From my later school days on, the formula that impresses me the most is
$$c=\frac{1}{\sqrt{\varepsilon_0\mu_0}}$$
I know how this can be derived, both from Maxwell Equations and more intuitively. I can read up on wikipedia what $\varepsilon_0$ (permittivity) and $\mu_0$ (permeability) are. I roughly know what Ampere, Volt, Ohm and Watt are and that already concludes my knowledge of electromagnetism, everything else from school I have forgotten already.
When I have speed, I know that $1\frac{m}{s}$ means one meter per second change in distance. When I have acceleration, I know that $1\frac{m}{s^2}$ means one $\frac{m}{s}$ per second change in acceleration. Acceleration is still intuitive from driving in cars. I know $1Pa=1\frac{F}{m^2}$ can be visualized as the pressure coming from (roughly) having 100g of chocolate rasped and spread out over a square meter, while $1bar$ is about a pack of sugar (1kg) being hold up in the air by my thumb (1cm²). Giving $Pa$ as $\frac{kg}{ms^2}$ bears no meaning I know of, but is only the short form of $\frac{kg}{m^2}\cdot\frac{m}{s^2}$ where the second factor basically just carries the factor 10 so I get from 100 gram to 1 Newton. I know that $1kcal$ is about the energy needed to heat water at normal pressure by one degree Celsius. My intuitive understanding of electromagnetic units is rather lacking.
So now I'm basically given $v=(\sqrt{\varepsilon\mu})^{-1}$
and I want an intuitive understanding why the units work out. I understand the equation in that way that e.g. if I were to take four times the permittivity and leave the permeability, I would get half the speed, just in terms of the equation. I'm aware I can't simply do that with the light speed equation above because all numbers involved are constants. But why do the units work out? I'm not asking if they do, I can see that. I'm asking why. And please no "There are no fundamental units" philosophy, I'm asking for an intuitive grasp of the units making sense just like my pressure example was graspable and made sense.
Best Answer
A quite different interpretation is as follows.
$ \epsilon = \frac 1{zc} \qquad \mu = \frac zc$
In a modern interpretation, $c$ represents the space-time parameter. Massless particles travel at this speed.
$z$ is a conversion from flux (i.e. transport of effect) to field (force per charge). The actual relation here is the photon continuity equation.
$E = cB = zH = zcD$, whence $\epsilon = 1/cz, \ \ \mu=z/c$.
Maxwell actually compared the speed of light to the Weber-Kohlrausch value. The latter represents the ratio of charge accumulated in a condenser, in esu, to the current delivering it, in emu. From this near equality, Maxwell concluded that light travels in the same medium as EM waves.
It is interesting that if gravitons are also massless, there is a graviton continuity equation as above. Oliver Heaviside explored this in 1893, from a conclusion, that if gravity travels at a finite speed, a co-gravitational force must exist too.