Question

Let $\left[ {{\varepsilon _0}} \right]$ denotes the dimensional formula of the permittivity of vacuum. If $M$= mass, $L$ = length, $T$ = time and $A$ = electric current, then
A. $\left[ {{\varepsilon _0}} \right] = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$
B. $\left[ {{\varepsilon _0}} \right] = \left[ {{M^{ - 1}}{L^2}{T^{ - 1}}{A^{ - 2}}} \right]$
C. $\left[ {{\varepsilon _0}} \right] = \left[ {{M^{ - 1}}{L^2}{T^{ - 1}}A} \right]$
D. $\left[ {{\varepsilon _0}} \right] = \left[ {{M^{ - 1}}{L^{ - 3}}{T^2}A} \right]$

Answer 1

Hint: The permittivity of vacuum $\left[ {{\varepsilon _0}} \right]$ and the mass, the length, time and electric current are related by Coulomb's law. By applying the force equation given in Coulomb’s law, the relation between the given parameters will be defined.
And substitute the SI unit of the term used in the relation of force by Coulomb’s law. The rearrange the equation in the form which gives the value of the permittivity of vacuum. Thus, the dimensional formula for the permittivity of vacuum can be derived.

Useful formula:
The coulomb’s law is given by,
$F = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Where, $F$ is the force of attraction or repulsion between the charges, ${\varepsilon _0}$ is the permittivity of vacuum or free space,${q_1}$ is the magnitude of first charge, ${q_2}$ is the magnitude of second charge and $r$ is the separation distance between the charges ${q_1}$ and ${q_2}$.

Newton’s second law of motion,
$F = ma$
Where, $F$ is the force on the body, $m$ is the mass of the body and $a$ is the acceleration of the body.

The current flows through the conductor,
$I = \dfrac{q}{t}$
Where, $I$ is the current flowing through the conductor, $q$ is the charge flowing in the conductor and $t$ is the time taken.

Given data:
$M$= mass
$L$ = length
$T$ = time
$A$ = electric current

Step by step solution:
By Coulomb’s law,
$F = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Rearranging the above equation, we get
${\varepsilon _0} = \dfrac{1}{{4\pi F}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}\;.............................................\left( 1 \right)$

By Newton’s second law of motion,
$F = ma\;...................................\left( 2 \right)$

The current flows through the conductor,
$I = \dfrac{q}{t}$
Rearranging the above equation, we get
$q = It\;..........................................\left( 3 \right)$
Substitute the equations (2) and (3) in equation (1), we get
${\varepsilon _0} = \dfrac{1}{{4\pi ma}} \times \dfrac{{{I_1}{t_1} \times {I_2}{t_2}}}{{{r^2}}}\;....................................\left( 4 \right)$
Substitute the SI units on RHS of above equation (4), we get
${\varepsilon _0} = \dfrac{1}{{kg \times m{s^{ - 2}}}} \times \dfrac{{As \times As}}{{{m^2}}}\;$
Since, $kg$ is unit of mass denoted by $M$, $m$ is unit of length denoted by $L$ and $s$ is unit of time denoted by $T$
Thus,
${\varepsilon _0} = \dfrac{1}{{M \times L{T^{ - 2}}}} \times \dfrac{{AT \times AT}}{{{L^2}}}\; \\\ {\varepsilon _0} = \dfrac{{{A^2}{T^2}}}{{M{L^3}{T^{ - 2}}}} \\\ {\varepsilon _0} = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2} \\\$

Hence, the option (A) is correct.

Note: The conventional unit of the permittivity of free space or vacuum is derived from Coulomb's law. While applying the conventional units, care should be taken more. The convention unit of mass is $M$, length is $L$, time is $T$ and current is $A$. Hence, don’t get confused with the SI units and the conventional units. Both are completely different from one another.