Question

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

Answer 1

Hint Applying the Coulomb’s law, first we can establish the relationship between the permittivity of free space and the force between two electrostatic charges. Using this relationship, we can determine the dimension of The permittivity.
Formula Used:
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ where $F$ is the force exerted by one static charge ${q_1}$ on another charge ${q_2}$ in free space and $r$ is the distance between the two charges, ${\varepsilon _0}$ being the permittivity of free space.

Complete step by step answer
The permittivity of free space is defined as the capability of an electric field to permeate free space or vacuum. Its value is approximately $8.85 \times {10^{ - 12}}F{m^{ - 1}}$ in SI units.
Now, we can determine its dimension with the help of its usage in the formula of Coulomb’s force.
The Coulomb’s law states that two stationary point charges repel or attract each other with a force $F$ which is directly proportional to the product of the charges and inversely proportional to the square of the distance $r$ between them.
The Coulomb’s law holds for point charges and is only valid for static charges.
The value of the Coulomb’s force is given as,
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ where $F$is the force exerted by one static charge ${q_1}$ on another charge ${q_2}$ in free space and $r$is the distance between the two charges.
Rearranging this equation we get,
${\varepsilon _0} = \dfrac{1}{{F4\pi }}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Now, the dimensions of force, charge and distance are $\left[ {{M^1}{L^1}{T^{ - 2}}} \right]$,$\left[ {{M^0}{L^0}{T^1}{A^1}} \right]$ and $\left[ {{M^0}{L^1}{T^0}{A^0}} \right]$ respectively.
Substituting the dimensions in place of the physical quantities gives,
${\varepsilon _0} = \dfrac{{{T^2}{A^2}}}{{{M^1}{L^1}{T^{ - 2}} \times {L^2}}} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$

Therefore the correct option is A.

Note The permittivity of free space as mentioned in the name, is only valid in free space. When the charges are present in a material medium having a dielectric constant of $K$, then the permittivity changes by a value $\varepsilon = K{\varepsilon _0}$ where $\varepsilon$ has the same dimension as that of ${\varepsilon _0}$. So in such a case, the force between the two charges decreases by a factor $K$.