Question

Let $\left[{\epsilon }_0\right]$ denote the dimensional formula of the permittivity of vacuum. If $M=\text{mass},L=\text{length},T=\text{time and}A=\text{electric current}$, then :

• A. $\left[{\epsilon }_0\right]=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$
• B. $\left[{\epsilon }_0\right]=\left[M^{-1}{L}^{-3}{T}^{2}A\right]$
• C. $\left[{\epsilon }_0\right]=\left[M^{-1}{L}^2{T}^{-1}{A}^{-2}\right]$
• D. $\left[{\epsilon }_0\right]=\left[M^{-1}{L}^2{T}^{-1}A\right]$

Answer 1

The correct option is A $\left[{\epsilon }_0\right]=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

Using Coulomb law,

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{R^2}$

$\Rightarrow {\epsilon }_0=\frac{q_1{q}_2}{4\pi F{R}^2}$

Substituting the units in the equation:

So, ${\epsilon }_0=\frac{C^2}{N.m^2}=\frac{[AT{]}^2}{\left[ML{T}^{-2}\right].\left[L^2\right]}=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

Hence, option $\left(B\right)$ is correct.