Let [ε₀] denote the permittivity of the vaccum, and [μ₀] denote the permeability of the vaccum. If M = mass, L= length, T = time and I= electrical current, then the dimensional formulae of ε₀ and μ₀ are

To find the dimensional formulae of the permittivity of free space (${\epsilon }_0$) and the permeability of free space (${\mu }_0$), we can derive them step-by-step from fundamental physical laws.

Step 1: Dimensional formula of Permittivity of vacuum (${\epsilon }_0$)
We can use Coulomb's Law, which defines the electrostatic force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ in a vacuum:
$F=\frac{1}{4\pi {\epsilon }_0}$

Rearranging the equation to solve for ${\epsilon }_0$ gives:
${\epsilon }_0=\frac{q_1{q}_2}{4\pi F{r}^2}$

Now, let's write the dimensional formula for each variable in terms of basic dimensions:
1. Electrical current $I=\frac{\text{Charge}\left(q\right)}{\text{Time}\left(T\right)}\Rightarrow \left[q\right]=\left[IT\right]$
2. Force $\left[F\right]=\left[ML{T}^{-2}\right]$
3. Distance $\left[r\right]=\left[L\right]$
4. The number $4\pi$ is dimensionless.

Substituting these dimensional formulae into the expression for ${\epsilon }_0$:
$\left[{\epsilon }_0\right]=\frac{\left[IT\right]\left[IT\right]}{\left[ML{T}^{-2}\right]\left[L^2\right]}$

Simplifying the expression:
$\left[{\epsilon }_0\right]=\frac{\left[I^2{T}^2\right]}{\left[M{L}^3{T}^{-2}\right]}=\left[M^{-1}{L}^{-3}{T}^4{I}^2\right]$

Step 2: Dimensional formula of Permeability of vacuum (${\mu }_0$)
We can use the relation between the speed of light in vacuum ($c$), permeability (${\mu }_0$), and permittivity (${\epsilon }_0$):
$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$

Squaring both sides:
$c^2=\frac{1}{{\mu }_0{\epsilon }_0}\Rightarrow {\mu }_0=\frac{1}{{\epsilon }_0{c}^2}$

The speed of light is velocity, which has the dimensional formula:
$\left[c\right]=\left[L{T}^{-1}\right]\Rightarrow \left[c^2\right]=\left[L^2{T}^{-2}\right]$

Substituting the dimensions of ${\epsilon }_0$ and $c^2$ in the formula for ${\mu }_0$:
$\left[{\mu }_0\right]=\frac{1}{\left[M^{-1}{L}^{-3}{T}^4{I}^2\right]\left[L^2{T}^{-2}\right]}$

Simplifying the denominator:
$\left[M^{-1}{L}^{-3}{T}^4{I}^2\right]\left[L^2{T}^{-2}\right]=\left[M^{-1}{L}^{-1}{T}^2{I}^2\right]$

Thus, we obtain:
$\left[{\mu }_0\right]=\frac{1}{\left[M^{-1}{L}^{-1}{T}^2{I}^2\right]}=\left[ML{T}^{-2}{I}^{-2}\right]$

Therefore, the dimensional formulae of ${\epsilon }_0$ and ${\mu }_0$ are $\left[M^{-1}{L}^{-3}{T}^4{I}^2\right]$ and $\left[ML{T}^{-2}{I}^{-2}\right]$, respectively.