$12.1.8.$ Using the law of electromagnetic induction and the connection of an alternating electric field with a magnetic field (see problem 11.6.1), prove that the wave propagation speed in a medium with permittivity $\varepsilon$ and permeability $\mu$ is equal to $c/\sqrt{\mu\varepsilon}$.
Solution
Let's write down Maxwell's equations for a homogeneous medium without free charges and currents: $$\text{div} \vec{E} = 0$$ $$\text{div} \vec{B} = 0$$ $$\text{rot} \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$ $$\text{rot} \vec{B} = \mu\mu_0\varepsilon\varepsilon_0 \frac{\partial \vec{E}}{\partial t}$$
Let a plane electromagnetic wave propagate along the $z$-axis. We direct the vector $\vec{E}$ along the $x$-axis ($E_x$) and the vector $\vec{B}$ along the $y$-axis ($B_y$). Then, in projections, the curl equations will take the form: $$\frac{\partial E_x}{\partial z} = -\frac{\partial B_y}{\partial t} \quad (1)$$ $$-\frac{\partial B_y}{\partial z} = \mu\mu_0\varepsilon\varepsilon_0 \frac{\partial E_x}{\partial t} \implies \frac{\partial B_y}{\partial z} = -\mu\mu_0\varepsilon\varepsilon_0 \frac{\partial E_x}{\partial t} \quad (2)$$
Let's differentiate equation (1) with respect to the coordinate $z$: $$\frac{\partial^2 E_x}{\partial z^2} = -\frac{\partial}{\partial z} \left( \frac{\partial B_y}{\partial t} \right) = -\frac{\partial}{\partial t} \left( \frac{\partial B_y}{\partial z} \right)$$
We have obtained the classical wave equation, which in its general form is written as: $$\frac{\partial^2 E_x}{\partial z^2} = \frac{1}{v^2} \frac{\partial^2 E_x}{\partial t^2}$$
Comparing the coefficients, we find the wave propagation speed $v$: $$\frac{1}{v^2} = \mu\mu_0\varepsilon\varepsilon_0 \implies v = \frac{1}{\sqrt{\mu\mu_0\varepsilon\varepsilon_0}}$$
Given that the speed of light in a vacuum is $c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$, we obtain the required relationship: $$v = \frac{c}{\sqrt{\mu\varepsilon}}$$