How was the value of vacuum permittivity originally found?

The vacuum permittivity appears originally in Maxwell's equations, used to describe electric fields. The permeability of vacuum was defined using Ampere's force law (itself derived from Biot-Savart law): If two current carrying wires were spaced one meter apart, both carrying one ampere of current, the force exerted against each of the wires would be exactly $$2*10^-7 N$$. This also defined the ampere. Therefore, the value of vacuum permeability was fixed to $$4π*10−7 H/m$$ by definition.

Using his laws, Maxwell was able to produce a wave equation with the speed:

$${\displaystyle c={1 \over {\sqrt {\mu _{0}\varepsilon _{0}}}}.}$$

This turned out to be the same as the speed of light (already measured using other means), so light was deduced to be an electromagnetic wave. Here $$c$$ was already known, $$\mu_{0}$$ was defined, but how was $$\epsilon_{0}$$ found? Was it experimentally determined? If so, how was it found in Maxwell's times? Today, also the speed of light is defined exactly, so $$\epsilon_{0}$$ also now has a defined value, but clearly this was not always the case.

The reason I'm asking this is because research into this turns up a huge amount of contradicting information and circular logic, so I want to be clear on this.

• I'm not 100% sure but I don't think $\mu_0$ would have been defined like that in Maxwell's day (and it no longer is either). – jacob1729 Jan 10 '19 at 17:47
• Good question, but it's probably more suited to the History of Science and Mathematics site. – PM 2Ring Jan 10 '19 at 17:48

786] The quantity $$V$$, in Art. 793, which expresses the velocity of propagation of electromagnetic disturbances in a non-conducting medium is, by equation (9), equal to $$\frac{1}{\sqrt{K\mu }}$$ ; If the medium is air, and if we adopt the electrostatic system of measurement, $$K = 1$$ and $$\mu =\frac{1}{{{v}^{2}}}$$ so that $$V = v$$, or the velocity of propagation is numerically equal to the number of electrostatic units of electricity in one electromagnetic unit. If we adopt the electromagnetic system $$K=\frac{1}{{{v}^{2}}}$$ and $$\mu =1$$, so that the equation $$V = v$$ is still true. On the theory that light is an electromagnetic disturbance, propagated in the same medium through which other electromagnetic actions are transmitted, $$V$$ must be the velocity of light, a quantity the value of which has been estimated by several methods. On the other hand, $$v$$ is the number of electrostatic units of electricity in one electromagnetic unit, and the methods of determining this quantity have been described in the last chapter. They are quite independent of the methods of finding the velocity of light. Hence the agreement or disagreement of the values of V and of v furnishes a test of the electromagnetic theory of light.