18.3 Coulomb’s Law

Coulomb's law describes the electrostatic force between two charged objects. It's one of the core equations in introductory physics, and understanding it well sets you up for everything from electric fields to circuits later on.

The key idea: charged objects push or pull on each other, and that force depends on how much charge each object has and how far apart they are. As charges get closer, the force grows dramatically. This explains why oppositely charged particles in atoms hold together, and why a charged balloon can cling to a wall.

Coulomb's Law

Electrostatic force vs distance relationship

Coulomb's law gives you the magnitude of the electrostatic force between two point charges (charges small enough that their size doesn't matter compared to the distance between them).

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

Here's what each variable means:

• $F$ = magnitude of the electrostatic force (in newtons, N)
• $k$ = Coulomb's constant = $8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2$
• $q_1$ and $q_2$ = the magnitudes of the two charges (in coulombs, C)
• $r$ = the distance between the centers of the two charges (in meters, m)

Two proportionality rules control how this force behaves:

• Directly proportional to the product of the charges. If you double either charge, the force doubles. If you double both charges, the force quadruples.
• Inversely proportional to the square of the distance. This is the famous inverse square relationship. If you double the distance, the force drops to one-fourth. Triple the distance, and the force drops to one-ninth.

That inverse square drop-off is steep. For example, moving two charges from 1 cm apart to 2 cm apart cuts the force to 25% of what it was. Move them to 3 cm apart, and you're down to about 11%.

Coulomb's constant $k$ is related to a more fundamental quantity called the permittivity of free space ($\varepsilon_0$) through the relationship $k = \frac{1}{4\pi\varepsilon_0}$. For this course, you'll mostly just use the value of $k$ directly.

Coulomb's law calculations

Here's a step-by-step process for solving Coulomb's law problems:

1. Identify the two charges ($q_1$ and $q_2$) and the distance ($r$) between them.
2. Convert all units to SI. Charges must be in coulombs and distance in meters. Watch out for microcoulombs: $1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C}$. Watch out for centimeters: $1 \, \text{cm} = 0.01 \, \text{m}$.
3. Plug values into the formula $F = k \frac{|q_1 q_2|}{r^2}$ using the absolute values of the charges to find the force magnitude.
4. Determine the direction. Opposite-sign charges attract; same-sign charges repel. The force acts along the line connecting the two charges.

Worked example: Find the force between $q_1 = 2 \, \mu\text{C}$ and $q_2 = -4 \, \mu\text{C}$ separated by $r = 0.05 \, \text{m}$.

• Convert charges: $q_1 = 2 \times 10^{-6} \, \text{C}$, $q_2 = 4 \times 10^{-6} \, \text{C}$
• Plug in: $F = (8.99 \times 10^9) \frac{(2 \times 10^{-6})(4 \times 10^{-6})}{(0.05)^2}$
• Calculate: $F = (8.99 \times 10^9) \frac{8 \times 10^{-12}}{2.5 \times 10^{-3}} = (8.99 \times 10^9)(3.2 \times 10^{-9}) = 28.8 \, \text{N}$
• Direction: The charges have opposite signs, so the force is attractive.

Note that the force itself is a vector quantity with both magnitude and direction. When you use Coulomb's law, the formula gives you the magnitude; you determine the direction from the signs of the charges and the geometry of the problem.

Electrostatic vs gravitational forces

Both Coulomb's law and Newton's law of gravitation follow an inverse square relationship, but the forces they describe differ enormously in strength.

At the atomic scale, electrostatic forces completely dominate. The electrostatic attraction between an electron and a proton in a hydrogen atom is roughly $2.3 \times 10^{39}$ times stronger than the gravitational pull between them. That's not a typo. The masses of subatomic particles are incredibly tiny, while their charges (on the order of the elementary charge, $e = 1.60 \times 10^{-19} \, \text{C}$) are large enough relative to those masses to produce enormous electrostatic forces.

So why does gravity seem to run the show at large scales? Because most everyday objects are electrically neutral, with nearly equal numbers of protons and electrons. Their net charge is zero or very close to it, so electrostatic forces mostly cancel out. Gravity, on the other hand, has no "negative mass" to cancel it. Every bit of mass attracts every other bit, and it adds up over large objects like planets and stars.

A comparison helps make this concrete: two 1 kg masses separated by 1 m attract each other gravitationally with a force of just $6.67 \times 10^{-11} \, \text{N}$. But if each of those objects carried a net charge of 1 C (which is a huge amount of charge), the electrostatic force between them would be $8.99 \times 10^{9} \, \text{N}$. That's a difference of about 20 orders of magnitude.

Electric fields and superposition

An electric field is the region around a charged object where another charge would experience a force. You can think of it as the "influence zone" of a charge. Rather than thinking about forces between specific pairs of charges, the field tells you what force any charge placed at a given point would feel.

The superposition principle says that when multiple charges are present, the net electric field at any point is the vector sum of the individual fields produced by each charge. You calculate the field from each charge separately, then add them as vectors (accounting for both magnitude and direction). This same principle applies to forces: the net force on a charge is the vector sum of the Coulomb forces from every other charge in the system.