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+### Statement
+
+$12.1.8.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$12.1.8.$ 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 $z$:
+$$\frac{\partial^2 E_x}{\partial z^2} = -\frac{\partial}{\partial z} \left( \frac{\partial B_y}{\partial t} \right) = -\frac{\partial}{\partial t} \left( \frac{\partial B_y}{\partial z} \right)$$
+
+Substitute equation (2) into this expression:
+$$\frac{\partial^2 E_x}{\partial z^2} = -\frac{\partial}{\partial t} \left( -\mu\mu_0\varepsilon\varepsilon_0 \frac{\partial E_x}{\partial t} \right) = \mu\mu_0\varepsilon\varepsilon_0 \frac{\partial^2 E_x}{\partial t^2}$$
+
+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:
+$$v = \frac{c}{\sqrt{\mu\varepsilon}}$$
+
+#### Answer
+$$v = \frac{c}{\sqrt{\mu\varepsilon}}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]