Question

The constant k in Coulomb’s law depends upon
A. nature of medium
B. system of units
C. intensity of charge
D. Both (a) and (b)

Answer 1

Hint: The value of k depends on the electric permittivity of the medium ($\varepsilon $). The electric permittivity of a medium depends on the nature of the medium. This means that its value is different for different mediums.

Complete step by step answer:
We know that any two bodies having mass exert a force of attraction on each other. This force of attraction is called the gravitational force or force of gravity. Similarly, any two charges in the universe exert a force on each other, which is called an electrostatic force. Unlike gravity, electrostatic force can be attractive as well as repulsive depending on the nature of the charges.

The electrostatic force between two point charges is given by Coulomb's law. According to the law, if there are two point charges (say ${{q}_{1}}$ and ${{q}_{2}}$) that are separated by a distance r, then the electrostatic that each one exert on the other charge is given as $F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$ …. (i)
Here, k is a proportionality constant called the Coulomb’s constant.
It is found that the value is constant only for a given medium. The value of k depends on the electric permittivity of the medium ($\varepsilon $). The electric permittivity of a medium depends on the nature of the medium. This means that its value is different for different mediums.
It is found that $k=\dfrac{1}{4\pi \varepsilon }$ …. (ii).
Therefore, we can say that the value of k depends on the nature of the medium (since k depends $\varepsilon $).

Hence, the correct option is A.

Note:According to Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges considered to be point charges. The value of k from vacuum (also known as free space) is equal to $9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$.