Physics And Measurement Question 223
Question: The speed of light in vacuum, c, depends on two fundamental constants, the permeability of free space, $ {\mu _{0}} $ d me permittivity of free space, $ {\varepsilon _{0}}. $ The speed of light is given by $ c=\frac{1}{\sqrt{{\mu _{0}}{\varepsilon _{0}}}} $ . The units of $ {\varepsilon _{0}} $ are $ {{N}^{-1}}C^{2}{{m}^{-2}} $ . The units for $ {\mu _{0}} $ are
Options:
A) $ k{{g}^{-1}}{{m}^{-1}}C^{2} $
B) $ kgm{{C}^{-2}} $
C) $ kgm{{s}^{-4}}{{C}^{-2}} $
D) $ k{{g}^{-1}}{{s}^{-3}}{{C}^{-2}} $
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Answer:
Correct Answer: B
Solution:
[b] We have, $ c=\frac{1}{\sqrt{{\mu _{0}}{\in _{0}}}}\Rightarrow {\mu _{0}}=\frac{1}{{\in _{0}}c^{2}} $
$ \therefore $
units of $ {\mu _{0}}=\frac{1}{unitsof{\in _{0}}c^{2}} $
$ =\frac{1}{{{N}^{-1}}C^{2}{{m}^{-2}}{{(m{{s}^{-1}})}^{2}}}=Ns^{2}{{C}^{-2}} $
$ =(kgm{{s}^{-2}})s^{2}{{C}^{-2}}=kgm{{C}^{-2}} $
