The value of ${\varepsilon _o}$(permittivity of the free space) is
A. $8.854 \times {10^{ - 12}}\dfrac{{coulomb}}{{N{m^2}}}$
B. $8.854 \times {10^{ - 12}}\dfrac{{N{m^2}}}{{coulomb}}$
C. $8.854 \times {10^{ - 12}}\dfrac{{coulomb^2}}{{N{m^2}}}$
D. $8.854 \times {10^{12}}\dfrac{{coulomb^2}}{{N{m^2}}}$

Hint:Hint: The given question is about the value of permittivity of the free space. This question is easy to solve as all the values given here in the options are the same and the units are different. So one can easily solve this problem if the concept of permittivity of the free space is known. In general permittivity is an electrical characteristic of a medium on account of which it absorbs the effect of charge due to which a force is applied on the other charge.

Complete step by step answer:Permittivity of the free space ${\varepsilon _o}$ is a constant that represents the capability of the vacuum to permit electric fields. Permittivity of the free space is related to the Coulomb’s constant and their relationship is given by the expression given below.
$C = \dfrac{1}{{4\pi {\varepsilon _o}}}$……….(1)
Now according to the Coulomb’s law we know that the expression for force (F) exerted by a charge of magnitude q on the other charge situated at a distance r having a charge of similar magnitude can be written as,
$F = C\dfrac{{{q^2}}}{{{r^2}}}$.............(2)
From equation (1) and equation (2) we have the following equation,
$F = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{{q^2}}}{{{r^2}}}$
Therefore, the permittivity of the free space is given by,
${\varepsilon _o} = \dfrac{1}{{4\pi F}}\dfrac{{{q^2}}}{{{r^2}}}$.........(3)
Put $F = 1N$, $r = 1m$ and $q = 1$ coulomb in equation (3),
${\varepsilon _o} = \dfrac{{{1^2}}}{{4\pi \times 1 \times {1^2}}}$
${\varepsilon _o}$ = $8.854 \times {10^{ - 12}}$ $\dfrac{{coulomb^2}}{{N{m^2}}}$
Hence, permittivity of the free space has a value of $8.854 \times {10^{ - 12}}\dfrac{{coulomb^2}}{{N{m^2}}}$.

Therefore, option C is the correct answer option.

Note:
-The unit of permittivity is $\;{{\mathbf{C}}^{\mathbf{2}}}{{\mathbf{N}}^{ - {\mathbf{1}}}}{{\mathbf{m}}^{ - {\mathbf{2}}}}$.
-Value of the permittivity depends upon the medium.
-The permittivity of free space is a very important factor which helps us to find the Coulomb force.