Physics And Measurement Question 111
Question: The dimensions of permittivity $ {\varepsilon _{0}} $ are
[MP PET 1997; AIIMS-2004; DCE-2003]
Options:
A) $ A^{2}T^{2}{{M}^{-1}}{{L}^{-3}} $
B) $ A^{2}T^{4}{{M}^{-1}}{{L}^{-3}} $
C) $ {{A}^{-2}}{{T}^{-4}}ML^{3} $
D) $ A^{2}{{T}^{-4}}{{M}^{-1}}{{L}^{-3}} $
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Answer:
Correct Answer: B
Solution:
$ F=\frac{1}{4\pi {\varepsilon _{0}}}\frac{q _{1}q _{2}}{r^{2}} $
$ \Rightarrow $ $ {\varepsilon _{0}}=\frac{|q _{1}||q _{2}|}{[F][r^{2}]}=\frac{[A^{2}T^{2}]}{[ML{{T}^{-2}}][L^{2}]}=[A^{2}T^{4}{{M}^{-1}}{{L}^{-3}}] $
