A) newton-metre 2 /coulomb2
B) coulomb/newton-metre
C) coulomb/newton-metre2
D) coulomb/newton-metre2
Correct Answer: D
Solution :
From the relation \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] Or \[4\pi {{\varepsilon }_{0}}=\frac{1}{F}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[\therefore \]Unit of \[{{\varepsilon }_{0}}=\frac{coulom{{b}^{2}},}{newton\text{ }metr{{e}^{2}}}\]\[=coulom{{b}^{2}}/newton\text{ }metr{{e}^{2}}\]
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