Physics - Unit & Dimension Question with Solution | TestHub

Force, $F=\frac{q_1{q}_2}{4\pi {\epsilon }_0{r}^2}$ or ${\epsilon }_0=\frac{q_1{q}_2}{4\pi r^{2}F}$

Charge $=$ current $\times$ time

Force $=$ mass $\times$ acceleration

Thus, dimensional formula of electric permittivity,

$\left[{\epsilon }_0\right]=\frac{\left(IT\right)\left(IT\right)}{\left({\mathrm{L}}^2\right)\left({\mathrm{MLT}}^{-2}\right)}$$={\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{I}}^2$

Since speed of light, $c=\frac{1}{\sqrt{{\epsilon }_0{\mu }_0}}$

Therefore, dimensional formula of magnetic permeability, $\left[{\mu }_0\right]=\frac{1}{\left[c^2{\epsilon }_0\right]}=\frac{1}{\left(L^2{T}^{-2}\right)\left({\mathrm{M}}^{-1}{\mathrm{L}}^{-3}{\mathrm{T}}^4{\mathrm{I}}^2\right)}$

$\Rightarrow \left[{\mu }_0\right]={\mathrm{MLT}}^{-2}{\mathrm{I}}^{-2}$