Why is $\epsilon_0$ in Coulomb’s law permittivity of free space if since the free space has nothing (unlike the old concept of ether)? Also why did Griffiths in his book said that “The constant $\epsilon_0$ is  called (ludicrously) the permittivity of free space”? What was he trying to say? Please lead. Thank you.
 A: 
Why is $ϵ_0$ in Coulomb’s law permittivity of free space if since the free space has nothing (unlike the old concept of ether)?

It's the consequence of how we've defined our unit of charge, the coulomb. In other systems of units the numeric value of $\varepsilon_0$ may be different. In particular, in Gaussian system it could be taken as $\varepsilon_0=\frac1{4\pi}$. There's no concept of "permittivity of free space" there at all, so the Coulomb's law has a simpler form, without any special constant.
A: The permittivity of free space was named when people still accepted the aether. One might think it ludicrously named since we keep the name when we have removed the concept by which it was named. I had grown used to the convention that in natural units $\epsilon_0 = 1$, meaning that it is just a conversion constant for units of charge to conventional units which refer to force. Now it depends on the fine structure constant, which is a measured quantity referring to the likelihood of the exchange of a photon between charged particles, not anything specifically to do with the vacuum or free space.
