The dimension of $\sqrt{\frac{\mu_0}{\epsilon_0}}$ is equal to that of :
( $\mu_0=$ Vacuum permeability and $\epsilon_0=$ Vacuum permittivity)

Question

The dimension of $\sqrt{\frac{\mu_0}{\epsilon_0}}$ is equal to that of : ( $\mu_0=$ Vacuum permeability and $\epsilon_0=$ Vacuum permittivity)

Aicharya · Accepted Answer

$\begin{aligned} & L=\frac{\mu_0 N A}{\ell} \ & C=\frac{A \epsilon_0}{d} \ & \frac{L}{C} \propto \frac{\mu_0}{\epsilon_0} \ & \sqrt{\frac{\mu_0}{\epsilon_0}} \propto \sqrt{\frac{L}{C}} \ & \frac{L}{C}=\frac{	au R}{(	au / R)}=R^2 \ & \sqrt{\frac{\mu_0}{\epsilon_0}}=R\end{aligned}$

The dimension of $\sqrt{\frac{\mu_0}{\epsilon_0}}$ is equal to that of : ( $\mu_0=$ Vacuum permeability and…

Solution