A) \[coulomb/newton-metre\]
B) \[newton-metr{{e}^{2}}/coulom{{b}^{2}}\]
C) \[coulom{{b}^{2}}/newton-metr{{e}^{2}}\]
D) \[coulom{{b}^{2}}/ewton-metre{{)}^{2}}\]
Correct Answer: C
Solution :
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 \[{{\varepsilon }_{0}}=\frac{coulomb\,\times coulomb}{newton-{{(metre)}^{2}}}=\frac{{{(coulomb)}^{2}}}{newton-{{(metre)}^{2}}}\] Methods 2: \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{Q}_{1}}{{Q}_{2}}}{{{r}^{2}}}\Rightarrow {{\varepsilon }_{0}}\propto \frac{{{Q}^{2}}}{F\times {{r}^{2}}}\] So, \[{{\varepsilon }_{0}}\]has units of \[coulom{{b}^{2}}/Newton-{{m}^{2}}\]
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