Question

$\frac{E^2}{\mu_0}$ has the dimensions of (where $E$ is electric field and $\mu_0$ is permeability of free space)

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

Answer 1

Answer: (C)

$\frac{E^{2}}{\mu_{0}} = \frac{\varepsilon_{0}E^{2}}{\varepsilon_{0}\mu_{0}}$
where $\varepsilon_0$ is the permittivity of free space
$\therefore $ Energy density (i.e. energy per unit volume)
of an electric field is $u_{E} = \frac{1}{2}\varepsilon_{0}E^{2}$
Speed of light in vacuum, $c= \frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}} $
$\therefore \frac{E^{2}}{\mu_{0}} = \frac{2u_{E}}{\frac{1}{c^{2}}} $
$= 2u_{E}c^{2} $
$\left[\frac{E^{2}}{\mu_{0}}\right] = \left[ML^{-1}T^{-2}\right]\left[LT^{-1}\right]^{2} $
$ = \left[MLT^{-4}\right]$