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**Hint**There can be two ways to approach the solution, first, if you know the dimensional formula of $\varepsilon $ (permittivity of free space), just inverse the dimensional formula since the relation between $k$ and $\varepsilon $ is given as:

$k = \dfrac{1}{{4\pi \varepsilon }}$

Other than that, you can use the relation between the force between two charged particles at rest, lying at a distance of $r$ from each other with charges ${q_1}$ and ${q_2}$ . The relation is:

$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$

Take the dimensional formula of charge as: ${I^1}{T^1}$

**Complete step by step answer**

As explained in the hint section of the solution to the asked question, we will approach the solution by finding out the dimensional formula of Coulomb’s Constant using the relation between force, distance and charges.

The relation is given as:

$F = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$

As we already know, the SI unit of force is $N$ and the dimensional formula is nothing but: $\left[ {{M^1}{L^1}{T^{ - 2}}} \right]$

On the other hand, the dimensional formula of the distance, as the denominator term on the right-hand side, $r$ is:

$\left[ {{L^1}} \right]$

The dimensional formula of charge is already told to be $\left[ {{I^1}{T^1}} \right]$

Now, if we perform transposition in the equation, we get:

$k = \dfrac{{F{r^2}}}{{{q_1}{q_2}}}$

Substituting the quantities with their dimensional formulae, we get;

$k = \dfrac{{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]{{\left[ {{L^1}} \right]}^2}}}{{\left[ {{I^1}{T^1}} \right]\left[ {{I^1}{T^1}} \right]}}$

If we simplify the above-mentioned term on the right-hand side of the equation, we get:

$k = \left[ {{M^1}{L^3}{T^{ - 4}}{I^{ - 2}}} \right]$

Hence, the dimensional formula of $k$ (Coulomb’s Constant) is: $\left[ {{M^1}{L^3}{T^{ - 4}}{I^{ - 2}}} \right]$

As we can see, this matches with the dimensional formula given in the option (C). Hence,

**option (C) is the correct answer to the question.**

**Note**A major mistake that students make is that they take the dimension of charge exactly as $I$ instead of ${I^1}{T^1}$ . Think of it as the equation of charge, $Q = IT$ Where, $Q$ is the charge passing through a particular cross-sectional area., $I$ is the current flowing through the cross-sectional area and $T$ is the time for which the current flowed through the cross-sectional area.

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