A) coulomb/newton-metre
B) newton - metre2/coulomb2
C) coulomb / newton metre
D) coulomb2/ (newton- metre)2
Correct Answer: C
Solution :
| Key Idea: Substitute the units for all the quantities involved in an expression written for permittivity of free space. By Coulomb's law, the electrostatic force |
| \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] |
| \[\Rightarrow \] \[{{\varepsilon }_{0}}=\frac{1}{4\pi }\times \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}F}\] |
| Substituting the units for q, r and F, we obtain unit of |
| \[{{\varepsilon }_{0}}\text{=}\frac{\text{constant }\!\!\times\!\!\text{ coulomb}}{\text{newton-(metre}{{\text{)}}^{\text{2}}}}\text{=}\frac{{{\text{(coulomb)}}^{\text{2}}}}{\text{newton-(metre}{{\text{)}}^{\text{2}}}}\] |
| \[={{C}^{2}}/N-{{m}^{2}}\] |
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