A) \[[M{{L}^{-2}}{{T}^{-2}}A]\]
B) \[[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}]\]
C) \[[ML{{T}^{-2}}A]\]
D) \[[M{{L}^{2}}{{T}^{-1}}{{A}^{2}}]\]
Correct Answer: B
Solution :
Key Idea: Putting the dimensions for quantities in the expression containing \[{{\varepsilon }_{0}}\]. From Coulombs law, two stationary point charges \[{{q}_{1}}\] and \[{{q}_{2}}\] attract/repel each other with a force F which is directly proportional to the product of charges and inversely proportional to the square of distance r between them That is, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[\Rightarrow \] \[{{\varepsilon }_{0}}=\frac{1}{4\pi }\frac{{{q}_{1}}{{q}_{2}}}{F\,{{r}^{2}}}\] \[\therefore \] Dimensions of permittivity \[{{}_{\text{0}}}\text{=}\frac{\text{dimensions}\,\text{of}\,{{\text{q}}^{\text{2}}}}{\text{dimensions}\,\text{of}\,\text{F }\!\!\times\!\!\text{ dimensions}\,\text{of}\,{{\text{r}}^{\text{2}}}}\] \[[{{\varepsilon }_{0}}]=\frac{[{{A}^{2}}{{T}^{2}}]}{[ML{{T}^{-2}}][{{L}^{2}}]}=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}]\]
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