Coulomb’s Law

Scalar Form

Suppose two point charges are separated by a certain distance. Then, according to Coulomb’s law, the electrostatic force, also known as Coulomb’s force, between the two charges is given by the following formula.

Where,

F: Electrostatic force

k: Coulomb’s constant

q₁, q₂: Magnitude of the point charges

r: Distance of separation between the charges

SI Unit: Newton or N

Dimension: [M L T^-2]

Because of the inverse relationship, this law is sometimes called Coulomb’s inverse square law.

Vector Form

The vector form of Coulomb’s law is given by,

Where,

Coulomb’s Law Constant

The constant in Coulomb’s law is known as Coulomb’s constant, which is denoted by k. Suppose, q₁ = 1, q₂ = 1, and r = 1, then,

Therefore, Coulomb’s constant is defined as the electrostatic force experienced by two unit charges when a unit distance separates them. It is expressed as follows.

In free space,

In a medium,

Where,

ε_o: Permittivity of free space or vacuum (= 8.85 x 10^-12 C²N^-1m^-2)

ε_r: Permittivity of the medium

The ratio of ε_r and ε_o is known as the relative permittivity of the medium or dielectric constant.

Dielectric Constant: κ = ε_r/ε_o

Putting the value of ε_o gives the value of k,

Coulomb’s Constant: 9 x 10⁹ Nm²C^-2

Using the expression for k, the Coulomb’s Law equation in free space is given by,

Coulomb’s law refers to the interaction between two point charges. For more than two charges, the force on each charge due to the presence of other charges should be taken into account. Since force is a vector quantity, the superposition principle must be applied.

According to the superposition of forces, all linear forces can be added vectorially. The net force is the vectorial sum of all the individual forces. Consider a system of n charges given by q₁, q₂, …, q_n, the force due to charges q₁ and q₂ is,

Where,

Similarly, the force between q₁ and any other charges are,

Therefore, the net force due to q₁ is,

• Applicable only to point charges at rest
• Applicable to cases where inverse-square law is obeyed
• It cannot be applied to irregular-shaped charges
• It cannot be applied to a continuous charge distribution

• To calculate the force between two charges
• To calculate the distance between two charges
• To study the motion of electrons around an orbit in an atom
• To calculate the electrostatic potential and hence the work done in moving a charge from one point to another
• To calculate the relative permittivity (dielectric constant) of a medium
• Inkjet printers use electrostatic forces to spray tiny ink droplets, which are charged when passing through a nozzle
• Photocopiers and laser printers use an electrostatic process in which an electrically charged drum coated with selenium attracts a blank piece of paper to print images

Other applications include power coating and electrostatic air cleaning

Problem 1. Two identical charges -6.25 nC are held apart at a separation distance of 42 cm. Determine the magnitude of the electrical force of repulsion between them.

Solution:

Given,

q₁ = 6.25 nC

q₂ = 6.25 nC

r = 42 cm = 0.42m

We have,

F = k q₁ q₂ /r²

   = (9 x 10⁹ Nm²C^-2 x 6.25 nC x 6.25 nC)/0.42² m²

   = 2 μN

Problem 2. The force between two identical charges separated by 2 cm is equal to 45 N. What is the magnitude of the two charges?

Solution:

Given,

F = 45 N

d = 2 cm = 0.02 m

q = q₁ = q₂

We have,

F = k q₁ q₂ /r²

   = k q²/r²

Or, q = sqrt (F r²/k)

          = sqrt (45 N x (0.02 m)²/ 9 x 10⁹ Nm²C^-2)

          = 1.41 μC

Problem 3. Two balloons with charges of +4 µC and -8 µC attract each other with a force of 0.0625 N. Determine the separation distance between the two balloons.

Solution:

Given,

F = 0.0625 N

q₁ = 4 μC = 4 x 10^-6 C

|q₂|= 8 μC = 8 x 10^-6 C

We have,

F = k q₁ q₂ /r²

Or, r = sqrt (k q₁ q₂ / F)

          = sqrt (9 x 10⁹ Nm²C^-2 x 4 x 10^-6 C x 8 x 10^-6 C/0.0625 N)

          = 2.15 m

Problem 4. Two neutrally charged bodies are separated by 0.5 cm. Electrons are stripped off from one body and placed on the second body until a force of 2 μN is generated between them. How many electrons were transferred between the bodies?

Solution.

Given,

q₁ = +ne

q₂ = -ne

e = 1.6 x 10^-19 C

r = 0.5 cm = 5 x 10 ^-3 m

F = 2 μN = 2 x 10^-6 N

We have,

F = k q₁ q₂ /r²

Or,|F| = k n²e² /r²

Or, n = sqrt (F r² / k e²)

          = sqrt (2 x 10^-6 N x (5 x 10 ^-3 m)² / (9 x 10⁹ Nm²C^-2 / (1.6 x 10^-19 C)²))

          = 6.5 x 10⁸