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Homework Help: Energy Carried by Electromagnetic Waves- why is it wrong?

  1. Nov 17, 2006 #1
    Energy Carried by Electromagnetic Waves- why is it wrong???

    Hi! I have this homework problem that I just don't understand why it's wrong.

    A monochromatic light source emits 110 W of electromagnetic power uniformly in all directions.

    (a) Calculate the average electric-field energy density 3.00 m from the source.

    (b) Calculate the average magnetic-field energy density at the same distance from the source.

    (c) Find the wave intensity at this location.

    First of all, I know that (a) and (b) will be the same. I have also calculate part (c), which is correct, and it is 0.9726 W/m^2.

    I have used various equations that would work for this type of problems.

    1. The Sav, or average Poynting vector, in terms of the magnetic and electric field. I always get the same answer, which is 6.4*10^-8. Then I use the energy density equation, which gives me 3*10^-9.

    2. I use Sav=cu....since I have Sav, and I know c, then it seemed logical to use this. My answer is the same, 3*10^-9.

    But its WRONG, and I don't understand WHY! PLEASE HELP ME.

  2. jcsd
  3. Nov 17, 2006 #2


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    First of all what does it mean that "A monochromatic light source emits 110 W of electromagnetic power uniformly in all directions."? I think it means that you have a source that radiates spherical waves and the average flux of the Poynting vectors through a spherical surface centered on the source is of 110W. In particular, taking the sphere to be of radius 3.00 m, this gives

    [tex]110=<\oint_{\mathbb{S}_3} \vec{S}(\vec{r}=3\hat{r})\cdot\hat{n}dA >= <\int_0^{2\pi}\int_{0}^{\pi} ||\vec{E}(\vec{r}=3\hat{r})\times\vec{H}(\vec{r}=3\hat{r})|| 3^2 sin\theta d\phi d\theta >=36\pi <E(\vec{r}=3\hat{r})H(\vec{r}=3\hat{r})>[/tex]

    But in a monochromatic plane wave propagating in the void, [itex]H=E/\mu_0 c=[/itex] so we have the equation

    [tex]\frac{110\mu_0 c}{36\pi } =<E(\vec{r}=3\hat{r})^2>[/tex]

    Now you can substitute that into the formula for the average mean electric energy:

    [tex]<u_E(\vec{r}=3\hat{r})>=\frac{\epsilon_0<E(\vec{r}=3\hat{r})^2>}{2} =\frac{110\epsilon_0\mu_0 c}{72\pi }=\frac{110}{72\pi c}\approx 1.62\times 10^{-9} [J/m^3][/tex]

    What is the correct answer?

    (I don't know what equations are given to you; if you write them I can help you sort out your problem using those equations)
    Last edited: Nov 17, 2006
  4. Nov 17, 2006 #3


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    I took a bit more simplistic approach and just calculated the energy density from the rate of energy flowing through the surface (power) and the speed of light. Looks to me like the total energy density is

    u = P/Ac = 3.24*10^-9 J/m³
  5. Nov 17, 2006 #4
    Actually...the equations that are given in the book are very general, and none of them involve integrals.

    I have used the third equatin below Poynting vector.


    And I also used Saverage/c=uaverage...but it's still wrong.
  6. Nov 17, 2006 #5
    I just divided the 3.24 by 2, which gave me the answer.

    The reason why I did not get it, I think, is because the book does not present it correctly.

    The Smax equation in the picture above is actually given as Sav in my book, so that's why the answer was not correct.

    Thank you so much!!!
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