I have been given the following equation for the speed of light:

\$V = \Large {1 \over {\sqrt {\mathstrut \epsilon \cdot \mu}}} = {1 \over \sqrt{\mathstrut \epsilon_0 \cdot \epsilon_r \cdot \mu_0 \cdot \mu_r}}\$


What is the difference between simply \$\epsilon\$ and \$\epsilon_0\$? Likewise with \$\mu\$ and \$\mu_0\$?

  • \$\begingroup\$ That link doesn't make much sense; "foo+bar"? \$\endgroup\$
    – ocrdu
    Nov 30, 2020 at 14:55

1 Answer 1


What is the difference between simply epsilon and epsilon_0?


V = 1/sqrt(epsilon.mu) = 1/sqrt(epsilon_0.epsilon_r.mu_0.mu_r)

If you look at your formula carefully it says "epsilon_0" (\$\epsilon_0\$) and "epsilon_r" (\$\epsilon_r\$).

  • \$\epsilon_0\$ is the absolute permittivity of free space in farads per metre. Also known as vacuum permittivity
  • \$\epsilon_r\$ is the relative permittivity of a material with \$\epsilon_0\$ as the reference.

Hence, \$\epsilon_0 \epsilon_r\$ is the absolute permittivity of the material in farads per metre.

Likewise with mu and mu_0?

"mu" is the magnetic permeability (in henries per metre) so, it's the same principle as above but substituting \$\mu\$ for \$\epsilon\$.

So, speed of light in a medium is: -

$$c = \dfrac{1}{\sqrt{\epsilon_0 \epsilon_r \mu_0\mu_r}}$$


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