Question

Using Maxwell's equation in vacuum, derive the wave equation for the x-component of the electric field vector associated with an electromagnetic wave

Answer 1

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

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

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

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

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

$\frac{\partial E}{\partial x }=-\frac{\partial B}{\partial t }$

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

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

$\frac{\partial^2 E}{\partial x^2 }=\mu_0\epsilon_0\frac{\partial^2 E}{\partial t^2 }$ ​