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Velocity of EM waves

  1. Nov 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the velocity of EM waves as a function of [tex]\epsilon_{0}[/tex] and [tex]\mu_{0}[/tex]

    2. The attempt at a solution

    [tex]E = E_{0}cos(kx-\omega t)[/tex]

    Using [tex]v= \frac{\omega}{k}[/tex]
     
  2. jcsd
  3. Nov 7, 2008 #2

    gabbagabbahey

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    When you used Maxwell's equations to derive the wave equation, you should have ended up with an answer to this :wink:
     
  4. Nov 7, 2008 #3
    No, I didn't. But I can substitute my expression for E into the wave equation. What is [tex]\vec{\nabla}^{2}E[/tex]?

    [tex]\frac{\partial^{2}E}{\partial x^{2}} + \frac{\partial^{2}E}{\partial t^{2}}[/tex]?
     
  5. Nov 7, 2008 #4

    gabbagabbahey

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    Last time I checked, Maxwell's equations were in terms of [itex]\epsilon_0[/tex] and [itex]\mu_0[/itex] not [itex]c[/itex]; so you should have ended up with a wave equation where the propagation speed is in terms of [itex]\epsilon_0[/tex] and [itex]\mu_0[/itex]...if you didn't, then you did something wrong....I think you should go back to that problem and show me your work.
     
  6. Nov 7, 2008 #5
    double
     
  7. Nov 7, 2008 #6
    Of course...

    So [tex]\vec{\nabla}^{2}E[/tex] = [tex]\frac{\partial^{2}E}{\partial x^{2}} + \frac{\partial^{2}E}{\partial y^{2}} + \frac{\partial^{2}E}{\partial z^{2}}[/tex] (only spatial dimension, not time)?
     
  8. Nov 7, 2008 #7

    gabbagabbahey

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    Did you even read my last post?
     
  9. Nov 7, 2008 #8
    Yes.[tex]\frac{1}{v^{2}} = \mu_{0}\epsilon_{0}[/tex], so [tex]\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}} = v[/tex]. That's what you meant, right?

    That would mean that (if I substitute my expression for E into the wave equation) [tex]\vec{\nabla}^{2}E = \frac{\partial^{2}E}{\partial x^{2}} + \frac{\partial^{2}E}{\partial y^{2}} + \frac{\partial^{2}E}{\partial z^{2}}[/tex].

    Can I also write [tex]\vec{\nabla}^{2}E = \frac{\partial^{2}E}{\partial \vec{r}^{2}}[/tex]?
     
    Last edited: Nov 7, 2008
  10. Nov 8, 2008 #9

    gabbagabbahey

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    Yes.

    First, the electric field is vector, not a scalar so this relation is incorrect...second what does this have to do with finding v...or anything else for that matter? :confused:
     
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