Question

The S.I unit of permittivity of free space \[({{\varepsilon }_{0}})\] 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, \[\text{F=}\dfrac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\], 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,
\[\text{F=}\dfrac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\], 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.
\[{{\varepsilon }_{0}}\text{=}\dfrac{{{q}_{1}}{{q}_{2}}}{4\pi F{{r}^{2}}}\]
To find the SI unit of permittivity of free space, we can substitute all the SI units of given quantities of the above equation.
\[\Rightarrow \dfrac{C.C}{N.{{m}^{2}}}\]
\[\Rightarrow {{C}^{2}}{{N}^{-1}}{{m}^{-2}}\]
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,
\[C=\dfrac{A{{\varepsilon }_{0}}}{d}\], where A is the area of the plates of the capacitor and d is the distance between the capacitor.
\[{{\varepsilon }_{0}}=\dfrac{Cd}{A}\]
So, the unit of electrical permittivity of free space will be,
\[\Rightarrow \dfrac{\text{Farad}\times m}{{{m}^{2}}}\]
So \[\text{Farad/meter}\] can also be used as the unit of electrical permittivity of free space.

The mathematical expression of permittivity is,
\[\text{Permittivity = }\dfrac{\text{Electric displacement}}{\text{Electric field density}}\]

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.