Answer to Question #104221 in Electricity and Magnetism for Akshay

Question

Using Maxwell’s equations in vacuum, derive the wave equation for the z-component
of the electric field vector associated with an electromagnetic wave.

Answer 1

$\overrightarrow{\nabla}\times\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t }$

$\overrightarrow{\nabla}\times\overrightarrow{E}=\mu_0\epsilon_0\frac{\partial \overrightarrow{E}}{\partial t }$

$\overrightarrow{\nabla}\times\overrightarrow{E}(z,t)\overrightarrow{i}=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial x } & \frac{\partial }{\partial y } & \frac{\partial }{\partial z } \\ \overrightarrow{E}(z,t) &0 & 0 \end{vmatrix}=\frac{\partial E}{\partial z }\overrightarrow{j}$

$\frac{\partial E}{\partial z }=-\frac{\partial B}{\partial t }$ (1)

$\overrightarrow{\nabla}\times\overrightarrow{B}(z,t)\overrightarrow{j}=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial x } & \frac{\partial }{\partial y } & \frac{\partial }{\partial z } \\ 0 & \overrightarrow{B}(z,t) & 0 \end{vmatrix}=-\frac{\partial B}{\partial z }\overrightarrow{i}$

$\frac{\partial B}{\partial z }=-\mu_0\epsilon_0\frac{\partial E}{\partial t }$

$\frac{\partial^2 E}{\partial z^2 }=-\frac{\partial}{\partial z }\frac{\partial B}{\partial t }=-\frac{\partial}{\partial t }\frac{\partial B}{\partial z }=-\frac{\partial}{\partial t }(-\mu_0\epsilon_0\frac{\partial E}{\partial t })=\mu_0\epsilon_0\frac{\partial^2 E}{\partial t^2 }$

$\frac{\partial^2 E}{\partial z^2 }=\mu_0\epsilon_0\frac{\partial^2 E}{\partial t^2 }$ Answer