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 :
Putting the dimensions for quantities in the expression containing \[{{\varepsilon }_{0}}\]. From Coulomb?s 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, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[\Rightarrow \,\,\,\,{{\varepsilon }_{0}}=\frac{1}{4\pi }\,\frac{{{q}_{1}}{{q}_{2}}}{F\,{{r}^{2}}}\] \[\therefore \,\, Dimensions of permittivity\] \[{{\varepsilon }_{0}}=\frac{\dimensions\,\,of\,\,{{q}^{2}}}{\dimensions\,\,of\,\,F\times \dimensions\,\,of\,\,{{r}^{2}}}\] \[\left[ {{\varepsilon }_{0}} \right]=\frac{\left[ {{A}^{2}}{{T}^{2}} \right]}{\left[ ML{{T}^{-2}} \right]\left[ {{L}^{2}} \right]}=\left[ {{M}^{-1}}{{L}^{-\,3}}{{T}^{4}}{{A}^{2}} \right]\]
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