Physics

# In terms of resistance $R$ and time $T$, the dimensions of ratio $\dfrac {\mu}{\epsilon}$ of the permeability $\mu$ and permittivity $\epsilon$ is:

$[R^{2}]$

##### SOLUTION
The electrostatic force can be given by:
$F_e=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$

So, permittivity is:
So, $\varepsilon_0=\dfrac{q^2}{4\pi r^2F_e}$

So, dimensions : $\varepsilon_0=\dfrac{[Q^2]}{[L^2][MLT^{-2}]}$

The dimensions of the permittivity $\varepsilon$ is $=[M^{-1}L^{-3}T^2Q^2]$

The magnetic force can be given by:
$F_m=\dfrac{\mu_0}{4\pi}\dfrac{q_{m_1}q_{m_2}}{r^2}$

So, permeability is:
So, $\mu_0=\dfrac{q_{m_1}q_{m_2}}{4\pi r^2F_m}$

The dimensions of the permeability $\mu$ is $=[M^{1}L^{1}Q^{-2}]$

So, the ratio of the two dimensions is:
$\dfrac{\mu}{\varepsilon}=\dfrac{[M^{1}L^{1}Q^{-2}]}{[M^{-1}L^{-3}T^2Q^2]}$

$\Rightarrow M^2L^4T^{-2}Q^{-4}$

The dimension of the resistance is $R=[ML^2T^{-1}Q^{-2}]$

comparing the dimension with the dimesion of resistance:
$\therefore \dfrac{\mu}{\varepsilon}=[R^2]$

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Single Correct Medium Published on 18th 08, 2020
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