Coulomb's Law (DP IB Physics: HL): Revision Note

Ann Howell

Written by: Ann Howell

Reviewed by: Caroline Carroll

Updated on

Coulomb's Law

  • All charged particles generate an electric field

    • This field exerts a force on charged particles which are nearby

  • The electric force between two charges is defined by Coulomb’s law, which states that:

The electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of their separation

  • This electric force can be calculated using the expression:

F = kq1q2r2

  • Where:

    • F = electric force (N)

    • q1, q2 = magnitudes of the charges (C)

    • r = distance between the centres of the two charges (m)

    • k = Coulomb constant (8.99 × 109 N m2 C–2)

  • Coulomb's law for two charges is analogous to Newton's law of gravitation for two masses

    • This means that electric and gravitational forces are very similar

    • For example, both forces follow an inverse square law with the separation between charge or mass

Electrostatic attraction between two charges

Coulomb's Law between Charges

The attractive electric force F between two point charges +q1 and −q2 with a separation of r is defined by Coulomb’s law

  • Coulomb's constant is given by:

k = 14πε0

  • Where ε0 is the permittivity of free space

    • ε0 = 8.85 × 10–12 C2 N–1 m–2 and refers to charges in a vacuum

    • The value of the permittivity of air is taken to be the same as ε0

    • All other materials have a higher permittivity ε > ε0

    • ε is a measure of the resistance offered by a material in creating an electric field within it

  • The value of k depends on the material between the charges

    • In a vacuum, k = 8.99 × 109 N m2 C–2

Repulsive & Attractive Forces 

  • Unlike the gravitational force between two masses which is only attractive, electric forces can be attractive or repulsive

  • Between two charges of the same type

    • The product q1q2 is positive, so the forces have positive signs

    • Positive forces mean the charges experience repulsion

  • For two opposite charges:

    • The product q1q2 is negative, so the forces have negative signs

    • Negative forces mean the charges experience attraction

Worked Example

An alpha particle is placed 2.0 mm from a gold nucleus in a vacuum.

Taking them as point charges, calculate the magnitude of the electric force acting between the nuclei.

  • Proton number of helium = 2

  • Proton number of gold = 79

Answer:

Step 1: Write down the known quantities

  • Separation between charges, r = 2.0 mm = 2.0 × 10–3 m

  • Elementary charge, e = 1.60 × 10–19 C (from the data booklet)

  • Coulomb constant, k = 8.99 × 109 N m2 C–2 (from the data booklet)

Step 2: Calculate the charges of the alpha particle and gold nucleus 

  • An alpha particle (helium nucleus) has 2 protons, hence, it has a charge of:

q1 = 2e = 2 × (1.60 × 10–19)

  • A gold nucleus has 79 protons, hence, it has a charge of:

q2 = 79e = 79 × (1.60 × 10–19)

Step 3: Write down Coulomb's law

F = kq1q2r2

Step 4: Substitute the values and calculate the magnitude of the electric force

F = (8.99×109)×2×79×(1.60×1019)2(2.0×103)2 = 9.1×1021 N (2 s.f.)

Examiner Tips and Tricks

You do not need to memorise the numerical value of the Coulomb's constant k or that of the permittivity of free space ε0. They will both be given in the data booklet.

Unless specified in the question, you should assume that charges are located in a vacuum.

You should note that Coulomb's law can only be applied to charged spheres whose size is much smaller than their separation. Only in this case, the point charge approximation is valid. You must remember that the separation r must be taken from the centres of the spheres.

You cannot use Coulomb's law to calculate the electrostatic force between charges distributed on irregularly-shaped objects.

Different Values of Permittivity

  • Permittivity is the measure of how easily an electric field can be generated in a certain material

  • The relative permittivity εr is sometimes known as the dielectric constant

  • For a given material, it is defined as:

The ratio of the permittivity of a material to the permittivity of free space

  • Relative permittivity can be expressed as:

εr = εε0

  • Where:

    • εr = relative permittivity

    • ε = permittivity of a material (F m−1)

    • ε0 = permittivity of free space (F m−1)

  • Relative permittivity has no units because it is a ratio of two values with the same unit

  • When there is a material between two charges, the Coulomb constant becomes

k = 14πε

  • In air, the relative permittivity is 1, so ε = ε0

  • In other materials, the Coulomb constant reduces as ε = εrε0

Examples of relative permittivity

  • Some values of relative permittivity for different insulators are shown in the table below:

Material

Relative Permittivity, εr

free space (vacuum)

1

air

1.0005

paper

4

polystyrene

3

ceramic

100 - 15 000

paraffin

2.3

pure water

80

Worked Example

Calculate the permittivity of a material that has a relative permittivity of 4.5 × 1011. State an appropriate unit for your answer.

Answer:

Step 1: Write down the relative permittivity equation

εr = εε0

Step 2: Rearrange for the permittivity of the material ε

ε = εrε0

Step 3: Substitute the values and calculate

ε = (4.5×1011)×(8.85×1012) = 3.98 = 4.0 F m−1 (2 s.f.)

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Build on this topic

Ann Howell

Author: Ann Howell

Expertise: Physics Content Creator

Ann obtained her Maths and Physics degree from the University of Bath before completing her PGCE in Science and Maths teaching. She spent ten years teaching Maths and Physics to wonderful students from all around the world whilst living in China, Ethiopia and Nepal. Now based in beautiful Devon she is thrilled to be creating awesome Physics resources to make Physics more accessible and understandable for all students, no matter their schooling or background.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.