Coulomb's law

Balloons that push apart

Rub two balloons on your hair and hang them from threads, side by side. They swing apart. Bring them closer, and the push grows. Move them far apart, and it almost vanishes.

How does the push depend on the charges, and on the distance? In 1785 a French engineer, Charles Coulomb, measured it. He hung a charged ball from a light rod on a thin wire. A second charged ball pushed the first one, and the wire twisted. The angle of the twist measured the force. His result is Coulomb's law.

Coulomb's law

Take two point charges. These are charged bodies much smaller than the distance between them, so their size does not matter. The force between them:

• acts along the straight line that joins them;
• is proportional to the product of the two charges;
• is inversely proportional to the square of the distance between them.

In symbols,

Here $F$ is the size of the force in newton (N), $q_1$ and $q_2$ are the charges in coulomb (C), and $r$ is the distance in metre (m). The constant $k$ is the same everywhere in a vacuum, and nearly the same in air:

The Greek letter $\varepsilon_0$ (epsilon zero) is the permittivity of free space, $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)$. Why write $k$ in such a strange way, with $4\pi$ at the bottom? It is a choice. Later formulas, such as Gauss's law, become simpler when $\varepsilon_0$ stands alone.

The direction of the force

The two charges pull or push on each other. Newton's third law says that the force on $q_1$ from $q_2$ and the force on $q_2$ from $q_1$ are equal in size and opposite in direction. A bigger charge does not push harder than a smaller one.

The simple way to find the direction is the rule that we know:

• Like charges repel, so each force points away from the other charge.
• Unlike charges attract, so each force points towards the other charge.

In vector form, the force on charge 1 from charge 2 is

where $\hat r_{21}$ is a vector of length one, pointing from charge 2 to charge 1. When $q_1 q_2$ is positive the force points away from charge 2, which is a push. When $q_1 q_2$ is negative the force points towards charge 2, which is a pull. And $\vec F_{21} = -\vec F_{12}$.

In a calculation, the easy way is to put in the sizes of the charges, find the size of the force, and then decide the direction from the signs.

How big is one coulomb?

Put two charges of $1$ C each, $1$ m apart. The force is $9.0 \times 10^9$ N. That is the weight of about $9 \times 10^{8}$ kg, nearly a million tonnes. No small body can hold a charge like this. So the charges that we meet are mostly $\mu\text{C}$ and nC.

A worked example

A charge of $+2.0\ \mu\text{C}$ and a charge of $-3.0\ \mu\text{C}$ are $0.30$ m apart in air. Find the force between them.

1. Change to SI units. $q_1 = 2.0 \times 10^{-6}$ C and $q_2 = 3.0 \times 10^{-6}$ C, as sizes.
2. The size. $F = k\dfrac{q_1q_2}{r^2}$. The top is $kq_1q_2 = 9.0 \times 10^9 \times 2.0 \times 10^{-6} \times 3.0 \times 10^{-6}\ \text{N m}^2 = 5.4 \times 10^{-2}\ \text{N m}^2$. The bottom is $r^2 = (0.30\ \text{m})^2 = 0.090\ \text{m}^2$. So $F = \dfrac{5.4 \times 10^{-2}\ \text{N m}^2}{0.090\ \text{m}^2} = 0.60\ \text{N}$.
3. The direction. The charges are unlike, so they attract. Each charge feels $0.60$ N towards the other.

What if the distance doubles? Then $r^2$ becomes four times larger, and the force falls to $\dfrac{0.60\ \text{N}}{4} = 0.15\ \text{N}$.

What if both charges double? The product becomes four times larger, so the force is $4 \times 0.60\ \text{N} = 2.4\ \text{N}$.

What if the $-3.0\ \mu\text{C}$ charge is changed to $+3.0\ \mu\text{C}$? The size stays $0.60$ N, but the force is now a push.

The inverse square law

Coulomb's law says that the force falls very fast with the distance. Double the distance, and the force is a quarter. Make the distance three times as large, and the force is a ninth. Halve the distance, and the force is four times as large.

Set the second charge to a negative value. What happens to the sign of the force? Now raise the dielectric constant $K$ to $80$. How much does the force fall?

The electric force and gravity

Newton's law of gravitation, $F = Gm_1m_2/r^2$, also has the form of an inverse square law. How do the two forces compare? Take the electron and the proton in a hydrogen atom, $5.3 \times 10^{-11}$ m apart.

• Electric: $F_e = \dfrac{ke^2}{r^2}$. The top is $9.0 \times 10^9 \times (1.6 \times 10^{-19})^2\ \text{N m}^2 = 2.3 \times 10^{-28}\ \text{N m}^2$. The bottom is $(5.3 \times 10^{-11}\ \text{m})^2 = 2.8 \times 10^{-21}\ \text{m}^2$. So $F_e = 8.2 \times 10^{-8}\ \text{N}$.
• Gravity: $F_g = \dfrac{Gm_em_p}{r^2} = 3.6 \times 10^{-47}\ \text{N}$, with $G = 6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2$ and the mass of the proton $m_p = 1.67 \times 10^{-27}$ kg.

The ratio is $\dfrac{F_e}{F_g} = 2.3 \times 10^{39}$. The distance cancels, because both forces go as $1/r^2$. So inside an atom, gravity does not count at all.

Then why do we feel the gravity of the earth, and not its electric force? Gravity only pulls, so the pulls of all the atoms add up. In ordinary matter, positive and negative charges are nearly equal in number, so their electric forces cancel almost completely.

The force in a medium

Everything so far was for charges in a vacuum or in air. When the space between the charges is filled with a material such as oil or water, the force is smaller:

The number $K$ is the dielectric constant of the medium, also called its relative permittivity. It has no unit. For a vacuum $K = 1$, and for air it is very nearly $1$. For a typical oil it is about $2$, and for water it is about $80$.

So water makes the force between two charges $80$ times smaller. This is why many salts dissolve in water. The ions in a crystal are held together by electric forces, and in water these forces are $80$ times weaker, so the ions come apart easily.

Practise with new numbers

Try it in Python

The program finds the force between two charges of the same size at four distances. It also checks that $Fr^2$ stays the same. Then it puts the charges in other media.

Common mistakes

• Using the distance in cm. Change cm to m first. $1\ \text{cm} = 10^{-2}$ m.
• Forgetting to square the distance. The force is in inverse proportion to $r^2$, not to $r$.
• Forgetting that $\mu$ means $10^{-6}$. A charge of $5\ \mu\text{C}$ is $5 \times 10^{-6}$ C.
• Saying that the larger charge pulls harder. The two forces are equal and opposite, always.
• Putting the signs of the charges into the size. Use the sizes to find the size of the force. Then use the rule (like repel, unlike attract) for the direction.
• Using $k = 9.0 \times 10^9$ for a charge in water. In a medium, divide the force by $K$.

Remember these

• Coulomb's law: $F = k\dfrac{q_1q_2}{r^2}$, along the line joining the charges.
• $k = \dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^9\ \text{N m}^2/\text{C}^2$ and $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)$.
• The two forces are equal and opposite. Like charges repel. Unlike charges attract.
• Double the distance: the force falls to a quarter. Double one charge: the force doubles.
• In a medium, $F = \dfrac{1}{K}\left(k\dfrac{q_1q_2}{r^2}\right)$. For water, $K \approx 80$.
• Between an electron and a proton the electric force is about $10^{39}$ times the gravitational force.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Two charges repel with a force $F$. The distance between them is made three times as large. What is the new force?

2. A charge of $8\ \mu\text{C}$ and a charge of $2\ \mu\text{C}$ are close to each other. Which of them feels the larger force, and by how much?

3. Find the force between a charge of $4\ \mu\text{C}$ and a charge of $5\ \mu\text{C}$ that are $0.20$ m apart in air.

4. Two charges in air attract with a force of $36$ N. They are put in water ($K = 80$) at the same distance. What is the force now?