
Find the resultant field from two charges along a line.
Compare electric and gravitational forces.

Pause and think

Both gravity and electric force follow an inverse-square law. Why does gravity dominate the motion of planets, while electric forces dominate inside atoms?

Section 2 of 14

Core idea

Start hereStronger than gravity by 10³⁹

The electric attraction between a proton and an electron is about $10^{39}$ times their gravitational attraction. Yet we notice gravity more, because large objects are almost perfectly neutral. Both forces follow the same inverse-square pattern.

The big idea

$F=\frac{Qq}{4\pi\varepsilon_0r^2}$ ; point-charge field $E=\frac Q{4\pi\varepsilon_0r^2}$ . Like charges repel.

1. The law

Field lines: a positive and a negative charge

Lines start on positive charge and end on negative. A small positive test charge would move along them.

F=\frac{Qq}{4\pi\varepsilon_0r^2}\qquad E=\frac Q{4\pi\varepsilon_0r^2}

$\frac1{4\pi\varepsilon_0}$: $8.99\times10^9$ N m² C⁻².
$r$: distance between centres.

Worked exampleTwo +2.0 μC charges, 0.30 m apart

Force; field from one at the other.

Goal: $F$ , $E$ .
Strategy: Coulomb's law; $E=F/q$ .

$F=\frac{8.99\times10^9(2\times10^{-6})^2}{0.30^2}=0.40\text{ N}$
Repulsive.
$E=\frac{0.40}{2\times10^{-6}}=2.0\times10^5\text{ N C}^{-1}$
Field.

Answer

0.40

2.0\times10^5

N C

^{-1}

✓ Check

E=\frac{8.99\times10^9\times2\times10^{-6}}{0.09}=2.0\times10^5

✓.

Worked exampleHydrogen atom

$r=5.3\times10^{-11}$ m.

Goal: Electric force; comparison with gravity.
Strategy: Ratio of the two laws.

$F_e=\frac{8.99\times10^9(1.6\times10^{-19})^2}{(5.3\times10^{-11})^2}=8.2\times10^{-8}\text{ N}$
Coulomb.
$F_g=\frac{Gm_pm_e}{r^2}\approx3.6\times10^{-47}\text{ N}$
Newton.

AnswerRatio ≈

2\times10^{39}

✓ CheckGravity negligible in atoms ✓.

Force between two charges falls as 1/r²

Double the distance, a quarter of the force; triple it, a ninth.

Common misconceptionForgetting to square r Inverse square: tripling

r

gives

\frac19

Summary

Inverse square.
$E=Q/(4\pi\varepsilon_0r^2)$ .

Permittivity: $\varepsilon_0$ : sets the strength of electric forces.

HistoryCoulomb's torsion balance. In

1785

Charles-Augustin de Coulomb measured the force between small charged spheres using a delicate torsion balance: a thin wire twisted by the force. He showed the force fell with the square of distance, the same law Newton found for gravity a century earlier.

Fun factYou never touch anything. When you "touch" a table, the electrons in your fingertip and the table repel strongly at atomic distances, so the atoms never really meet. The force you feel when pressing is Coulomb repulsion. In a sense, you hover about

10^{-10}

m above everything you sit on.

Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free account Sign in