Electrostatics

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Senior 1 · Physics

Electrostatics

5 lessons

Lesson 4 of 5

Coulomb's Law

24 min

How strong is the push?

Hang two balloons, both rubbed on hair, from threads that meet at one point. They swing apart. Rub them more and they swing further apart; move the threads' top points closer together and they swing out more strongly again. So the electric force between two charges depends on how much charge there is and on how far apart the charges are. In 1785 Charles-Augustin de Coulomb measured exactly how, with a balance so sensitive that it could weigh a force far smaller than the weight of a grain of sand.

What the force depends on

Figure 4.1: The electric forces on two point charges act along the line joining them. They are equal in size and opposite in direction.

Measurements like Coulomb's give the pattern in the table. Start with a force $F$ between charges $q_{1}$ and $q_{2}$ a distance $r$ apart, and change one thing at a time:

Change	New force
double $q_{1}$	$2 F$
double both $q_{1}$ and $q_{2}$	$4 F$
double the distance $r$	$\frac{F}{4}$
triple the distance $r$	$\frac{F}{9}$
halve the distance $r$	$4 F$

The force is proportional to each charge, and inversely proportional to the SQUARE of the distance.

Definition

Coulomb's law: the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

F = k\frac{q_1 q_2}{r^2}

where $F$ is the force in newtons (N), $q_{1}$ and $q_{2}$ are the charges in coulombs (C), $r$ is the distance between their centres in metres (m), and $k = 9.0 \times 10^{9}\,\mathrm{N\,m^2\,C^{-2}}$ is Coulomb's constant.

The force is a repulsion for like charges and an attraction for unlike charges. The two charges push or pull each other with forces of the same size (Newton's third law).

Worked example 1: two positive charges

Charges of $+2.0\,\mu\mathrm{C}$ and $+3.0\,\mu\mathrm{C}$ are $30\,\mathrm{cm}$ apart. Find the force between them.

Convert to SI units: $q_1 = 2.0 \times 10^{-6}\,\mathrm{C}$ , $q_2 = 3.0 \times 10^{-6}\,\mathrm{C}$ , $r = 0.30\,\mathrm{m}$ .

F = k\frac{q_1 q_2}{r^2} = \frac{(9.0 \times 10^{9})(2.0 \times 10^{-6})(3.0 \times 10^{-6})}{(0.30)^2} = \frac{0.054}{0.090} = 0.60\,\mathrm{N}

Both charges are positive, so the force is a repulsion of $0.60\,\mathrm{N}$ on each.

Worked example 2: changing the distance

Two charged spheres attract each other with a force of $12\,\mathrm{N}$ . What is the force when the distance between them is (a) doubled; (b) halved?

(a) $r$ becomes $2 r$ , so $r^{2}$ becomes $4 r^{2}$ : $F = \frac{12\,\mathrm{N}}{4} = 3.0\,\mathrm{N}$ .

(b) $r$ becomes $\frac{r}{2}$ , so $r^{2}$ becomes $\frac{r^2}{4}$ : $F = 12\,\mathrm{N} \times 4 = 48\,\mathrm{N}$ .

The inverse-square curve

Figure 4.2: The force between two charges of 0.20 μC at different distances. The force falls quickly as the charges move apart, but never quite reaches zero.

Worked example 3: finding the distance

Two equal charges of $4.0\,\mu\mathrm{C}$ repel each other with a force of $0.90\,\mathrm{N}$ . How far apart are they?

Rearrange Coulomb's law for $r^{2}$ :

r^2 = \frac{k q_1 q_2}{F} = \frac{(9.0 \times 10^{9})(4.0 \times 10^{-6})^2}{0.90} = \frac{0.144}{0.90} = 0.16\,\mathrm{m^2}

r = \sqrt{0.16} = 0.40\,\mathrm{m}

Worked example 4: inside a hydrogen atom

In a hydrogen atom, one electron is about $5.3 \times 10^{-11}\,\mathrm{m}$ from the proton. Find the force between them.

F = \frac{(9.0 \times 10^{9})(1.6 \times 10^{-19})^2}{(5.3 \times 10^{-11})^2} = 8.2 \times 10^{-8}\,\mathrm{N}

An attraction (unlike charges). It looks small, but the electron's mass is only $9.11 \times 10^{-31}\,\mathrm{kg}$ : for such a light particle it is an enormous force, and it is what holds atoms together.

Warning

Common mistakes. (1) Forgetting to SQUARE the distance. (2) Leaving charges in $\mu\mathrm{C}$ or nC, or distances in cm: change everything to C and m first. (3) Putting the minus sign of a negative charge into the formula and then puzzling over a negative force: use the sizes of the charges, and decide attraction or repulsion from like or unlike signs.

Tip

Inverse square: distance $\times 2$ gives force $\div 4$ ; distance $\times 3$ gives force $\div 9$ ; distance $\div 2$ gives force $\times 4$ .

Where Coulomb forces matter

Coulomb's law describes the force that holds every electron in its atom and binds atoms into the solids around you: when you lean on a desk, it is the electric repulsion between the electrons in your hand and in the wood that stops your hand going through. Because the force grows so quickly as charges come closer, a charged comb picks up paper only when it is very near. Explore it with PhET: Coulomb's Law (version 1.0.29): change the charges and the distance and read the force.

Coulomb's law (Khan Academy). Watch how the force depends on the product of the charges and on the square of the distance, and follow the worked calculation.

✦

Practice

State Coulomb's law in words, and write it as a formula with each symbol and unit.
Charges of $+5.0\,\mu\mathrm{C}$ and $-4.0\,\mu\mathrm{C}$ are $0.20\,\mathrm{m}$ apart. Find the force between them and say whether it attracts or repels.
Two small spheres carry $3.0\,\mathrm{nC}$ and $6.0\,\mathrm{nC}$ and are $3.0\,\mathrm{cm}$ apart. Find the force between them.
The force between two charges is $20\,\mathrm{N}$ . Find the new force when (a) the distance is doubled; (b) one charge is tripled; (c) both charges are doubled AND the distance is doubled.
Charges of $2.0\,\mu\mathrm{C}$ and $8.0\,\mu\mathrm{C}$ repel with a force of $1.6\,\mathrm{N}$ . How far apart are they?
Uwase says: "The $8.0\,\mu\mathrm{C}$ charge pushes the $2.0\,\mu\mathrm{C}$ charge four times as hard as the small charge pushes the big one." Evaluate her statement.

Show answersHide answers

The force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them: $F = k\frac{q_1 q_2}{r^2}$ , with $F$ in N, $q_{1}$ and $q_{2}$ in C, $r$ in m and $k = 9.0 \times 10^{9}\,\mathrm{N\,m^2\,C^{-2}}$ .
$F = \frac{(9.0 \times 10^{9})(5.0 \times 10^{-6})(4.0 \times 10^{-6})}{(0.20)^2} = \frac{0.18}{0.040} = 4.5\,\mathrm{N}$ , an attraction (unlike charges).
$F = \frac{(9.0 \times 10^{9})(3.0 \times 10^{-9})(6.0 \times 10^{-9})}{(0.030)^2} = 1.8 \times 10^{-4}\,\mathrm{N}$ , a repulsion.
(a) $\frac{20}{4} = 5.0\,\mathrm{N}$ ; (b) $3 \times 20 = 60\,\mathrm{N}$ ; (c) the charges give $\times 4$ and the distance gives $\div 4$ , so the force is still $20\,\mathrm{N}$ .
$r^2 = \frac{(9.0 \times 10^{9})(2.0 \times 10^{-6})(8.0 \times 10^{-6})}{1.6} = 0.090$ , so $r = 0.30\,\mathrm{m}$ .
She is wrong. Coulomb's law gives ONE force, $F = k\frac{q_1 q_2}{r^2}$ , which acts on each charge; by Newton's third law the two forces are equal in size and opposite in direction.

Key points

Coulomb's law: $F = k\frac{q_1 q_2}{r^2}$ , with $k = 9.0 \times 10^{9}\,\mathrm{N\,m^2\,C^{-2}}$ .
The force doubles if one charge doubles, and falls to a quarter if the distance doubles (inverse square).
Like charges repel and unlike charges attract; the two forces are equal and opposite.

Sources and credits

18.3: Coulomb's Law, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Drawing: Bcoulomb.png by Charles-Augustin de Coulomb (1785), Wikimedia Commons, in the public domain.
Coulomb's Law (version 1.0.29). Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC BY 4.0 (https://phet.colorado.edu)
Video: Coulomb's law | Physics | Khan Academy, Khan Academy, embedded from YouTube.

The diagrams, the graph, the worked examples and the practice exercises are written by Ganzaa.

1How strong is the push?

2What the force depends on

3The inverse-square curve

4Where Coulomb forces matter

How strong is the push?

What the force depends on

The inverse-square curve

Where Coulomb forces matter