Coulomb’s law
The inverse square law
In 1785, Charles Augustin de Coulomb measured the force between two small charged metal spheres. He used a torsion balance: the force twists a fine wire, and the angle of twist tells you the force.
How do you change a charge by a known amount? Touch a charged sphere with an identical uncharged one. They share the charge equally, so each now has q by 2. Touch again with a fresh sphere: q by 4.
Coulomb found that 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. It acts along the line joining them.
So F equals k times q₁ q₂ over r². In SI units, k is written as 1 over 4πε₀, and its value is 9 × 10⁹ newton metre squared per coulomb squared.
ε₀ is the permittivity of free space: 8.854 × 10⁻¹² coulomb squared per newton metre squared. The law, as written here, is for charges in vacuum.
Watch the inverse square. Double the distance, and the force falls to one quarter. Make it three times, and the force is one ninth. The graph of F against r falls steeply.
How big is a coulomb? Two charges of 1 coulomb each, 1 metre apart, push with 9 × 10⁹ newtons. That is the weight of nearly a million tonnes.
In vector form: the force on q₂ due to q₁ is F₂₁ equals k q₁ q₂ over r₂₁², times r̂₂₁. Here r̂₂₁ is the unit vector pointing from q₁ to q₂.
The signs take care of the direction. If q₁ q₂ is positive, the force on q₂ points away from q₁: repulsion. If it is negative, it points towards q₁: attraction.
And the force on q₁ due to q₂ is exactly opposite: F₁₂ equals minus F₂₁. Coulomb’s law obeys Newton’s third law.
Coulomb’s law at work
Example one. A charge of plus 2 microcoulombs and a charge of minus 3 microcoulombs are 30 centimetres apart. What is the force between them?
F equals 9 × 10⁹ times 2 × 10⁻⁶ times 3 × 10⁻⁶, divided by 0.3 squared. The top is 5.4 × 10⁻²; the bottom is 0.09. So F is 0.6 newton, and it is attractive, because the signs are unlike.
Remember: put r in metres, not centimetres, and the charges in coulombs.
Example two. How strong is the electric force compared with gravity? Take an electron and a proton a distance r apart. Both forces go as 1 over r², so their ratio does not depend on r.
The ratio is k e² divided by G mₑ mₚ. Put in the numbers: 9 × 10⁹ times 1.6 × 10⁻¹⁹ squared, over 6.67 × 10⁻¹¹ times 9.1 × 10⁻³¹ times 1.67 × 10⁻²⁷. It comes to about 2.3 × 10³⁹.
The electric force is about 10³⁹ times stronger than gravity! Gravity wins for planets only because large bodies are almost exactly neutral.
At a distance of 1 ångström, 10⁻¹⁰ metre, the force between them is 9 × 10⁹ times 2.56 × 10⁻³⁸, over 10⁻²⁰: about 2.3 × 10⁻⁸ newton. This gives the electron an acceleration F by m of about 2.5 × 10²² metres per second squared.
Example three. Two identical small spheres A and B, each with charge q, repel with a force F. An identical uncharged sphere C touches A, then touches B, and is taken away.
After touching A, both A and C have q by 2. C then touches B: together they have q plus q by 2, which is 3q by 2, so each gets 3q by 4.
The new force is k times q by 2 times 3q by 4, over r². That is 3 by 8 of k q² over r², so the force becomes 3F by 8.
Try it: two charges
- Drag either charge. The forces on the two charges are always equal and opposite.
- Halve the distance (from 6 cm to 3 cm): the force becomes four times bigger.
- Click a charge and change its value with the slider. Make both charges the same sign: attraction turns into repulsion.
- Check a readout yourself with F = 9 × 10⁹ × q₁q₂ / r², with r in metres.
Double and halve
Two point charges attract each other with a force F. Both charges are doubled and the distance between them is halved. What is the new force?
- F
- 4F
- 8F
- 16F
Answer
D. F ∝ q₁q₂ / r². Doubling both charges multiplies q₁q₂ by 2 × 2 = 4. Halving r divides r² by 4, which multiplies F by another 4. So the new force is 4 × 4 = 16F.
Work it out
Charges of +2 μC and −3 μC are 30 cm apart in vacuum. What is the force between them? (k = 9 × 10⁹ N m² C⁻²)
- 0.6 N, attractive
- 0.6 N, repulsive
- 0.18 N, attractive
- 6 × 10⁻⁵ N, attractive
Answer
A. F = kq₁q₂/r² = 9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶ / (0.3)² = 5.4 × 10⁻² / 0.09 = 0.6 N. Unlike charges attract. (0.18 N forgets to square r; 6 × 10⁻⁵ N uses r = 30 in centimetres.)
Summary
- Coulomb’s law:
F = k |q₁q₂| / r² - Constant:
k = 1/4πε₀ = 9 × 10⁹ N m² C⁻² - Permittivity of free space:
ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻² - Vector form:
F₂₁ = k q₁q₂ r̂₂₁ / r₂₁² = −F₁₂
- The force between two point charges is proportional to the product of the charges and to 1/r², and acts along the line joining them.
- Like charges repel, unlike charges attract; the two forces are equal and opposite.
- Use SI units: charges in coulombs, distances in metres.
- The electric force between an electron and a proton is about 10³⁹ times the gravitational force.