5.1.2 Coulomb's Law
By the end of this lesson you can:
- Use Coulomb's law to find the force between two charges
- Explain how force changes with charge and distance
- Add electric forces from several charges as vectors
We commonly say that opposite electric charges attract and like charges repel. More precisely, charged objects exert an electrostatic force on one another. Because this force can be measured, a key question arises: How does the magnitude of this force depend on the size of the charges and the distance between them?
In the late 18th century, French physicist Charles-Augustin de Coulomb investigated this interaction using a precise instrument called a torsion balance. He discovered that the electrostatic force between two point charges follows a fundamental rule:
- Directly proportional to charge: The force increases as the magnitude of either charge (q1 or q2) increases.
- Inverse-square relationship: The force decreases rapidly as the distance (r) between the charges increases—specifically, inverse to the square of the distance (1⁄r2).
This mathematical relationship is known as Coulomb's Law:
Where:
- F is the magnitude of the electrostatic force in Newtons (N)
- q1 and q2 are the amounts of charge in Coulombs (C)
- r is the separation distance between the centers of the charges in meters (m)
- k is Coulomb's constant (≈ 8.99 × 109 N·m2/C2)
Much like Newton's Law of Universal Gravitation, Coulomb's Law shows how fundamental forces operate across space. However, while gravity is always attractive, the electrostatic force can be either attractive or repulsive depending on the signs of the interacting charges.
This video examines this and it is referred to as Coulombs Law.
Practice Problems: Coulomb's Law
Use F = k|q1q2| / r2, where Coulomb's constant k = 8.99 × 109 N m2 C-2. Like charges repel. Unlike charges attract.
Given: q1 = 5.0 × 10-5 C, F = 300 N, r = 15 cm = 0.15 m
Rearrange: q2 = Fr2 / (kq1)
Calculation: q2 = (300 × 0.152) / (8.99 × 109 × 5.0 × 10-5) = 6.75 / (4.495 × 105) = 1.50 × 10-5 C
Given: F = 0.30 N, r = 75 cm = 0.75 m, q1 = q2 = q
Formula: F = kq2 / r2, so q = √(Fr2 / k)
Calculation: q2 = (0.30 × 0.752) / (8.99 × 109) = 0.16875 / (8.99 × 109) = 1.877 × 10-11 C2
q = √(1.877 × 10-11) = 4.33 × 10-6 C
Given: q1 = q2 = 2.5 × 10-6 C, F = 0.50 N
Rearrange: r2 = k|q1q2| / F
Calculation: r2 = (8.99 × 109 × (2.5 × 10-6)2) / 0.50 = (8.99 × 109 × 6.25 × 10-12) / 0.50 = 0.0562 / 0.50 = 0.112 m2
r = √0.1124 = 0.335 m
Given: q1 = +8.0 μC, q2 = -2.0 μC, r = 0.20 m
Step 1 (Share the charge): The spheres are identical, so when they touch the total charge is shared equally. Qtotal = +8.0 μC + (-2.0 μC) = +6.0 μC, so each sphere now has q′ = +3.0 μC = 3.0 × 10-6 C.
Step 2 (Force): F = kq′2 / r2 = (8.99 × 109 × (3.0 × 10-6)2) / (0.20)2 = 0.08091 / 0.04 = 2.02 N
Direction: Both spheres now have the same sign, so they repel.
Given: q1 = +2.0 × 10-6 C at 0, q2 = -4.0 × 10-6 C at 0.10 m, q3 = +3.0 × 10-6 C at 0.30 m
Step 1 (Force of q2 on q1): Unlike charges attract, so q1 is pulled towards q2 (the +x direction). F21 = (8.99 × 109 × 2.0 × 10-6 × 4.0 × 10-6) / (0.10)2 = 7.19 N
Step 2 (Force of q3 on q1): Like charges repel, so q1 is pushed away from q3 (the -x direction). F31 = (8.99 × 109 × 2.0 × 10-6 × 3.0 × 10-6) / (0.30)2 = 0.599 N
Step 3 (Add the forces as vectors): Fnet = +7.19 N - 0.599 N = +6.59 N