Question

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

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

Answer 1

Answer: (A)

${ϵ}_0=\left[M^{-1}{L}^{-3}{T}^2{A}^2\right]$

Using Coulomb law, $F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{R^2}$

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

Substituting the units in the equation:

So, ${ϵ}_0=\frac{C^2}{N.m^2}$

$=\frac{[AT{]}^2}{ML{T}^{-2}.L^2}=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$