A simplified derivation of $E_o=B_o C$ , without maxwell's wave equation I recently came across the following equation in my physics course
$$E_o=B_o *c$$ ,where $E_o$ is the amplitude of the electric field, $B_o$ that of the magnetic field, and $c$ the speed of light
while searching for a derivation, I came across this question on if the equation $E_o=B_o C$ ,is always true, however , it doesn't answer my question.I would really appreciate it if someone could give me a simple proof using atmost only elementary calculus
 A: The density of energy in the electric field in MKS units is given by:
$$u_E = \frac{\epsilon_0}{2} \mathbf{E}^2.$$
The density of energy in the magnetic field is
$$u_B = \frac{1}{2\mu_0} \mathbf{B}^2.$$
Now, there are two facts that you would need to study the wave equations to know what these imply.

*

*In general for waves the kinetic and potential energy densities average out over one period to be equal.

*The magnetic field feeds into the potential energy part and the electric field is the kinetic energy part for electromagnetic waves.

The details of which energy is kinetic and which is potential actually end up being unimportant, but that they are related in this way is all we need.
So, let's take the ratio of our energy densities, using the electric and magnetic amplitudes because we averaged both over one wave period.
\begin{align}
   \frac{\langle u_E\rangle}{\langle u_B\rangle} &= \frac{\mu_0\epsilon_0 E_0^2}{B_0^2} \\
&= 1
\end{align}
Now, solve for $E_0$:
\begin{align}
   E_0 &= \frac{1}{\sqrt{\mu_0\epsilon_0}}B_0.
\end{align}
If you haven't already learned this, then now is as good a time as any to learn that $c=1/\sqrt{\mu_0\epsilon_0}$.
Granted, with the number of facts I have thrown at you without deriving them it may not help, but the first fact should be pretty plausible even without derivation. I can sketch where the derivation comes from, though. The wave equation is what you get when you have an infinite number of simple harmonic oscillators connected to each-other. The end result is that each wavelength acts like an independent simple harmonic oscillator. You can derive that the energy in a simple harmonic oscillator just shifts back and forth between kinetic and potential, and spends about an equal amount of time as each. The only calculus facts needed in this derivation are the derivatives of sine and cosine, and how to integrate $\sin^2$ and $\cos^2$.
I can also motivate why the electric field is kinetic energy and why the magnetic field is potential. The key is looking at how to write them in terms of the vector potential and electric potential. First, one of Mawell's equations says that there is no magnetic charge. In calculus, we write that like this: $\nabla \cdot \mathbf{B} = 0$. Heuristically, you can think of this as saying that magnetic fields can't point into or out of some point, they can only swirl around. There is another mathematical fact that says any vector field that acts this way can be written as the curl of another vector field,
$$\mathbf{B} = \nabla \times \mathbf{A}.$$
The only important part of that equation is that it says that the magnetic energy density only has space derivatives in it - no time derivatives. Any energy that doesn't have time derivatives in it is a potential energy.
Second, another of Maxwell's equations says that curling electric fields can only exist if there are changing magnetic fields, and vice-versa. When you combine that with Coulomb's law, you get that we can write
$$\mathbf{E} = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t},$$
where this $\mathbf{A}$ is the same $\mathbf{A}$ as we use to calculate the magnetic field. The key fact this shows is that the electric field energy is the only part of the energy with time derivatives in it. That's why it's the kinetic energy. Now, there are also space derivatives in $\mathbf{E}$, but that actually isn't important for electromagnetic waves: we can separate that piece out so that it doesn't participate.
A: Let's make it easier by finding solutions in the form of a  wave moving in the x direction  with z polarisation
$$\nabla^2  \vec{E}  = \mu_0\epsilon_0 \frac{\partial^2\vec{E}}{\partial t^2}$$
$$[\nabla^2 {E_{x}}]\hat i + [\nabla^2 {E_{y}}]\hat j + [\nabla^2 {E_{z}}]\hat k  = \mu_0\epsilon_0 \frac{\partial^2\vec{E}}{\partial t^2}$$
For simplicity let's assume that  my $\vec{E}$ field only points in the $\hat k$ direction $\vec{E} = E_{z} \hat k$ meaning our equation reduces to:
$$ \nabla^2 {E_{z}}   = \mu_0\epsilon_0 \frac{\partial^2E_{z}}{\partial t^2}$$
$$\frac{\partial^2 E_{z}}{\partial x^2} + \frac{\partial^2 E_{z}}{\partial y^2} + \frac{\partial^2 E_{z}}{\partial z^2} = \mu_0\epsilon_0 \frac{\partial^2E_{z}}{\partial t^2}$$
Assuming the wave only is dependant on x;
$$\frac{\partial^2 E_{z}}{\partial x^2} = \mu_0\epsilon_0 \frac{\partial^2E_{z}}{\partial t^2}$$
To solve this equation:
Assume there exists some solution in the form
$E_{z} =E_{0}e^{i(kx-\omega t)}$
Why I assume this is unimportant, if we substitute this solution into our equation, we can show that the equation is satisfied given a "dispersion relation" holds.
Substituting this into the equation, we find that in order for our solution to satisfy the wave equation, we need:
$$v_{p}= \frac{\omega}{k} = \frac{1}{\sqrt{\mu_0\epsilon_{0}}}$$
This firstly, proves the phase velocity of   the wave is a number that matches the measured speed of light.
To prove your desired relation  we need to use faradys law.
$$\nabla × \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$
Taking the curl of our new found E field, integrating to find the B field, we find that
$$|\vec{B}| c = |\vec{E}|$$
I'll let you to do the math for that part since it is important to try.
This derivation only holds for a plane wave, this is a condition about em waves,and not the EM field in general
P.s the real of our trial solution is also a solution, and obeys the same properties as our complex wave
