The unit of pemtiftivity of free space ε₀ is

The correct option is coulomb²/newton-metre².

To find the unit of the permittivity of free space (${\epsilon }_0$), we can use Coulomb's Law, which describes the electrostatic force between two point charges.
Coulomb's Law is given by the formula:

$F=\frac{1}{4\pi {\epsilon }_0}$

where:
- $F$ is the electrostatic force between the charges (measured in newtons, N)
- $q_1$ and $q_2$ are the magnitudes of the charges (measured in coulombs, C)
- $r$ is the distance between the centers of the two charges (measured in metres, m)
- ${\epsilon }_0$ is the permittivity of free space
- $4\pi$ is a dimensionless constant

Rearranging the equation to solve for the permittivity of free space (${\epsilon }_0$), we get:

${\epsilon }_0=\frac{q_1{q}_2}{4\pi F{r}^2}$

Now, we substitute the SI units for each of the physical quantities into this rearranged equation to find the unit of ${\epsilon }_0$:

$\text{Unit of}{\epsilon }_0=\frac{\text{coulomb}\cdot \text{coulomb}}{\text{newton}\cdot {\text{metre}}^2}$

Simplifying the expression gives:

$\text{Unit of}{\epsilon }_0=\frac{{\text{coulomb}}^2}{\text{newton}\cdot {\text{metre}}^2}$

Therefore, the SI unit of permittivity of free space is coulomb²/newton-metre² (or ${\text{C}}^2{\text{N}}^{-1}{\text{m}}^{-2}$).