Question

Derive the wave equation for the z-component of the electric field of an electromagnetic
wave and show that the wave propagates in free space with the speed of light .

Answer 1

Let the electric field vector is in the z direction, magnetic field vector is in the direction of the y axis. so the propagation of the wave is in the direction of the x direction, and electric and magnetic field vectors are in the function of x and t.

So, $\overrightarrow{E}(z,t)=E(z,t) \hat{k}$

Magnetic field $\overrightarrow{B}(y,t)=B(y,t)\hat{j}$

Now, as per the maxwell's equation for the free space

$\nabla . E=0$ and $\nabla.B=0$

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

and $\nabla\times B=\mu_o\epsilon_o\dfrac{\partial E}{\partial t}$

Now, taking the curl of the electric field vector

So, $-\dfrac{\partial B}{\partial t}=\dfrac{\partial E}{\partial z}$ ------(i)

$-\dfrac{\partial B}{\partial z}=-\mu_o\epsilon_o\dfrac{\partial E}{\partial z}$

Similarly for the magnetic field

now taking the partial derivative of the equation (i)

$\dfrac{\partial^2 E}{\partial z^2}=-\dfrac{\partial (\partial B)}{\partial t\partial z}=\mu_o\epsilon_o\dfrac{\partial^2 E}{\partial z^2}$ -------(ii)

We know that the general wave equation for the traveling wave.

$\dfrac{\partial^2\psi}{\partial x^2}=\dfrac{1}{\nu}\dfrac{\partial^2\psi}{\partial t^2}$ ------(iii)

From the equation (ii) and (iii)

$c=\dfrac{1}{\sqrt{\mu_o\epsilon_o}}=2.997\times 10^8m/sec$