# Where does Coulomb's constant, $k_e$, come from?

I'm familiar with Coulomb's law, $$\vec{F} = k_e\frac{q_1q_2}{r^2}$$, and know how to apply it (having used it extensively in practice, on exams, etc). I find that the $$\frac{q_1q_2}{r^2}$$ "part" of Coulomb's law is at least somewhat intuitive, however I've never quite understood precisely where Coulomb's constant, $$k_e$$, comes from.

I know that it can be measured experimentally, of course, but where do we get that $$k_e = \frac{1}{4\pi \epsilon_0}$$ and, further, how to we get $$\epsilon_0 = \frac{1}{\mu_oc^2}$$?

I'm looking for both a "rigorous" derivation, and intuition (where possible).

• – rob
Jan 9, 2021 at 5:05
• The very short answer is that $k_e$ (or $\epsilon_0$) is assumed to be a fundamental constant of nature. It cannot be derived and must be found experimentally. (This is true even at the level of quantum field theory. There is a coefficient in the quantum electrodynamics Lagrangian, when not expressed in natural units, that has to be determined experimentally.) Jan 9, 2021 at 5:17

Note that Coulomb’s Law is experimentally determined, though because Maxwell’s equations describe all electromagnetic phenomena you can consider doing it this way (so you can see how the constant $$k$$ comes about):

Consider this equation which is one of Maxwell’s equations,

$$\oint_S \vec E \cdot \vec{dS} = \frac{Q}{\epsilon_0}$$

where this surface integral $$S$$ is taken over a closed sphere surrounding the charge $$Q$$ then

$$E (4\pi r^2) = \frac{Q}{\epsilon_0}$$

and given that the electric field $$E$$ is defined as the force per unit charge ie.,

$$E = \frac{F}{q}$$

then we can write

$$F = \frac{Qq}{4\pi \epsilon_0 r^2}$$

which is Coulomb’s Law.

The vacuum permitivity constant is a measure of how much a vacuum permits an electric field. Most problems we assume that there is nothing else in the space between the charges we are studying, which is why we use $$\epsilon_0$$. In cases where we do, we use the dielectric constant $$\epsilon$$ which is a measure of how much the medium permits an electric field.

When we manipulate Maxwell’s equations, we end up with the electromagnetic wave equation and as you can see in the link

$$v = \frac{1}{\sqrt{\mu \epsilon }}$$

where $$v$$ is the speed of an electromagnetic wave in a region with permeability $$\mu$$ and permitivity $$\epsilon$$. Of course if we are in a vacuum this becomes

$$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$

where $$c$$ is the speed of light in a vacuum.