Question

The unit of permittivity of free space, $\varepsilon \circ $ is
(A) $\mathop {coulomb/newton - metre}\nolimits^{} $
(B) $\mathop {newton - metre}\nolimits^2 /\mathop {coulomb}\nolimits^2 $
(C) $\mathop {\mathop {coulomb}\nolimits^2 /newton}\nolimits^{} - \mathop {metre}\nolimits^2 $
(D) $\mathop {\mathop {coulomb}\nolimits^2 /(newton - \mathop {metre)}\nolimits^{} }\nolimits^2 $

Answer 1

Hint: permittivity of resistance to the electric field. Generally, permittivity of free space is represented by 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 $F = \dfrac{{\mathop q\nolimits_1 \mathop q\nolimits_2 }}{{4\pi \in \circ \mathop r\nolimits^2 }}$ , where $\mathop q\nolimits_1 and\mathop q\nolimits_2 $are two charges and r is the distance between two charges.

Complete step by step solution
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 of 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 $\mathop {8.85 \times 10}\nolimits^{ - 12} Farad/meter$
According to coulomb’s law, the force between two charges can be written as,
$F = \dfrac{{\mathop q\nolimits_1 \mathop q\nolimits_2 }}{{4\pi \in \circ \mathop r\nolimits^2 }}$ , where $\mathop q\nolimits_1 and\mathop q\nolimits_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.
$ \in \circ = \dfrac{{\mathop q\nolimits_1 \mathop q\nolimits_2 }}{{4\pi \mathop {Fr}\nolimits^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}}{{\mathop {N.m}\nolimits^2 }}$
$ \Rightarrow \mathop C\nolimits^2 \mathop N\nolimits^{ - 1} \mathop m\nolimits^{ - 2} $

So, the correct option is D.

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.