Question

S.I unit of permittivity is:
A) \[{C^2}{m^2}{N^2}\]
B) \[{C^2}{m^ - }^2{N^{ - 1}}\]
C) \[{C^2}{m^2}{N^{ - 1}}\]
D) \[{C^{ - 1}}{m^2}{N^{ - 2}}\]

Answer 1

Hint: To get the correct answer to this question we can write a formula where the required quantity is appeared and use the known S.I. units which means the international system of units, all the other quantities we can get the units of our required quantity.

Complete step by step answer:
Here we will be using coulomb’s law in electrostatics
$F = \dfrac{{{q_1} \times {q_2}}}{{4\pi { \in _0} \times {r^2}}}$
Where we know that,
F= force between charges q₁ and q₂
S.I. unit of force: Newton=N
q₁ and q₂ are the charges
S.I. unit of charge = Coulomb=C
r = distance between the charges
S.I. unit of distance: meter=m
${ \in _0}$=permittivity
4 and π are constants so and are unitless
${ \in _0} = \dfrac{{{q_1} \times {q_2}}}{{F \times 4\pi \times {r^2}}}$
Units of permittivity \[{\text{ = }}\left[ {\dfrac{{{\text{unit of charge }}{{\text{q}}_1} \times {\text{unit of charge }}{{\text{q}}_2}}}{{{\text{unit of force}} \times {{\left( {{\text{unit of distance}}} \right)}^{\text{2}}}}}} \right]\]
= ${C^2} \times {N^{ - 1}} \times {m^{ - 2}}$
Hence, (B) is the correct option.

Additional Information: Permittivity is the property of every material, which measures the opposition offered against the formation of an electric field. It is represented by the Greek alphabet ϵ. It tells the material to store potential energy under influence of an electric field. In electromagnetism, the absolute permittivity, often simply called permittivity and denoted by the Greek letter ε, is a measure of the electric polarizability of a dielectric.

Note:
Another method:
We can use the formula for finding the capacitance of a capacitor to find the unit of permittivity
We know that $C = { \in _0} \times A/d$
Where C= capacitance of a capacitor
Units of capacitance=farad
Farad in units=\[{s^4} \cdot {A^2} \cdot {m^{ - 2}} \cdot k{g^{ - 1}}\]
We know that A = Ampere “unit of the current “
Since current = charge /per unit time
So, A = C/s, Where “C“ is unit of charge
Farad=\[\]\[{s^4} \cdot {C^2} \cdot {m^{ - 2}} \cdot k{g^{ - 1}}{s^{ - 2}}\]
“d” is the distance between plates
Unit of d = m
”A” is the area of the plate
Units of A = m2
${ \in _0} = C \times d/A$
Calculating the SI unit of ε:
Required units = \[\dfrac{{unit{\text{ }}of{\text{ }}C \times unit{\text{ }}of{\text{ }}d}}{{unit{\text{ }}of{\text{ }}A}}\]
= \[{s^4} \cdot {C^2} \cdot {m^{ - 2}} \cdot k{g^{ - 1}}{s^{ - 2}} \times m/{m^2}\]
= $({s^2}{m^{ - 1}}k{g^{ - 1}}){C^2}/{m^2}$
Where $({s^2}{m^{ - 1}}k{g^{ - 1}}) = {N^{ - 1}}$
So the unit of permittivity =${C^2}{N^{ - 1}}{m^{ - 2}}$
Hence, Option (B) is the correct answer.