The short answer: Because \(k\) is Coulomb’s constant, defined by \(k=\dfrac{1}{4\pi\varepsilon_0}\), and using the measured permittivity of free space gives \(k\approx8.99\times10^9\), which is rounded to \(9\times10^9\) for simplicity.
Explanation
- Coulomb’s law for the electrostatic force between two point charges is usually written as
$$F=k\frac{q_1q_2}{r^2}.$$
The constant \(k\) is related to the permittivity of free space \(\varepsilon_0\) by
$$k=\frac{1}{4\pi\varepsilon_0}.$$
- The measured value of the vacuum permittivity is \(\varepsilon_0\approx8.854187817\times10^{-12}\text{ F/m}.\) Plugging this into the formula gives
$$k=\frac{1}{4\pi(8.854187817\times10^{-12})}\approx8.98755179\times10^9\text{ N·m}^2\text{/C}^2.$$
- For most textbook and classroom work this is rounded to a convenient value:
$$k\approx9\times10^9\text{ N·m}^2\text{/C}^2.$$
The rounding makes calculations easier and the small difference (about 0.14%) is negligible in typical problems.
Why that particular number?
- The numerical value comes from the chosen SI units (meter, kilogram, second, ampere) and the experimentally measured \(\varepsilon_0\).
- Conceptually, \(\varepsilon_0\) and hence \(k\) arise from the way the electric field behaves in free space (Maxwell’s equations). The value is not arbitrary — it is determined by measurements and the SI system.
If you’d like, I can show the numeric substitution step-by-step or explain how the value connects to other constants (e.g., the speed of light and magnetic constant).