In SI units the constant $k$ in Coulomb's law $F = k\frac{q_1 q_2}{r^2…

Question

In SI units the constant $k$ in Coulomb's law $F = k\frac{q_1 q_2}{r^2}$ is approximately

Accepted Answer

C. $9 	imes 10^{9}\ \mathrm{N\,m^2/C^2}$

1. $k = \frac{1}{4\pi\varepsilon_0}$ where $\varepsilon_0$ is the permittivity of free space.
2. Substituting $\varepsilon_0 = 8.854 	imes 10^{-12}\ \mathrm{C^2/N\,m^2}$ gives $k \approx 8.99 	imes 10^{9}$.
3. Rounded, $k = 9 	imes 10^{9}\ \mathrm{N\,m^2/C^2}$. The other options are off by factors of $10^3$ each way, or have the wrong sign of the exponent.
_Source: syllabus Physics Ch 1 "Electric Charges and Fields", §1.6 Coulomb's Law_

Answer

$9 	imes 10^{6}\ \mathrm{N\,m^2/C^2}$

Answer

$9 	imes 10^{-9}\ \mathrm{N\,m^2/C^2}$

Answer

$9 	imes 10^{12}\ \mathrm{N\,m^2/C^2}$

In SI units the constant $k$ in Coulomb's law $F = k\frac{q_1 q_2}{r^2}$ is approximately

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