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}}] $



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