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.

  • 1
    $\begingroup$ How is $1/\epsilon$ easier than $\epsilon$? What does easier mean? However, as for the questions, the dielectric and magnetic constant are chosen in the SI so that the Coulomb force only presents one $k$ and their product is the inverse of the speed of light. $\endgroup$
    – gented
    Jul 26, 2016 at 22:11
  • $\begingroup$ With easier I mean more logical. It wouldn't make sense to define a constant as the inverse of something if there is no reasons for doing it. $\endgroup$
    – Kevin
    Jul 26, 2016 at 22:15
  • $\begingroup$ The main historical reasons are the ones I quoted above (this said, I still don't see how direct multiplication is more "logical" than multiplication by the 1 over the inverse). $\endgroup$
    – gented
    Jul 26, 2016 at 22:19
  • $\begingroup$ You should wait until you learn more electrostatics. You've literally seen one place where this constant appears. One can have a lot of debates back and forth about which unit systems are best (see pi vs. tau, SI vs. cgs) but the question doesn't make sense directed at just one equation. $\endgroup$
    – knzhou
    Jul 26, 2016 at 22:25
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    $\begingroup$ @Kevin Why would computers have anything to do with it? The theory of electromagnetism predates the invention of computers by a century. The existence of electromagnetism as a separate force predates computers by billions of years. $\endgroup$
    – knzhou
    Jul 26, 2016 at 22:34

1 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 ...

  • $\begingroup$ We do not deduce the differential form of Maxwell equations, we deduce the integral expressions and then derive their correspondence differential parts applying Stoke's theorem. $\endgroup$
    – gented
    Jul 26, 2016 at 22:44
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    $\begingroup$ @GennaroTedesco if you feel that that makes any difference, be my guest $\endgroup$
    – Sanya
    Jul 26, 2016 at 22:46

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