Question

The electrostatic force between two point charges $q_1$ and $q_2$ at separation '$r$' is given by $F{\text{ }} = {\text{ }}\dfrac{{K{q_1}{q_2}}}{{{r^2}}}$
The constant K:
A) Depends on the system of units only
B) Depends on the medium between the charges only
C) Depends on both the system of units and the medium between the charges
D) Is independent of both the system of units and the medium between the charges

Answer 1

Hint
K is a constant for only a specific medium. K depends on the permittivity of the medium which is different for different media. Value of k is dependent on the units as well.

Complete step-by-step answer
K is a constant and its value is $8.98755 \times {10^9}\;kg{\text{ }}{m^3}{s^ - }^2{C^ - }^2$
It is expressed in terms of as:
$K = \dfrac{1}{{4\pi \epsilon }}$
Here the only variable is, and it is the measure of the lines of force that can pass through the medium and is known as permittivity of a medium. It depends on the medium according to the dielectric coefficient of that medium. Therefore,
$\epsilon = k{\epsilon _0}$
Where k is the dielectric coefficient of the medium and is known as permittivity of free space or vacuum and its value is $8.85418782{\text{ }} \times {\text{ }}{10^{ - 12}}\;{m^{ - 3}}\;k{g^{ - 1}}\;{s^4}\;{A^2}$. In the CGS system, the dimension reduces to unit less and thus $k$ also becomes unit less. Therefore, the value of constant depends on the dielectric of the medium and the system of units that we are using.
So, the option with the correct answer is option C.

Note
Usually one assumes that constants are dimensionless. But that isn’t true. In this question we have to take care of that very specifically. Higher the value of dielectric coefficient, lower will be the value of electrostatic constant, lesser will be the force between the 2 charges.