Poynting vector and electric field

In summary, the Poynting vector is a way to calculate the power per unit area in an electromagnetic wave. It can be written in terms of the electric field and magnetic field, and assuming they are perpendicular, the equation becomes ##S =\frac{1}{\mu_0 c}E^2##. The book gives a different form for the Poynting vector, which can be explained by the fact that it represents the time average power. This is due to the fact that the electric and magnetic fields are completely in phase with each other in an EM wave. This can be seen through the application of Maxwell's Equations, where the solutions for both electric and magnetic fields are of the same phase and travel with
  • #1
Kaguro
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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 ?
 
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  • #2
Kaguro said:
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?
 
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  • #3
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?
 
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  • #4
Kaguro said:
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!

Kaguro said:
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.

1606495552597.png

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).
 
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  • #5
That was very clearly explained.
Thank you very much!
 
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Related to Poynting vector and electric field

1. What is the Poynting vector and how is it related to the electric field?

The Poynting vector is a mathematical quantity that describes the direction and magnitude of the flow of energy in an electromagnetic field. It is directly related to the electric field, as it is calculated by taking the cross product of the electric and magnetic field vectors.

2. What is the physical significance of the Poynting vector?

The Poynting vector represents the rate at which electromagnetic energy is being transferred through a given area. It is a useful tool for understanding the flow of energy in electromagnetic systems, such as antennas and transmission lines.

3. How is the Poynting vector related to the intensity of an electromagnetic wave?

The magnitude of the Poynting vector at a given point is equal to the intensity of the electromagnetic wave at that point. This means that the Poynting vector can be used to calculate the amount of energy carried by an electromagnetic wave.

4. Can the Poynting vector be used to determine the direction of energy flow in an electromagnetic field?

Yes, the direction of the Poynting vector indicates the direction of energy flow in an electromagnetic field. The vector points in the direction of the energy flow, which is perpendicular to both the electric and magnetic field vectors.

5. How is the Poynting vector used in practical applications?

The Poynting vector is used in a variety of practical applications, such as in the design and analysis of antennas, transmission lines, and electromagnetic shielding. It is also used in the study of electromagnetic radiation and its effects on biological systems.

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