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:

$F = k \frac{q_{1} q_{2}}{r^{2}}$

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:

$k = \frac{1}{4 π ε_{0}}$

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

$F = k \frac{q_{1} q_{2}}{r^{2}}$

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

$F = (8.99 \times 10^{9}) \times \frac{2 \times 79 \times {(1.60 \times 10^{- 19})}^{2}}{{(2.0 \times 10^{- 3})}^{2}} = 9.1 \times 10^{- 21}$ N (2 s.f.)

Exam Tip

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 easy it is to generate an electric field in a certain material
The relativity permittivity ε_r 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

Relativity permittivity can be expressed as:

$ε_{r} = \frac{ε}{ε_{0}}$

Where:
- ε_r = 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

$k = \frac{1}{4 π ε}$

In air, the relative permittivity is 1, so $ε = ε_{0}$
In other materials, the Coulomb constant reduces as $ε = ε_{r} ε_{0}$

Examples of Relative Permittivity

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

Material	Relative Permittivity, ε_r
free space (vacuum)	1
air	1.00054
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

$ε_{r} = \frac{ε}{ε_{0}}$

Step 2: Rearrange for permittivity of the material ε

$ε = ε_{r} ε_{0}$

Step 3: Substitute the values and calculate

$ε = (4.5 \times 10^{11}) \times (8.85 \times 10^{- 12}) = 3.98 = 4.0$ F m⁻¹ (2 s.f.)

Coulomb's Law (HL IB Physics)

Revision Note

Coulomb's Law

Electrostatic attraction between two charges

Repulsive & Attractive Forces

Worked example

Exam Tip

Different Values of Permittivity

Examples of Relative Permittivity

Worked example

Author: Ann H