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

Given :

Maxwell's equations in vaccum.

To find :

Equation for the z-component of the electric field vector associated with an electromagnetic wave.

Solution :

According to maxwell's third equation : The relation between Electric field and Magnetic field is :

$\nabla\times E=-\frac{\partial B}{\partial t }.</p><p></p><p></p><p>$

(equation 1)

With respect to electric field :

$\nabla\times E=- \mu _{0} \epsilon_{0} ( \frac{\partial E}{\partial t })</p><p> </p><p></p><p>$

In z direction by matrix form , we get the above equation in the form as :

$∇×E(z,t)i= \left| \begin{array}{cc} \hat i & \hat j & \hat k \\ \frac{ \partial}{ \partial x} &\frac{ \partial}{ \partial y} & \frac{ \partial}{ \partial z}\\ E(z,t)& 0& 0 \\ \end{array} \right| = \frac {\partial E}{ \partial \: z} \hat j</p><p></p><p>$

as we know that :

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

then

$∇×B(z,t)i= \left| \begin{array}{cc} \hat i & \hat j & \hat k \\ \frac{ \partial}{ \partial x} &\frac{ \partial}{ \partial y} & \frac{ \partial}{ \partial z}\\ 0&B(z,t) & 0 \\ \end{array} \right| = - \frac {\partial B}{ \partial \: z} \hat i</p><p></p><p>$

Putting the value of B, we get :

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

On double differentiating E with respect to z direction,

we get :

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

So finally after all the calculations done,

The wave equation for the z-component of the electric field vector associated with an electromagnetic wave is :

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

Answer.