Definition, Formula, Application, Limitation & Examples

Coulomb’s Law

Coulomb’s Law: According to this law, “The electrical force acting between two charged objects is equal to the product of such charged particles and inversely proportional to the square of the distance between them. The force acting can be attractive or repulsive. Such force is also called an electrostatic force. It quantifies the amount of force between the two charged particles or objects. This concept was first formulated by French Physicist Charles-Augustin de Coulomb in the year 1785. This force acts along a straight line. If the charged object has the same sign then the force acting is repulsive and if the charged object has the opposite sign then the force acting is attractive in nature.

Coulomb’s Law Formula

As we know, coulomb’s law is directly proportional to the product of two charged objects and inversely proportional to the distance between them. Mathematically, it can be written as-

F ∝ q1q2/d²

F = 1/4π€° x q1q2/d²

Where,

After removing the proportion we get a proportionality constant of 1/4π€° which is equal to the value of 9×10^9 N-m²/C².

q1 and q2 are the two charged bodies.

d² is the square of the distance between the two bodies.

Units, Dimensions and Value of €°

€° = 1/4πF x q1q2/d²

As SI Unit of charge is coulomb(C)

Therefore,

Unit of €° = 1CC/ Nm² = C²N^-1m^-2

Dimensions of €° = [M^-1L^-3T^4A^2]

Value of €°= 8.85 x 10^-12 C²N^-1m^-2

Vector Form of Coulomb’s Law

There are two types of quantities, that is, scalar and vector quantities. Scalar quantity determines the magnitude only whereas Vector quantity determines the direction and magnitude. As we know force is the vector quantity that has both magnitude and direction. Hence Coulomb’s law is in vector form.

Let us assume that there are two charges q1 and q2 respectively. Since both the charges are of the same sign, the force acting will be a repulsive force between them. Let the force on the q1 charge due to q2 be F12 and the force on the q2 charge due to the q1 charge be F21. The force on q1 due to q2 is equal and opposite to the force acting on q2 due to q1. Thus, Coulomb’s law obeys Newton’s third law of motion.

F12 = -F21

Therefore, the force acting in vector form is F12 = 1/4π€° x q1q2/d² = -F21

Application of Coulomb’s Law

The application of Coulomb’s law is-

• In order to calculate the distance and force between the two charges.
• Coulomb’s law also helps to calculate the electric field.

Limitations of Coulomb’s Law

The limitation of Coulomb’s law is-

• The charged object must have a spherically symmetric distribution (e.g. point charges, or a charged metal sphere).
• The charges should not overlap (e.g. they must be distinct point charges).
• The charge body must be stationary with respect to the other body.

Solved Example of Coulomb’s Law

Example1: How is the force between two charges affected when each charge is doubled and the distance between them is also doubled?

Solution: As we know, F ∝ q1q2/d²

Therefore, F becomes (2) (2)/(2)² time = 1 time, that is, Force remains the same.

Example2: Two charged particles having charge 2.0 x 10^-8 C each is joined by an insulating string of length 1m and the system is kept on a smooth horizontal table. Find the tension in the string.

Solution: Here q1 = q2 = 2.0 x 10^-8 C , r = 1m

Tension in the string act as the force of repulsion(F) between the two charges.

According to Coulomb’s law, F = 1/4π€° x q1q2/d² = 9 x 10^9 (2 x 10^-8) (2 x 10^-8)/ 1² = 3.6 x 10^-6 N

Example 3: Two-point charges, q1 = +12 μC and q2 = 8μC are separated by a distance r = 20 cm. Calculate the magnitude of the electric force.

Solution: Here, q1 = +12 μC = 12 x 10^-6 C

q2 = 8μC = 8 x 10^-6C

r = 20 cm = 0.20m

According to Coulomb’s law, F = 1/4π€° x q1q2/d²

F = (9×10^9) (12 x 10^-6) (8 x 10^-6)/ (0.20)²

F = 25.92 N