Categories: Education, Physics, Science


COULOMB’S LAW is also the related topic to the electrostatics.


We know that a force of attraction or repulsion acts between
two charged bodies. How is this force affected when the
quantity of the charge on the two bodies or the distance
between them is changed? In order to find the answers of
these  questions,  a  French  scientist  Charles  Coulomb
(1736–1806)  in  1785  experimentally  established  the
fundamental law of electric force between two stationary
charged particles.

Coulomb’s Law: The force of attraction or repulsion between
two point charges is directly proportional to the product of
the quantity of charges and inversely proportional to the
square of the distance between them. Therefore

Fq1q2…….. (13.1)
F=1/r2…….. (13.2)

Combining Eqs. (13.1) and (13.2), we get

F=K(q1q2/r2)……..  (13.3)

Eqution (13.3) is known as Coulomb’s law.
where F is the force between the two charges and is called
the Coulomb force, q1  and q2  are the quantities of two

charges and r is the distance between the centre of two
charges (Fig. 13.12-a,b). K is the constant of proportionality
given by

k =1⁄ 4π ε0

The value of k depends upon the medium between the two
charges and the system of units in which F, q, and r are
measured, ε0is the permittivity of free space.

Now if the medium between the two charges is air then the

value of K in SI units will be 9 ×109Nm2C-2.
Coulomb’s law is true only for point charges whose sizes are
very small as compared to the distance between them. Like
other forces, electric forces also obey Newton’s third law.


Example 13.1:
Two bodies are oppositely charged with
500 µC and 100 µC. Find the force between the two charges if
the distance between them in air is 0.5m.
 Solution:Given, q1 = 500 µC = 500 × 10-6 C , q2 = 100 µC = 100 ×
10-6C,  Distance between charges r = 0.5m
Substituting these values in equation of Coulomb’s law, we

F=q1q2/4Πεor2=9× 106Nm2C-2×500×10-6C×100×10-6C/(0.5m)2


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