In the previous article, you found of the speed of the plane electromagnetic wave, and hence the speed of light. In this article, we again find the speed of the electromagnetic wave but after the derivation of the electromagnetic wave equation.

The Figure 1 shows an electromagnetic wave travelling in positive x-direction. The electric and magnetic fields are perpendicular to each other and both fields are perpendicular to the direction of the propagation of the wave, therefore you know that the electromagnetic wave is a transverse wave.

Note that both electric and magnetic fields change with time as opposed to the case in plane electromagnetic wave in the previous article, that is both fields are the functions of position $x$ and time $t$.

First we apply Faraday's law (Maxwell's third equation) to our wave. In Figure 1, you can see that at time $t$ the electric field at position $x$ at the left side of the rectangle $da$ is $E(x, t)$ and the electric field at the right side of the rectangle $bc$ at the position $x+\Delta x$ is $E(x+\Delta x, t)$. You know the expression of the Faraday's law is

$\oint \vec E \cdot d\vec l = -\frac{d\Phi_B}{dt} \tag{1} \label{1}$

Now let's evaluate the left hand side of the above equation. In the rectangle $abcd$ in Figure 1, the sides $ab$ and $cd$ do not contribute to the integral $\oint \vec E \cdot d\vec l$ because they are perpendicular to the electric field. Only the sides $bc$ and $da$ contribute to the integral. So, you know from the plane electromagnetic wave that we choose the direction of area vector towards positive z-direction, and therefore we integrate in anticlockwise direction around the rectangle as suggested by the right hand rule and Lenz's law as we did before in plane electromagnetic wave. If $h$ is the height of the rectangle, we have

$\vec E \cdot d\vec l = -E(x, t)h + E(x+\Delta x , t)h \tag{2} \label{2}$

Now it's time to evaluate the right hand side of the Equation \eqref{1}. Let's consider $\Delta x$ is small enough that the $\vec B$ can be considered uniform in the rectangle. Therefore,

$\frac{d\Phi_B}{dt} = \frac{\partial B(x,t)}{\partial t}h\Delta x \tag{3} \label{3}$

So, using Equations \eqref{2} and \eqref{3}, the Equation \eqref{1} becomes

$E(x, t)h + E(x+\Delta x , t)h = \frac{\partial B(x,t)}{\partial t}h\Delta x$

If you divide the above equation by $\Delta x$ on both sides and rearrange, you get

$\frac{ E(x+\Delta x , t) - E(x, t)}{\Delta x} = - \frac{\partial B(x,t)}{\partial t}$

If you take the limit of $\Delta t \to 0$, you can rewrite the above equation as

$\frac{\partial E(x, t)}{\partial x} = - \frac{\partial B(x,t)}{\partial t} \tag{4} \label{4}$

If the magnetic field changes with time, there must be the electric field that changes with position. Next we move to the Ampere-Maxwell's law (the fourth Maxwell's equation).

The Figure 3 again shows a rectangle but this time it is parallel to the xz-plane. The fourth Maxwell's equation (without conduction current) is

$\oint \vec B \cdot d\vec l = \mu_0\epsilon_0\frac{d\Phi_E}{dt} \tag{5} \label{5}$

Now we evaluate the left hand side of the Equation \eqref{5}, that is $\oint \vec B \cdot d\vec l$. Here the sides $ab$ and $cd$ (perpendicular to $\vec B$) do not contribute to the integral, only the sides $bc$ and $da$ (parallel or antiparallel to $\vec B$) contribute to the integral (integrating anticlockwise around the rectangle), therefore

$\vec B \cdot d\vec l = B(x, t) h - B(x+\Delta x, t)h \tag{6} \label{6}$

Similar to we did above, we consider that $\Delta x$ is small enough that the electric field can be considered uniform in the rectangle, so the change in electric flux is

$\frac{d\Phi_E}{dt} = \frac{\partial E (x, t)}{\partial t} \Delta x \, h \tag{7} \label {7}$

Using Equations \eqref{5}, \eqref{6} and \eqref{7}, you get

$- B(x+\Delta x, t)h + B(x, t)h = \mu_0 \epsilon_0 \frac{\partial E(x, t)}{\partial t} \Delta x\, h$

After further simplification (dividing both sides by $\Delta x \, h$) and taking the limit of $\Delta x$ approaches zero, that is $\Delta x \to 0$, you get

$-\frac{\partial B(x, t)}{\partial x} = \mu_0 \epsilon_0 \frac{\partial E(x, t)}{\partial t} \tag{8} \label{8}$

Let's differentiate Equation \eqref{4} with respect to $x$ and \Equation \eqref{8} with respect to $t$, so that we can remove the magnetic field part, and solving you'll get

$\frac{\partial^2 E(x, t)}{\partial x^2} = \mu_0\epsilon_0 \frac{\partial^2 E(x, t)}{\partial t^2} \tag{9} \label{9}$

This is the electromagnetic wave equation. You can get the similar equation for the magnetic field (you need to differentiate Equation \eqref{4} with respect to $t$ and Equation \eqref{8} with respect to $x$ and then you can remove the electric field part by solving the equations) which is

$\frac{\partial^2 B(x, t)}{\partial x^2} = \mu_0\epsilon_0 \frac{\partial^2 B(x, t)}{\partial t^2} \tag{10} \label{10}$

Both Equations \eqref{9} and \eqref{10} represent the electromagnetic wave equation one represented by the electric field and another by the magnetic field (electromagnetic wave is the combination of both electric and magnetic fields and both fields are perpendicular to each other and both are perpendicular to the direction of propagation of the wave). The Maxwell's equations reveal the fact that both waves sustain each other to form an entity to travel in space.

In the mechanical wave case we derived the wave equation which was

$\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2} \tag{11} \label{11}$

Now you can compare Equations \eqref{9} or \eqref{10} with Equation \eqref{11} to find the speed of the electromagnetic wave, that is

$v = \frac{1}{\sqrt{\mu_0\epsilon_0}} \tag{12} \label{12}$

which is the same as we found in the case of plane electromagnetic wave in the previous article. Here we considered the wave travelling in vacuum not in a material (dielectric), and the above expression gives the speed of the wave in vacuum $c$, that is $v = c$. If the wave is propagating in a dielectric, you need to replace $\mu_0$ by $\mu$ and $\epsilon_0$ by $\epsilon$ in the above expression, that is $v = 1/\sqrt{\mu \epsilon}$.