Physics

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


ANSWER

$$[R^{2}]$$


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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