Let $$[\epsilon_0]$$ denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then :

We start from Coulomb’s law, which gives the electrostatic force between two point charges in vacuum:

$$F \;=\; \dfrac{1}{4\pi\epsilon_0}\,\dfrac{q_1\,q_2}{r^{2}}.$$

Here $$F$$ is force, $$q_1$$ and $$q_2$$ are the magnitudes of the two charges, $$r$$ is the separation between them and $$\epsilon_0$$ is the permittivity of free space.

For dimensional analysis we may take both charges to have the same magnitude $$q$$, so the equation becomes

$$F \;=\; \dfrac{1}{4\pi\epsilon_0}\,\dfrac{q^{2}}{r^{2}}.$$

Re-arranging to isolate $$\epsilon_0$$ we have

$$4\pi\epsilon_0\,F\,r^{2} \;=\; q^{2},$$

and therefore

$$\epsilon_0 \;=\; \dfrac{q^{2}}{4\pi\,F\,r^{2}}.$$

The numerical factor $$4\pi$$ is dimensionless, so it does not influence the dimensional formula. Hence the dimensional symbol of $$\epsilon_0$$ is simply the ratio

$$[\epsilon_0] \;=\; \dfrac{[q]^{2}}{[F]\,[r]^{2}}.$$

Now we write the dimensions of each quantity in terms of the fundamental symbols $$M,\;L,\;T,\;A$$ provided in the question:

• Electric charge: $$[q] = [A\,T] = A\,T.$$

• Force: $$[F] = [M\,L\,T^{-2}] = M\,L\,T^{-2}.$$

• Distance: $$[r] = [L] = L.$$

Substituting these into the ratio we get

$$[\epsilon_0] = \dfrac{(A\,T)^{2}}{\bigl(M\,L\,T^{-2}\bigr)\,(L)^{2}}.$$

We now expand the powers explicitly:

$$[\epsilon_0] = \dfrac{A^{2}\,T^{2}}{M\,L\,T^{-2}\;L^{2}}.$$

Combining the like dimensions in the denominator first gives

$$M\,L\,L^{2} = M\,L^{3}, \qquad T^{-2} \text{ remains as is.}$$

So the expression becomes

$$[\epsilon_0] = \dfrac{A^{2}\,T^{2}}{M\,L^{3}\,T^{-2}}.$$

Now we bring together the time dimensions in the numerator and denominator. Dividing by $$T^{-2}$$ is equivalent to multiplying by $$T^{2}$$, hence

$$T^{2} \times T^{2} = T^{4}.$$

Thus we have

$$[\epsilon_0] = A^{2}\,T^{4}\,M^{-1}\,L^{-3}.$$

Writing the exponents in the conventional order $$M,\;L,\;T,\;A$$ gives

$$[\epsilon_0] = [M^{-1}\,L^{-3}\,T^{4}\,A^{2}].$$

This matches option D in the list provided.

Hence, the correct answer is Option D.