A) \[\text{ }\!\![\!\!\text{ }{{\text{M}}^{\text{-1}}}{{\text{L}}^{\text{-2}}}{{\text{A}}^{\text{2}}}{{\text{T}}^{\text{2}}}\text{ }\!\!]\!\!\text{ }\]
B) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{-2}}}{{\text{A}}^{\text{-2}}}{{\text{T}}^{4}}\text{ }\!\!]\!\!\text{ }\]
C) \[\text{ }\!\![\!\!\text{ }{{\text{M}}^{-1}}{{\text{L}}^{-3}}{{\text{A}}^{\text{2}}}{{\text{T}}^{4}}\text{ }\!\!]\!\!\text{ }\]
D) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{3}}{{\text{A}}^{\text{-2}}}{{\text{T}}^{-4}}\text{ }\!\!]\!\!\text{ }\]
Correct Answer: C
Solution :
From Coulomb's law, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[\therefore \] \[{{\varepsilon }_{0}}=\frac{1}{4\pi F}\,\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[\therefore \] Dimensions of \[{{\varepsilon }_{0}}=\frac{1}{[ML{{T}^{-2}}]}\times \frac{{{[AT]}^{2}}}{[{{L}^{2}}]}\] \[=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}]\] [AT]2
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