A) \[[{{A}^{2}}{{M}^{-1}}{{L}^{-3}}{{T}^{4}}]\]
B) \[[{{A}^{2}}{{M}^{-1}}{{L}^{3}}{{T}^{0}}]\]
C) \[[A{{M}^{-1}}{{L}^{-3}}{{T}^{4}}]\]
D) \[[{{A}^{2}}{{M}^{0}}{{L}^{-3}}{{T}^{4}}]\]
Correct Answer: A
Solution :
From Coulombs law, the force of attraction/repulsion between two point charges q, q separated by distance\[r\]is \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{r}^{2}}}\] \[\Rightarrow \] \[{{\varepsilon }_{0}}=\frac{1}{4\pi }.\frac{{{q}^{2}}}{F{{r}^{2}}}\] where\[{{\varepsilon }_{0}}\]is electric permittivity. Dimensions of \[{{\varepsilon }_{0}}=\frac{{{[AT]}^{2}}}{[ML{{T}^{-2}}][{{L}^{2}}]}\] \[{{\varepsilon }_{0}}=[{{A}^{2}}{{M}^{-1}}{{L}^{-3}}{{T}^{4}}]\] Note: The constant\[{{\varepsilon }_{0}}\](epsilon zero) is called permittivity of free space and its vaiue is \[8.85\times {{10}^{-12}}\]\[Couloum{{b}^{2}}/newton\text{ }metr{{e}^{2}}\].
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