Poynting vector and electric field

[Kaguro]

Homework Statement

Given that total power emitted by a source is P. Find the electric field strength at a distance r from the spherically symmetric source.

Relevant Equations

$$\vec S=\frac{1}{\mu_0} \vec E \times \vec B$$
$$c=\frac{1}{\sqrt{\mu_0 \epsilon_0}} $$
E=B*c

The Poynting vector $$\vec S=\frac{1}{\mu_0} \vec E \times \vec B$$ gives the power per unit area. If I need this in terms of electric field only,I should be able to write B=E/c (for EM wave)
Assuming they're perpendicular, $S =\frac{1}{\mu_0 c}E^2$. Now, $c=\frac{1}{\sqrt{\mu_0 \epsilon_0}} \Rightarrow c^2 \epsilon_0 = \frac{1}{\mu_0}$

So, $S =c \epsilon_0 E^2$

I am given the total power emitted as P. I need to find E at distance r. So, $P= 4 \pi r^2 S$. And so I find,
$$\frac{P}{4 \pi r^2}= c \epsilon_0 E^2$$

But in the book the Poynting vector is given as: $S =\frac{c \epsilon_0 E^2}{2}$
So... I need to assume that magnetic field contributes to the total power separately, and when asked for electric field, I should not convert EB as E^2/c ?

 

[baseballfan_ny]

But in the book the Poynting vector is given as: S=cϵ0E22
So... I need to assume that magnetic field contributes to the total power separately, and when asked for electric field, I should not convert EB as E^2/c ?

Careful with the problem. Given that the book is giving you $S = \frac {c\varepsilon_0E^{2}} {2}$, it is likely that the question is giving you the time average of the Poynting Vector and P is the time average power. Do you know why the time average of the Poynting vector would be 1/2?

 

[Kaguro]

I think because it contains a sine squared. The average value of sine square is half. That's where the 1/2 comes from.

My question: do the electric and magnetic fields both have 0 phase difference? So that at any time either both are maximum or both are minimum?

 

[baseballfan_ny]

I think because it contains a sine squared. The average value of sine square is half. That's where the 1/2 comes from.

Yes, exactly!

My question: do the electric and magnetic fields both have 0 phase difference? So that at any time either both are maximum or both are minimum?

Yes, that's correct! E and B are completely in phase with each other. It's a bit complicated to prove that, but the idea is that when you apply Maxwell's Equations in free space (no charges or no currents), the only two non-trivial equations you end up with will be Faraday's Law and the Ampere-Maxwell Law. You can apply both of these on a closed loop in the EM wave, and you will end up with the following two equations:
$$\frac {\partial^2 E} {\partial x^2} = \mu_0\epsilon_0 \frac {\partial^2 E} {\partial t^2}$$
$$\frac {\partial^2 B} {\partial x^2} = \mu_0\epsilon_0 \frac {\partial^2 B} {\partial t^2}$$

Both of these correspond to the well known equation for a wave traveling with speed v:
$$\frac {\partial^2 f} {\partial x^2} = \frac {1} {v^2} \frac {\partial^2 f} {\partial t^2}$$

The solution to the wave equation (a second order differential equation) is known to be of the form $A\sin(k(x \pm vt)) = A\sin(kx \pm \omega t)$.

From that, you should be able to tell that the solutions to the above differential equations for Maxwell's Equations to be satisfied have to both be of the same phase and are of the form:
$$ E(x,t) = E_0\sin(kx \pm \omega t) $$
$$ B(x,t) = B_0\sin(kx \pm \omega t) $$

You should also be able to see how for both differential equations to be satisfied as the wave equation $v = c = \frac {1} {\sqrt{\mu_0\epsilon_0}}$

This implies that both E and B are completely in phase with each other, traveling in space with speed c.

https://brainly.in/question/14841109

A simple diagram of an EM wave also indicates that you are correct, they peak at the exact same times (in phase with each other). An understanding of this can follow from just looking at Faraday's Law and the Ampere-Maxwell Law in the case where there are no charges and currents. What will a changing magnetic flux cause? An electric field. What will the flux of that electric field cause? A magnetic field. And the process continues to cycle through, everything completely in phase with each other (as you correctly mentioned).

 

[Kaguro]

That was very clearly explained.
Thank you very much!