How do we get a expression for speed of light $\frac{1}{\sqrt{\mu\epsilon}}$ using Maxwell's equation? [duplicate]

Question

This question already has an answer here:

• Derivation of the speed of light using the integral forms of Maxwell's Equations 3 answers

I am a $12^{th}$ class student in India. I am quite new to these interesting concepts. And, I think I've learnt about electrostatics, magnetism, Maxwell's equations very thoroughly. But, my books doesn't convey satisfactory information about how $c$ is calculated. It just simply that: using Maxwell's equations, $c=\frac{1}{\sqrt{\mu\epsilon}}$.

Also, I've been reading many questions asked about light and photons on this site, stating even the electric wave and magnetic wave are out of phase (which sometimes seems obvious to me), but I lack systemetic resources to learn about them.

I just need some help like how the velocity was calculated or even some resources.

^{Probably, this part is more opinion based but can someone please also suggest some good resources to learn these things (free). I heavily rely on sites Khanacademy}

[Clarification after two answers used the differential form of Maxwell’s equations:]

The main problem is the equations taught to me are different which I find elsewhere:

A nice answer to this question thnx to @Elio Fabri

Answer 1

You can start from Maxwell's equations with no source in free space (a vacuum).

$\begin{array}{r}{\vec{\nabla} \cdot \vec{E}=0} \\ {\vec{\nabla} \cdot \vec{B}=0} \\ {\vec{\nabla} \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}} \\ {\vec{\nabla} \times \vec{B}=\mu_{0} \epsilon_{0} \frac{\partial E}{\partial t}}\end{array}$

Taking the curl of the third and forth equations give us 2 equations that show that Maxwell's equations in free space generate $\vec{E}$ and $\vec{B}$ fields that obey the wave equation. Comparing to the wave equation,

$\frac{\partial^{2} \Psi}{\partial x^{2}}=\frac{1}{v^{2}} \frac{\partial^{2} \Psi}{\partial t^{2}}$

we can see that

$\frac{1}{v^2} = \mu_0 \epsilon_0$

so that finally

$v = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$

Answer 2

You find the speed of light by using Maxwell's equations to derive wave equations for the electric and magnetic fields in which you discover that their speed of propagation is $1/\sqrt{\mu_0\epsilon_0}$. The details are below.

Start with Maxwell's equations in the absence of any charge density or current density:

$$\nabla\cdot\mathbf{E}=0$$ $$\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}$$ $$\nabla\cdot\mathbf{B}=0$$ $$\nabla\times\mathbf{B}=\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$

You see from the fourth and second equations how a changing electric field induces a magnetic field, and a changing magnetic field induces an electric field, so it seems plausible that they can "keep each other going", with a wave as a result.

To see this, first take the curl of the second and fourth equations:

$$\nabla\times(\nabla\times\mathbf{E})=-\frac{\partial}{\partial t}\nabla\times\mathbf{B}$$

$$\nabla\times(\nabla\times\mathbf{B})=\mu_0\epsilon_0\frac{\partial}{\partial t}\nabla\times\mathbf{E}$$

Now use the second and fourth equations to replace the curls on the right-hand-sides:

$$\nabla\times(\nabla\times\mathbf{E})=-\mu_0\epsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}$$

$$\nabla\times(\nabla\times\mathbf{B})=-\mu_0\epsilon_0\frac{\partial^2\mathbf{B}}{\partial t^2}$$

You see that we now have one equation with just $\mathbf{E}$ and one with just$\mathbf{B}$.

There is a mathematical identity for the curl of a curl of any vector field $\mathbf{V}$:

$$\nabla\times(\nabla\times\mathbf{V})=\nabla(\nabla\cdot\mathbf{V})-\nabla^2\mathbf{V}$$

Using this, and the first and third Maxwell equations to get rid of the divergence term, we get

$$\nabla^2\mathbf{E}=\mu_0\epsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}$$

$$\nabla^2\mathbf{B}=\mu_0\epsilon_0\frac{\partial^2\mathbf{B}}{\partial t^2}$$

These are three-dimensional wave equations for each component of the two fields, with the wave velocity equal to $1/\sqrt{\mu_0\epsilon_0}$.