Solution of electromagnetic wave equations in non conducting media

The electromagnetic wave equations in non-conducting media :

For electric field vector:

$\nabla^{2} \overrightarrow{E}=\frac{1}{v^{2}} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} \qquad(1)$

For magnetic field vector:

$\nabla^{2} \overrightarrow{B}=\frac{1}{v^{2}} \frac{\partial^{2} \overrightarrow{B}}{\partial t^{2}} \qquad(2)$

The wave equation of electric field vector:

$\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$

The wave equation of magnetic field vector:

$\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$

Now the solution of electromagnetic wave for electric field vector.

Differentiate with respect to $t$ of equation $(3)$

$\frac{\partial \overrightarrow{E}}{\partial t}=i \omega E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

Again differentiate with respect to $t$ of above equation:

$\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}=i^{2} \omega^{2} E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

$\frac{\partial^{2} \overrightarrow{E}}{\partial^{2} t}=- \omega^{2} \overrightarrow{E}(\overrightarrow{r},t)$

Now substitute the value of the above equation in equation$(1)$

$\nabla^{2} \overrightarrow{E}=\frac{-\omega^{2}}{v^{2}} \overrightarrow{E}(\overrightarrow{r},t)$

$\nabla^{2} \overrightarrow{E}=-(\frac{\omega}{v})^{2} \overrightarrow{E}(\overrightarrow{r},t)$

$\nabla^{2} \overrightarrow{E}=-\alpha^{2} \overrightarrow{E}(\overrightarrow{r},t) \qquad (\because \alpha=\frac{\omega}{v} )$

Where $\alpha$ - Wave propagation Constant

$\nabla^{2} \overrightarrow{E} + \alpha^{2} \overrightarrow{E}(\overrightarrow{r},t)=0 $

This is the solution of the electromagnetic wave equation in non-conducting media for the electric field vector.

Now the component form of the above equation:

$(\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} +\frac{\partial^{2}}{\partial z^{2}})(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) \\ =- \alpha^{2}(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) \qquad(5)$

If the wave is propagating along $z$ direction. Then for uniform-plane electromagnetic waves-

$\frac{\partial}{\partial x}=\frac{\partial}{\partial y}=0$

$\frac{\partial^{2}}{\partial x^{2}}=\frac{\partial^{2}}{\partial y^{2}}=0$

$E_{z}=0$

Now the equation $(5)$ can be written as:

$\frac{\partial^{2}}{\partial x^{2}} (\hat{i}E_{x}+\hat{j}E_{y})=- \alpha^{2}(\hat{i}E_{x}+\hat{j}E_{y})$

Now separate the above equation in $x$ and $y$ components so

$\frac{\partial^{2} E_{x}}{\partial z^{2}}=- \alpha^{2}E_{x}$

$\frac{\partial^{2}E_{y}}{\partial z^{2}} =- \alpha^{2}E_{y}$

The solution of electromagnetic wave for magnetic field vector can find out by following the above method.

Therefore $x$ and $y$ components of the solution of the electromagnetic wave equation for magnetic field vector can be written as. i.e.

$\frac{\partial^{2} B_{x}}{\partial z^{2}}=- \beta^{2}B_{x} \qquad \left( \because \beta=\frac{\omega}{v} \right)$

$\frac{\partial^{2}B_{y}}{\partial z^{2}} =- \beta^{2}B_{y}$