In a sourceless medium in which \(\mathbf{J}=0\) and \(\rho_{v}=0\), assume a rectangular coordinate system in which \(\mathbf{E}\) and \(\mathbf{H}\) are functions only of \(z\) and \(t .\) The medium has permittivity \(\epsilon\) and permeability \(\mu .(a)\) If \(\mathbf{E}=E_{x} \mathbf{a}_{x}\) and \(\mathbf{H}=H_{y} \mathbf{a}_{y}\), begin with Maxwell's equations and determine the second-order partial differential equation that \(E_{x}\) must satisfy. \((b)\) Show that \(E_{x}=E_{0} \cos (\omega t-\beta z)\) is a solution of that equation for a particular value of \(\beta .(c)\) Find \(\beta\) as a function of given parameters.

The first thing to do is write down the relevant Maxwell's equations for this problem. We have: 1. Ampere’s Law: \(\nabla \times \mathbf{H} = \epsilon \frac{\partial \mathbf{E}}{\partial t}\) 2. Faraday’s Law: \(\nabla \times \mathbf{E} = -\mu \frac{\partial \mathbf{H}}{\partial t}\) Since the medium is sourceless (\(\rho_v = 0\) and \(\mathbf{J} = 0\)), we don't need the rest of Maxwell's equations.

We are given that \(\mathbf{E}=E_{x} \mathbf{a}_{x}\) and \(\mathbf{H}=H_{y} \mathbf{a}_{y}\). Using the given coordinate system, we can write the curl equations as: 1a. \(\nabla \times \mathbf{H} = (\frac{\partial H_z}{\partial y} - \frac{\partial H_y}{\partial z})\mathbf{a}_x - \frac{\partial H_z}{\partial x}\mathbf{a}_y + (\frac{\partial H_y}{\partial x} - \frac{\partial H_x}{\partial y})\mathbf{a}_z = \epsilon \frac{\partial \mathbf{E}}{\partial t}\) 2a. \(\nabla \times \mathbf{E} = (\frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z})\mathbf{a}_x - \frac{\partial E_z}{\partial x}\mathbf{a}_y + (\frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y})\mathbf{a}_z = -\mu \frac{\partial \mathbf{H}}{\partial t}\)

Since \(\mathbf{E}\) and \(\mathbf{H}\) only depend on \(z\) and \(t\), we can simplify the curl equations: 1b. \(( - \frac{\partial H_y}{\partial z})\mathbf{a}_x = \epsilon \frac{\partial \mathbf{E}}{\partial t}\) 2b. \(( - \frac{\partial E_x}{\partial z})\mathbf{a}_z = -\mu \frac{\partial \mathbf{H}}{\partial t}\)

Now we go back to the equation we derived in step 3. To obtain the second-order partial differential equation for \(E_x\), we differentiate equation 2b with respect to \(z\) and then substitute equation 1b into the resulting equation: Differentiate 2b with respect to \(z\): \(-\frac{\partial^2 E_x}{\partial z^2}\mathbf{a}_z = -\mu \frac{\partial^2 \mathbf{H}}{\partial z \partial t}\) Substitute 1b: \(-\frac{\partial^2 E_x}{\partial z^2}\mathbf{a}_z = -\mu \frac{\partial}{\partial z} (\frac{1}{\epsilon} \frac{\partial \mathbf{E}}{\partial t})\) Now isolate \(E_x\) and get rid of the vector notation: \(-\frac{\partial^2 E_x}{\partial z^2} = -\mu \frac{1}{\epsilon} \frac{\partial^2 E_x}{\partial t \partial z}\) Finally, differentiate the above equation again with respect to \(z\) to get the required second-order partial differential equation: \(\frac{\partial^2 E_x}{\partial z^2} = \mu \epsilon \frac{\partial^2 E_x}{\partial t^2}\)

We need to show that \(E_x=E_{0} \cos (\omega t-\beta z)\) is a solution to the partial differential equation we derived in step 4: Substitute \(E_x\) into the left-hand side of the equation: \(-\beta^2 E_0 \cos(\omega t - \beta z) = -\beta^2E_x\) Substitute \(E_x\) into the right-hand side of the equation: \(\omega^2\mu\epsilon E_{0} \cos(\omega t - \beta z) = \mu\epsilon\omega^2E_x\) Comparing both equations, we conclude that \(E_x=E_{0} \cos (\omega t-\beta z)\) is a solution if \(\beta^2 = \omega^2\mu\epsilon\).

As derived in step 5, \(\beta^2 = \omega^2\mu\epsilon\). To obtain \(\beta\) as a function of given parameters: \(\beta = \sqrt{\omega^2\mu\epsilon}\) This concludes the solution to the problem.