Question

The dimension of permittivity of vacuum
${\text{A}}{\text{. }}\left[ {{M^{ - 2}}{L^{ - 3}}{T^4}{A^2}} \right]$
${\text{B}}{\text{. }}\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^3}} \right]$
${\text{C}}{\text{. }}\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$
${\text{D}}{\text{. }}\left[ {{M^{ - 1}}{L^{ - 5}}{T^4}{A^2}} \right]$

Answer 1

Hint – The permittivity of the vacuum is given by the formula ${F_e} = \dfrac{1}{{4\pi { \in _ \circ }}}.\dfrac{{{q^2}}}{{{r^2}}}$ , which implies ${ \in _ \circ } = \dfrac{{{q^2}}}{{({F_e}) \times {{(r)}^2}}}$ , then put the dimension of the force, charge and distance to find the dimension of permittivity.

Formula used - ${F_e} = \dfrac{1}{{4\pi { \in _ \circ }}}.\dfrac{{{q^2}}}{{{r^2}}}$.

Complete step-by-step solution -
We have been asked to find the dimension of permittivity.
So, we know the formula, ${F_e} = \dfrac{1}{{4\pi { \in _ \circ }}}.\dfrac{{{q^2}}}{{{r^2}}}$
Therefore, the permittivity will be, ${ \in _ \circ } = \dfrac{{{q^2}}}{{({F_e}) \times {{(r)}^2}}}$
Here, ${ \in _ \circ }$ is the permittivity of the vacuum, ${F_e}$ is the force and r is the distance.
Now, we can write charge as $q = i \times t$ , here i is the current and t is the time.
So, the dimension of q is-
$[q] = [AT]$
Also, force is mass * acceleration.
So, dimension of force is ${F_e} = ML{T^{ - 2}}$ , as dimension of mass is M and acceleration is $L{T^{ - 2}}$
Also dimension of distance is $[r] = L$
So, keeping all these values, we get-
${ \in _ \circ } = \dfrac{{{{[AT]}^2}}}{{[ML{T^{ - 2}}]{{[L]}^2}}} = \dfrac{{{{[AT]}^2}}}{{[M{L^3}{T^{ - 2}}]}} = [{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}]$
Therefore, the dimension of permittivity is $[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}]$
Hence, the correct option is C.

Note – Whenever such types of questions appear then first write the formula to find the permittivity of vacuum and then write the dimensions of the terms in the formula, as mentioned in the solution to obtain the dimensional formula of permittivity of vacuum.