Question 52: Units , Dimensions and Error Analysis

Question

If $\mu_0$ and $\epsilon_0$ are the permeability and permittivity of free space, respectively, then the dimension of $\left(\frac{1}{\mu_0 \epsilon_0}\right)$ is :

Accepted Answer

The expression $\frac{1}{\mu_0 \epsilon_0}$ is related to the speed of light $ c $, given by the equation:
$ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} $
Therefore, we have:
$ \frac{1}{\mu_0 \epsilon_0} = c^2 $
The speed of light $ c $ has the dimensions of $	ext{L T}^{-1}$, where $	ext{L}$ is the dimension of length and $	ext{T}$ is the dimension of time. Thus, when squared, the dimensions become:
$ c^2 = (	ext{L T}^{-1})^2 = 	ext{L}^2 	ext{T}^{-2} $
Hence, the dimensions of $\frac{1}{\mu_0 \epsilon_0}$ are $	ext{L}^2 	ext{T}^{-2}$.

Answer

$\mathrm{T}^2 / \mathrm{L}$

Answer

$\mathrm{L}^2 / \mathrm{T}^2$

Answer

$\mathrm{T}^2 / \mathrm{L}^2$

Answer

$\mathrm{L} / \mathrm{T}^2$

Units , Dimensions and Error Analysis

If \mu_0 and \epsilon_0 are the permeability and permittivity of free space, respectively, then the dimension of \left(\frac{1}{\mu_0 \epsilon_0}\right) is :

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