Let [{varepsilon _0}] denotes the dimensional | vedclass

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Question

Dimension formula of $\varepsilon_0$

$\varepsilon_0=\frac{1}{4 \pi F } \frac{ v _1 v _2}{ r ^2}\left( F = MLT ^{-2}\right)$

$\varepsilon_0=\frac

Vedclass · Accepted Answer

$[{\varepsilon _0}] = {M^{ - 1}}{L^{ - 3}}{T^4}{I^2}$

Vedclass · Answer

Dimension formula of $\varepsilon_0$

$\varepsilon_0=\frac{1}{4 \pi F } \frac{ v _1 v _2}{ r ^2}\left( F = MLT ^{-2}ight)$

$\varepsilon_0=\frac{1}{ MLT ^{-2}} \frac{ LAT 	imes AT }{1^2}$

$= M ^{-1} L ^{-3} t ^4 A ^2$

Dimension of $\mu_0$

$N / A ^2$ or $WbA A ^{-1} m ^{-1}$

$=[ M ][ L ][ T ]^{-2}[ A ]^{-2}$

Let $[{\varepsilon _0}]$ denotes the dimensional formula of the permittivity of the vacuum and $[{\mu _0}]$ that of the permeability of the vacuum. If $M = {\rm{mass}}$, $L = {\rm{length}}$, $T = {\rm{Time}}$ and $I = {\rm{electric current}}$, then

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