Solution:
For an EM wave in vacuum, \(c = \frac{E_0}{B_0}\) and \(c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}\). Substituting \(\frac{B_0}{E_0} = \frac{1}{c}\) into \(mu_0 \varepsilon_0 = \frac{B_0}{E_0 c}\) gives \(mu_0 \varepsilon_0 = \frac{1}{c^2}\), which is true.
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