Coulomb’s Law is a fundamental law of electrostatics. It describes the electrostatic interaction between the charged particles. The law basically states that the force between charges is proportional to the product of their charges and inversely proportional to the square of distance between them. More precisely, we can say that:

The force between two point charges are

  • Directly proportional to the magnitude of each charge
  • Is always in the direction of the separation vector connecting their centres
  • It is inversely proportional to the square of the separation between their centres 

In free space

F = q1q2/ 4π∈0 r2

Here ∈0 is the permittivity of the medium 

In material medium

F = q1q2/ 4π∈r2

As explained in the previous case εi stands for relative permittivity of the medium in vector form. 

Coulomb’s Law
Some of the important highlights of this law are listed below:
  • The law is based on physical observation and is the fundamental law of electrostatics
  • The force is an action-reaction pair
  • The direction of force is always along the line joining the two charges
  • Electrical force between two point charges is independent of presence or absence of other charges
  • If the two charges are of same sign, then their product is obviously positive and hence the force vector is directed with the separation vector. Such an action is repulsive
  • Similarly, if the two charges have opposite signs, their product is negative and the force vector is against the separation vector which denotes an attractive reaction.
The following video will further provide you more information on the law:

There exists some force of interaction between the charged particles but it acts over some distance of separation. Whether we consider the case of a plastic tube attracting paper bits or the repulsion between two same charged balloons, there are always two charges with some distance between them. The strength of interaction depends to a large extent on these three variables.

The quantitative expression that describes the influence of these three variables on electric force is known as the Coulomb’s law. In the form of an equation, the law can be written as

F= k.Q1.Q2/d2

Where Q1 represents the quantity of charge on object 1 and Q2 stands for the quantity of charge on object 2 in coulombs. These two quantities are generally expressed as ‘+’ or ‘-‘ which denote positive and negative charge. While negative charge denotes the presence of an excess number of electrons, the positive charge stands for a shortage of electrons. In terms of force, the negative sign represents a attractive force, the positive sign stands for a repulsive force. The symbol‘d’ represents the distance of separation between the two objects and ‘k’ is the proportionality constant called as the Coulomb’s law constant. This constant is affected by the medium of immersion of charged objects. In particular, for air the value of this k equals

9.0 x 109 N • m2 / C2.

If the medium of propagation or immersion is water then the constant k can also be reduced by a factor of 80. It is clearly evident form the mathematical expression of the coulomb’s law that when the units of k will be substituted into the equation, the units of charge and distance will get cancelled and ultimately, we will be left with Newton as the unit of force. The equation of the law clearly describes the force acting between the objects when they are assumed to be point charges. Although the charge is evenly distributed all throughout the sphere, the centre of the sphere can be assumed to be carrying all of the charge.

Mathematically, the net force value will be found to be positive if both Q1 and Q2 are of same charges whether both negative or both positive. On the contrary, if one of the charges is positive and other is negative, then the net charge would be negative.   

Let us discuss some of the problems based on Coulomb’s Law for IIT JEE.

Example: A particle ‘A’ having a charge of 2 × 10-6C and a mass of 100g is fixed at the bottom of a smooth inclined plane of inclination 30°. Where should another particle B, having same charge and mass be placed on the incline so that it may remain in equilibrium?

Solution: First of all draw the F.B.D. of the masses.
            For equilibrium ∑F = 0
            N = mg cos30° 

Electric Field Intensity: It is the force experienced by a unit positive charge placed at a point in an electric field

Example: Two particles A and B having charges 8 x10-6 C and –2 x10-6C respectively are held fixed with a separation of 20 cm. Where a third charged particle should be placed so that it does not experience a net electric force?

Solution: As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B.
Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. The situation is shown in figure. Suppose BC=x and the charge on C is Q

Superposition principle:    
Force experienced by a given charge in the field of a number of point charges is the vector sum of all the forces.

Lines of Force:
• Lines of force originate from positive charges and terminate on negative charges
• Lines of force originate or terminate perpendicular to the surface
• Tangent to the lines of force at any point gives the direction of the electric field.
• Lines of force do not intersect. 

askIITians offers wide ranging study material containing various questions on Coulomb’s Law. It also contains numerous free questions based on electric line of forces. It is vital for the JEE aspirants to practice previous year’s IIT questions on Coulomb’s Law and the study material of Coulomb’s law of askIITians is flooded with such types of questions.   

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