Question

The S.I unit of permittivity of free space $$({\epsilon }_0)$$ is:
A. $$C{N}^{-1}-m^{-1}$$
B. $$N{m}^{-2}{C}^{-2}$$
C. $$C^2{N}^{-2}{m}^{-2}$$
D. $$C^2{N}^{-1}{m}^{-2}$$

Answer 1

Hint: Permittivity is the resistance to the electric field. Generally, permittivity of free space is represented by $$\text{Farad/meter}$$. Here the options are in the terms of charge, force and length. To find that unit, we can use Coulomb’s law. Coulomb’s law can be written as, $$\text{F=}{\displaystyle \frac{q_1{q}_2}{4\pi {\epsilon }_0{r}^2}}$$, where $$q_1$$ and $$q_2$$ are two charges and r is the distance between two charges.

Complete Step by Step Answer:
Permittivity is a property of a material that can tell about the resistance of a material against the formation of an electric field. It is defined as the amount of charge required for the generation of one unit of electric flux in a specific medium. It depends upon the property of the medium. Generally, a charge will yield more electric flux in a low permittivity medium than the high permittivity medium.
Permittivity of the vacuum or free space is the lowest possible permittivity. It is treated as a physical constant and it is known as an electric constant. It has a value of $$8.85\times {10}^{-12}\text{Farad/meter}$$.
According to Coulomb’s law, the force between two charges can be written as,
$$\text{F=}{\displaystyle \frac{q_1{q}_2}{4\pi {\epsilon }_0{r}^2}}$$, where $$q_1$$ and $$q_2$$are two charges and r is the distance between two charges.
We can alter this equation to find the electric constant or permittivity of free space.
$${\epsilon }_0\text{=}{\displaystyle \frac{q_1{q}_2}{4\pi F{r}^2}}$$
To find the SI unit of permittivity of free space, we can substitute all the SI units of given quantities of the above equation.
$$\Rightarrow {\displaystyle \frac{C.C}{N.m^2}}$$
$$\Rightarrow C^2{N}^{-1}{m}^{-2}$$
So, the correct option is D.

Additional Information:
According to electrostatics, this unit can be further modified and can be written as Farad per meter.
Capacitance of a capacitor can be written as,
$$C={\displaystyle \frac{A{\epsilon }_0}{d}}$$, where A is the area of the plates of the capacitor and d is the distance between the capacitor.
$${\epsilon }_0={\displaystyle \frac{Cd}{A}}$$
So, the unit of electrical permittivity of free space will be,
$$\Rightarrow {\displaystyle \frac{\text{Farad}\times m}{m^2}}$$
So $$\text{Farad/meter}$$ can also be used as the unit of electrical permittivity of free space.

The mathematical expression of permittivity is,
$$\text{Permittivity = }{\displaystyle \frac{\text{Electric displacement}}{\text{Electric field density}}}$$

Note: Permittivity is actually the measurement of resistance to an electric field. Don’t confuse it with that name. It doesn’t mean the ability to permit. Relative permittivity is a ratio of permittivity of a medium to the permittivity of free space. Hence it doesn’t have units.