[Physics] Uncertainty of permittivity of vacuum

electromagnetismmetrologyphysical constantssi-unitsspecial-relativity

Question: The value of permittivity of vacuum, $\epsilon_0$, is given with absolutely no uncertainty in NIST

Why is this the case?

More details:

The permeability of vacuum can be given by

$$\mu_0=\frac{1}{\epsilon_0 c^2}$$

which comes from the definition of a magnetic field in special relativity, where we solve the problem of a wire with electrons flowing in, and we calculate the force exerted on an external charge moving with some velocity (for details, refer to the book Electricity and Magnetism, E. M. Purcell), and we define a new field called "magnetic field" with the form of Lorentz force, where the magnetic field is

$$B=\frac{I}{2\pi \epsilon_0 c^2 r}$$

where $I$ is the current due to the flow of electrons in the wire, and $r$ is the distance of the external charge from the wire. And there we get the definition of $\mu_0$ that makes $B$:
$$B=\frac{\mu_0 I}{2\pi r}$$

and there starts the concept "magnetism".

Why am I giving this detailed example? Because I wouldn't like to get the answer that $\mu_0$ has no error, and that's why $\epsilon_0$ has no error, and then we fall into circular logic. So I expect a reason which is independent of $\mu_0$.

lthough you might not like to hear it, the answer really DOES lie in the definition of $$\mu_0$$ (and $$c$$). $$\mu_0$$ is defined to be exactly $$4\pi *10^{-7}\ \text{H m}^{-1}$$. Similarly, $$c$$ is defined as exactly $$299792458\ \text{ms}^{-1}$$. It immediately follows from the relation $$\epsilon_0=\frac{1}{\mu_0 c^2}$$ that $$\epsilon_0$$ also has no uncertainty.