Question

Write the dimensions of 1/u0e0

Answer 1

Explanation:

Velocity of light in vacuum = 1/√(μoεo)

[LT-1] = [1/√(μoεo)]

[L2T-2] = [1/(μoεo)]

Answer 2

$\dfrac{1}{\mu_o\epsilon_o}=(LT^{-1})^2$

Explanation:

We need to find the dimensional formula of $\dfrac{1}{\mu_o\epsilon_o}$. We know that $\mu_o$ is the permeability and $\epsilon_o$ is the permitivity of the free space.

The electric force is given by :

$F=\dfrac{1}{4\pi \epsilon_o}\dfrac{q_1q_2}{r^2}$

$\epsilon_o=\dfrac{q_1q_2}{4\pi Fr^2}$

Putting dimension formula of all the quantities, we get :

$\epsilon_o=\dfrac{(AT^{-1})^2}{[MLT^{-1}][L^2]}$

$\epsilon_o=[M^{-1}L^{-3}T^4A^2]$

The dimension of $\mu_o$is,  [μ₀] = [M¹L¹T⁻²A⁻²]

So, the dimensions of  $\dfrac{1}{\mu_o\epsilon_o}$ is,

$\dfrac{1}{\mu_o\epsilon_o}=\dfrac{1}{[MLT^{-2}A^{-2}][M^{-1}L^{-3}T^4A^2]}$

$\dfrac{1}{\mu_o\epsilon_o}=(LT^{-1})^2$

Hence, this is the required solution.

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