Physics

# The dimension of permittivity of vacuum

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

##### SOLUTION
The permittivity of the vacuum can be given by:
$F_e=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$
$permitivity\ (\varepsilon_0)= \dfrac{(q)^2}{(F_e)\times(r)^2}$

$=\dfrac{[TA]^2}{[MLT^ {-2}] [L]^2}$

$=\dfrac{[TA]^2}{[ML^3T^ {-2}]}$

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

$= [M^{ -1} L^ {-3} T^4 A^2]$
This is the required solution.

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Single Correct Medium Published on 18th 08, 2020
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