Coulomb's law

Before this

This leads to

Electric field strength, E = F/q and E = V/d
The nuclear atom — Rutherford's experiment · coming soon
Lattice structures

A force far stronger than gravity

A hydrogen atom is one proton and one electron. The proton pulls on the electron in two ways. Gravity pulls it, because both have mass. The electric force pulls it, because they have opposite charges. Which pull is bigger?

It is not close. The electric pull is about $2 \times 1 0^{39}$ times the gravitational pull. That is a 2 followed by thirty-nine zeros. Gravity inside an atom is so weak that chemists and physicists ignore it completely.

But then a puzzle appears. If the electric force is so much stronger, why does gravity hold the Moon in orbit and keep you on the ground? Why does the electric force not rule the universe? This lecture finds the law of the electric force, uses it to work out that $1 0^{39}$ , and then answers the puzzle.

Coulomb's twisted wire

By the 1760s, several scientists suspected that the electric force got weaker with distance in the same way as gravity. Joseph Priestley suggested this in 1767. But a suspicion is not a measurement. The forces between small charged objects are tiny, and nobody could measure them well.

In 1785, the French engineer Charles-Augustin de Coulomb reported a way to do it. He hung a light rod from a very thin wire. On the end of the rod was a small charged ball. A second charged ball, fixed in place, pushed the first one away, and the push twisted the wire. Coulomb had shown that the twist of such a wire is proportional to the force on it. So the angle of twist measured the force.

In his first test, the balls settled 36 degrees apart on his scale. He then twisted the wire further to push the balls together, to half that distance. To hold them there, the twist had to rise to 144 degrees: four times as much. Halving the distance had multiplied the force by four. A third test, at about half the distance again, gave about four times again.

This is the pattern of an inverse square law. Historians have since rebuilt his balance and argued about how he obtained such clean numbers, because the experiment is very hard to repeat. But the law he stated has been tested in many other ways since, with great precision, and it holds.

Two point charges $q_{1}$ and $q_{2}$ , a distance $r$ apart, exert forces on each other along the line joining them. The size of each force is

$F = \frac{k q _{1} q _{2}}{r ^{2}}$ , where $k \approx 8.99 \times 1 0^{9} N m^{2} / C^{2}$ .

Like charges repel. Unlike charges attract.

Read each part of the law with its meaning.

$q_{1} q_{2}$ on top. Double either charge, and the force doubles. Double both, and it becomes four times as big.
$r^{2}$ underneath. Double the distance, and the force falls to a quarter. Triple it, and it falls to a ninth. This is the twisted-wire result.
$k$ , the Coulomb constant. It turns the units into newtons. In most calculations at this level you can use $k \approx 9.0 \times 1 0^{9} N m^{2} / C^{2}$ . Use the value your question gives.
$r$ is between the centres, in metres. A distance given in centimetres must be changed to metres first.

In this lecture, put the sizes of the charges into the formula, without their signs. The formula then gives the size of the force. Decide the direction separately, from "like charges repel, unlike charges attract". Some books keep the signs in, and read a positive answer as repulsion and a negative answer as attraction. Both methods give the same physics.

You will also see the law written with a different constant:

F = \frac{q _{1} q _{2}}{4 π ε _{0} r ^{2}}

Here $ε_{0} \approx 8.85 \times 1 0^{- 12} C^{2} / (N m^{2})$ is called the permittivity of free space. The two forms are the same law, because $k = 1/ (4 π ε_{0})$ . Check it: $1 \div (4 π \times 8.85 \times 1 0^{- 12}) \approx 8.99 \times 1 0^{9}$ . The $4 π$ is there for a reason you have already met. The area of a sphere is $4 π r^{2}$ , and the force spreads out over spheres around the charge.

The two charges always feel forces of the same size, in opposite directions, even when one charge is four times the other. Like charges push apart; unlike charges pull together.

Stop and think

A charge of +4 μC and a charge of +1 μC sit near each other. Which one feels the bigger force?

Show the answer

Neither. Both feel the same size of force. Look at the formula: $q_{1} q_{2}$ is the same number whichever charge you call $q_{1}$ . So the force on the small charge equals the force on the big one. This is Newton's third law: the two forces are a pair, equal in size and opposite in direction. The small charge may move more, if it is lighter, but the force on it is not bigger.

How fast does the force fall?

The inverse square is the part of the law that matters most, and the part students most often get wrong. Use the explorer to feel it before you calculate with it.

Explore — change the distance between two charges

The left charge is +2 μC. Choose the right charge, then change the distance r with the slider or the − / + buttons. Watch the force, and the point moving on the graph. Try 10 cm, then 20 cm.

Right charge:

Distance between the centres: 10 cm

This table gives the force between a +2 μC charge and a second charge, for several distances, using $k = 9.0 \times 1 0^{9} N m^{2} / C^{2}$ .

Distance r	Force with +2 μC	Force with +4 μC
5 cm	14.4 N	28.8 N
10 cm	3.6 N	7.2 N
15 cm	1.6 N	3.2 N
20 cm	0.9 N	1.8 N

Each time the distance doubles (5 → 10 → 20 cm), the force falls to a quarter. Doubling one charge doubles every force. With −2 μC instead of +2 μC, the sizes are the same as in the first column, but the force is attractive.

From 10 cm to 20 cm, the force falls from 3.6 N to 0.9 N, a quarter. The graph is not a straight line. It falls steeply at short distances and flattens out far away. Swap +2 μC for +4 μC and every force doubles, so the whole curve is twice as high.

Why the square? The gravitation lecture showed the reason. Anything that spreads out equally in all directions from a point is spread over the surface of a sphere. A sphere twice as wide has four times the area. The electric fields lecture gives the same picture with field lines. A charge sends out a fixed number of lines. At twice the distance, those lines are spread over four times the area, so they are four times further apart, and the field is a quarter as strong.

That is it. That is the whole idea. The $r^{2}$ is not a separate fact about electricity. It is a fact about space, which has three dimensions.

How much weaker is the field of one charge?

The last lecture showed that the field lines around a single charge spread out, so the field gets weaker with distance. It promised to find exactly how much weaker. Coulomb's law now gives the answer.

Put a small test charge q at a distance r from a charge Q. Coulomb's law gives the force on it: $F = k Qq / r^{2}$ . The field strength is the force per coulomb of test charge, $E = F / q$ . So the q cancels:

E = \frac{k Q}{r ^{2}}

The field of a single charge falls with the square of the distance, exactly like the force. For example, at 10 cm from a +2 μC charge, $E = 9.0 \times 1 0^{9} \times 2 \times 1 0^{- 6} \div 0.1 0^{2} = 1.8 \times 1 0^{6}$ N/C. At 20 cm it is a quarter of that, $4.5 \times 1 0^{5}$ N/C. The next lecture works with this result in detail, together with the uniform field between plates.

The same law also settles a question from the last lecture. Two charges, +1 and +4, have a neutral point between them, nearer the +1. Now we can find exactly where. At the neutral point the two fields are equal, so $1/ x^{2} = 4/ (d - x)^{2}$ , where d is the distance between the charges and x is the distance from the +1. Taking square roots, $d - x = 2 x$ , so $x = d /3$ . The neutral point is one-third of the way from the +1 to the +4.

Worked examples

Worked Example 1 — The force between two charges

A charge of +3.0 μC and a charge of −2.0 μC are 10 cm apart. Find the size of the force between them, and say whether it is attractive or repulsive. Use $k = 9.0 \times 1 0^{9} N m^{2} / C^{2}$ .

Think first. Change to coulombs and metres: 3.0 μC = $3.0 \times 1 0^{- 6}$ C, 2.0 μC = $2.0 \times 1 0^{- 6}$ C, 10 cm = 0.10 m.

F = \frac{k q _{1} q _{2}}{r ^{2}}

Substitute: $F = 9.0 \times 1 0^{9} \times 3.0 \times 1 0^{- 6} \times 2.0 \times 1 0^{- 6} \div 0.1 0^{2}$ .

The top is $9.0 \times 3.0 \times 2.0 = 54$ , times $1 0^{9} \times 1 0^{- 6} \times 1 0^{- 6} = 1 0^{- 3}$ . So the top is $5.4 \times 1 0^{- 2}$ . The bottom is $0.1 0^{2} = 1.0 \times 1 0^{- 2}$ . So $F = 5.4$ N.

The charges have opposite signs, so the force is attractive. Each charge is pulled towards the other with 5.4 N.

Check. 5.4 N is roughly the weight of a 550 g bag of rice. That is a large force for such small charges, which is a first hint of how strong electric forces are.

Worked Example 2 — Electric force against gravity in a hydrogen atom

In a hydrogen atom, take the proton–electron distance as $5.3 \times 1 0^{- 11}$ m. The charges are $\pm 1.60 \times 1 0^{- 19}$ C. The masses are $1.67 \times 1 0^{- 27}$ kg (proton) and $9.11 \times 1 0^{- 31}$ kg (electron). Compare the electric force with the gravitational force. Use $G = 6.67 \times 1 0^{- 11} N m^{2} / kg^{2}$ .

Think first. Both laws have $r^{2}$ underneath, with the same r. So the ratio does not depend on r at all.

Electric: $F_{E} = 9.0 \times 1 0^{9} \times (1.60 \times 1 0^{- 19})^{2} \div (5.3 \times 1 0^{- 11})^{2} \approx 8.2 \times 1 0^{- 8}$ N.

Gravitational: $F_{G} = 6.67 \times 1 0^{- 11} \times 1.67 \times 1 0^{- 27} \times 9.11 \times 1 0^{- 31} \div (5.3 \times 1 0^{- 11})^{2} \approx 3.6 \times 1 0^{- 47}$ N.

Ratio: $F_{E} / F_{G} \approx 8.2 \times 1 0^{- 8} \div 3.6 \times 1 0^{- 47} \approx 2.3 \times 1 0^{39}$ .

Check. Since the $r^{2}$ cancels, the ratio is $k e^{2} / (G m_{p} m_{e})$ , which gives the same $2.3 \times 1 0^{39}$ . The number depends on which particles you choose. For two protons, which are much heavier than an electron, the ratio is about $1.2 \times 1 0^{36}$ . It is still enormous.

Worked Example 3 — Finding a distance

Two equal charges of +2.0 μC repel each other with a force of 0.90 N. How far apart are they?

Think first. Rearrange the law for $r^{2}$ before putting numbers in.

r^{2} = \frac{k q _{1} q _{2}}{F}

Substitute: $r^{2} = 9.0 \times 1 0^{9} \times 2.0 \times 1 0^{- 6} \times 2.0 \times 1 0^{- 6} \div 0.90 = 0.040 m^{2}$ .

So $r = 0.040 = 0.20$ m, or 20 cm.

Check. This matches the explorer: two +2 μC charges at 20 cm feel 0.9 N. ✓

Worked Example 4 — Three charges in a line

Charge A (+2.0 μC), charge B (+1.0 μC) and charge C (−3.0 μC) lie on a straight line. B is 0.10 m to the right of A, and C is 0.20 m to the right of B. Find the resultant force on B.

The forces on B. A (positive) pushes B away, to the right. C (negative) pulls B towards it, also to the right. So the two forces add.

Think first. Find each force on B separately, with its direction. Then add them as vectors along the line. Take right as positive.

From A: $F = 9.0 \times 1 0^{9} \times 2.0 \times 1 0^{- 6} \times 1.0 \times 1 0^{- 6} \div 0.1 0^{2} = 1.8$ N. A and B are both positive, so A pushes B to the right.

From C: $F = 9.0 \times 1 0^{9} \times 3.0 \times 1 0^{- 6} \times 1.0 \times 1 0^{- 6} \div 0.2 0^{2} = 0.675$ N. C is negative and B is positive, so C pulls B to the right, towards C.

Resultant: $1.8 + 0.675 = 2.475$ N, so about 2.5 N to the right.

Check. Use ratios. C has 1.5 times A's charge, but it is twice as far from B. So its force is $1.5 \div 2^{2} = 0.375$ times A's force: $0.375 \times 1.8 = 0.675$ N. ✓ Ratios are a quick way to check a calculation.

Back to the puzzle: why gravity rules the sky

The electric force between two particles is enormously stronger than gravity. Yet the Earth holds the Moon by gravity, not by electricity. The reason is that charge comes in two kinds, and mass comes in only one.

Every atom has as many electrons as protons. So a large object, such as a planet, has almost exactly equal amounts of positive and negative charge. Its electric pushes and pulls on another object nearly cancel. Mass never cancels. Every kilogram of the Earth attracts every kilogram of the Moon, and all those pulls add up.

So gravity wins over large distances not because it is stronger, but because nothing cancels it. Inside atoms, where positive and negative charges are close together and not cancelled, the electric force wins easily. This is why chemistry is about electric forces, and astronomy is about gravity.

Three traps

Before the traps, find the mistake in a student's working.

Spot the error — a force between two charges

One line of this working contains the mistake that the rest follows from. Tap the line you think is wrong. You can try again as often as you like.

This working finds the force between two charges, and one line contains the mistake.

Question: charges of +3.0 μC and +5.0 μC are 30 cm apart. Find the force between them. Use k = 9.0 × 10⁹ N m²/C².
q₁ = 3.0 × 10⁻⁶ C and q₂ = 5.0 × 10⁻⁶ C.
The error: "r = 30, so r² = 900." The distance must be in metres: r = 0.30 m, so r² = 0.090 m².
The rest of the working then gives 1.5 × 10⁻⁴ N, which is 10 000 times too small. The correct answer is F = 0.135 ÷ 0.090 = 1.5 N, repulsive.

Trap 1 — Forgetting to square

Doubling the distance does not halve the force. It divides it by $2^{2} = 4$ . When you work with ratios, write the square every time: "r × 3, so F × 1/9".

Trap 2 — Centimetres and microcoulombs

Coulomb's law needs coulombs and metres. A distance in centimetres used as metres makes $r^{2}$ ten thousand times too big. A charge in μC used without $1 0^{- 6}$ makes it a million times too big. Convert every quantity before you substitute.

Trap 3 — "The bigger charge feels the bigger force"

The forces on the two charges are always equal in size and opposite in direction. The product $q_{1} q_{2}$ does not care which charge is which. Different charges may accelerate differently if their masses differ, but the forces are equal.

Try it yourself

Work each one on paper first. Use $k = 9.0 \times 1 0^{9} N m^{2} / C^{2}$ . Open a hint only when you are stuck.

Charges of +4.0 μC and +5.0 μC are 20 cm apart. Find the force between them, and say whether it is attractive or repulsive.

Hint 1
Change μC to C and cm to m before anything else.

Hint 2
$F = 9.0 \times 1 0^{9} \times 4.0 \times 1 0^{- 6} \times 5.0 \times 1 0^{- 6} \div 0.2 0^{2}$ .

Full answer
The top is $9.0 \times 4.0 \times 5.0 \times 1 0^{- 3} = 0.18$ . The bottom is $0.2 0^{2} = 0.040$ . So $F = 0.18 \div 0.040 = 4.5$ N. Both charges are positive, so the force is repulsive. Check: $4.5 \times 0.040 = 0.18$ . ✓
Two charges repel with a force of 18 N. The distance between them is tripled, and one of the charges is doubled. What is the new force?

Hint 1
Do not look for the charges. Work with factors: what does each change do to F on its own?

Hint 2
Doubling a charge multiplies F by 2. Tripling r multiplies F by $1/ 3^{2} = 1/9$ .

Full answer
New force = $18 \times 2 \times \frac{1}{9} = 4$ N, still repulsive. Check: the overall factor is $2/9 \approx 0.22$ , and $4 \div 18 \approx 0.22$ . ✓
Two equal charges, 10 cm apart, repel each other with a force of 3.6 N. Find the size of each charge.

Hint 1
With equal charges, $q_{1} q_{2} = q^{2}$ . Rearrange the law for $q^{2}$ .

Hint 2
$q^{2} = F r^{2} / k = 3.6 \times 0.1 0^{2} \div 9.0 \times 1 0^{9}$ .

Full answer
$q^{2} = 3.6 \times 0.010 \div 9.0 \times 1 0^{9} = 4.0 \times 1 0^{- 12} C^{2}$ , so $q = 2.0 \times 1 0^{- 6}$ C, which is 2.0 μC each. The charges have the same sign (they repel), but the force alone cannot say whether both are positive or both negative. Check: this is the explorer's +2 μC and +2 μC at 10 cm, which gave 3.6 N. ✓
Two small spheres, each of mass 1.0 kg, are 1 m apart. Each is given the same charge q. How big must q be for the electric repulsion to equal the gravitational attraction? Use $G = 6.67 \times 1 0^{- 11} N m^{2} / kg^{2}$ and $e = 1.60 \times 1 0^{- 19}$ C. How many extra electrons is that?

Hint 1
Set $k q^{2} / r^{2} = G m^{2} / r^{2}$ . The $r^{2}$ cancels, so the distance does not matter.

Hint 2
$q^{2} = G m^{2} / k$ , so $q = m G / k$ . Then divide q by the charge on one electron.

Full answer
$q = 1.0 \times 6.67 \times 1 0^{- 11} \div 9.0 \times 1 0^{9} \approx 8.6 \times 1 0^{- 11}$ C. In electrons: $8.6 \times 1 0^{- 11} \div 1.60 \times 1 0^{- 19} \approx 5.4 \times 1 0^{8}$ electrons. Check: $k q^{2} \approx 9.0 \times 1 0^{9} \times 7.4 \times 1 0^{- 21} \approx 6.7 \times 1 0^{- 11}$ N, the same as $G m^{2}$ at 1 m. ✓ A kilogram of ordinary matter contains roughly $1 0^{26}$ electrons. So an imbalance of about one electron in $1 0^{17}$ or $1 0^{18}$ is enough to match gravity completely. That is why objects must be almost perfectly neutral for gravity to win.

The bigger idea

Coulomb's law and Newton's law of gravitation have the same shape: a constant, the product of two "amounts", and $r^{2}$ underneath. The square comes from the three dimensions of space, so any influence that spreads out evenly from a point must follow it. Physicists have tested the electric inverse square law with extreme precision, and it holds.

The two laws differ in what the "amount" is. Mass has one sign, so gravity always adds up. Charge has two signs, so electric forces can cancel. That one difference decides the shape of the world. Electric forces hold atoms, molecules and crystals together, and they explain almost all of chemistry. Gravity, too weak to matter in an atom, takes over at the size of planets and stars, because there nothing cancels it.

Key concepts in this lesson

Coulomb's law gives the size of the force between two point charges as $F = k q_{1} q_{2} / r^{2}$ , with $k \approx 8.99 \times 1 0^{9} N m^{2} / C^{2}$ , which is the same as $1/ (4 π ε_{0})$ .
The force acts along the line joining the charges: like charges repel and unlike charges attract.
The force is proportional to each charge and inversely proportional to the square of the distance, so doubling the distance gives a quarter of the force.
The two charges always feel forces of equal size in opposite directions, whatever their sizes, by Newton's third law.
The field of a single charge is $E = k Q / r^{2}$ , and electric forces dominate inside atoms while gravity dominates at large scales because charges cancel and masses do not.