Why $\epsilon_0 = 1/(4\pi k)$ instead of $\epsilon_0 = 4\pi k$? [closed]

Question

In my physics book, they consider a positive point charge $q$ and a spherical surface of radius $r$ centered on this positive charge. Then the flux through this surface is given by $$\Phi = \iint \vec{E} \cdot d\vec{A} = \iint EdA\cos\theta$$ Since $\vec{E}$ and $\vec{A}$ are parallel, $\cos\theta=1$, so the integral become just $$\iint EdA$$ Furthermore, $E$ is constant on every point so you get $$\iint EdA = E \iint dA = E(4\pi r^2)$$ where the last step follows from the fact that the surface is a sphere. Then according to Gauss's law $E=kq/r^2$ we get $$\Phi = \left(\frac{kq}{r^2}\right)(4\pi r^2) = 4\pi kq$$ Then finally they define a new constant $\epsilon_0$ which is equal to $1/(4\pi k)$ so that we get $$\iint\vec{E}\cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}$$ Now the real question: why is $\epsilon_0 = 1/(4\pi k)$

Why $\epsilon_0 = 1/(4\pi k)$ instead of $\epsilon_0 = 4\pi k$? Why use the inverse (reciprocal)? Whouldn't be $$\iint\vec{E}\cdot d\vec{A} = q_{enclosed} \epsilon_0$$ an "easier" and better choice?

Thanks in advance.

Edit: With easier I mean a more logical choice.

Edit 2: I was under the impression that $\epsilon_0$ is defined to make Gauss's law simpler. I voted this question to close.

Answer 1

If we look at nature from an empirical side, we can deduce the four Maxwell equations from experiments as: $$\nabla\vec{E}=4\pi k \rho$$ $$\nabla \times \vec{E} = - k'' \frac{\partial}{\partial t}\vec{B}$$ $$\nabla\vec{B}=0 $$ $$\nabla \times \vec{B} =4 \pi k' \vec{j} + \frac{k'}{k} \frac{\partial}{\partial t}\vec{E} $$ The choice of the value of those constants $k, k', k''$ determines the unit system we use - or respectively, if we have a unit system, we will know which values to take. There are different choices that can legitimately be made, see e.g. the SI-system or cgs. All of those have different advantages or disadvantages. For a more general outline, I recommend Jackson's electrodynamics book.
The advantage of SI units, $4\pi k=1/\varepsilon_0$, $k''=1$ and $k'=k c^{-2}$, is that we can use normal units of charge etc. which we are probably used to already.
It is unfortunate that your book chooses to just motivate this by Gauss' law only ...