Question

What is the dimensional formula for permittivity of free space?

• A. $\left[M^{-1}{L}^{-3}{T}^0{A}^2\right]$
• B. $\left[M^{-1}{L}^3{T}^0{A}^2\right]$
• C. $\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$
• D. $\left[M^{-1}{L}^{-3}{T}^4\right]$

Answer 1

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

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

(Coulomb's law)

$\left[F\right]=\left[ML{T}^{-2}\right]$
$\left[r^2\right]=\left[L^2\right]$

Charge $q=\frac{I}{t}$
$\left[q\right]=\left[AT\right]$

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

Hence, $\left[{ϵ}_o\right]=\frac{{\left[AT\right]}^2}{\left[ML{T}^{-2}\right]\left[L^2\right]}$

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