A) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{-1}}}{{\text{T}}^{\text{2}}}\text{A }\!\!]\!\!\text{ }\]
B) \[\text{ }\!\![\!\!\text{ }{{\text{M}}^{2}}{{\text{L}}^{\text{-3}}}{{\text{T}}^{\text{2}}}\text{A }\!\!]\!\!\text{ }\]
C) \[\text{ }\!\![\!\!\text{ }{{\text{M}}^{-1}}{{\text{L}}^{\text{-3}}}{{\text{T}}^{\text{4}}}{{\text{A}}^{2}}\text{ }\!\!]\!\!\text{ }\]
D) \[\text{ }\!\![\!\!\text{ }{{\text{M}}^{2}}{{\text{L}}^{\text{3}}}{{\text{T}}^{\text{-2}}}\text{A }\!\!]\!\!\text{ }\]
Correct Answer: C
Solution :
From Coulombs law, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[\therefore \] \[{{\varepsilon }_{0}}=\frac{1}{4\pi F}\cdot \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[\therefore \] Dimension of\[{{\varepsilon }_{0}}=\frac{1}{[ML{{T}^{-2}}]}\times \frac{{{[AT]}^{2}}}{{{[L]}^{2}}}\] \[=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}]\]
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