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Parametric representation of a vector electric field

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data

    See Attachment

    2. Relevant equations

    None I can think of

    3. The attempt at a solution

    I'm fairly certain that phi_yx is zero
    Also I tried factoring out the cos and splitting up the equation into it's respective components, but to no avail. Am I even going about this correctly?
     

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    Last edited: Oct 23, 2011
  2. jcsd
  3. Oct 24, 2011 #2

    vela

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    The last bit is a typo. It should be [itex]\phi_{xy} = \phi_x[/itex].

    I'm not sure what you mean by "splitting up the equation into its respective components" since the equation you're supposed to show is true is a scalar equation.
     
  4. Oct 25, 2011 #3
    Thanks for the reply Vela! Actually I think I figured it out. Maybe you or someone else can check my work? I assumed that it the phi's were not typos, but they could very well be. I went ahead and attached my work.
     

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  5. Oct 25, 2011 #4

    vela

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    Your expressions for Ex and Ey are slightly wrong. For example,
    [tex]E_x = A_x\cos(kz-\omega t+\phi_x) = A_x\cos(-r+\phi_x) = A_x\cos(r-\phi_x)[/tex]since cosine is an even function. You have an extra negative signs floating around.

    You must have made a mistake somewhere in the middle because the cross term has the wrong sign. The sign doesn't flip the way you did it because, again, [itex]\cos (-\theta)=\cos\theta[/itex].

    You actually made it more complicated by keeping the E's and A's around. You know that
    \begin{align*}
    \frac{E_x}{A_x} &= \cos(\phi_x-r) \\
    \frac{E_y}{A_y} &= \cos(r) \\
    \end{align*}so you could have just written the lefthand side in terms of cosines and then shown it simplifies down to the righthand side.
     
  6. Oct 30, 2011 #5
    Ah that I forgot about that cosine property and I found a convenient sign mistake half way through the problem. Thanks Vela!
     
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