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

Answer 1

The correct option is A

$\left[{ϵ}_0\right]=\left[M^{–1}{L}^{-3}{T}^4{A}^2\right]$

According to Coulomb,s law,

$f=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$

$\therefore {ϵ}_0=\frac{1}{4\pi }\frac{q_1{q}_2}{F{r}^2}$

$\left[{ϵ}_0\right]=\frac{\left[AT\right]\left[AT\right]}{\left[ML{T}^{-2}\right][L{]}^2}=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$