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Maxwell's Equation

  1. Jan 16, 2006 #1
    Hi,
    I'm new here. I have a question. How do you verify that: E = E(max)cos(kx-wt) is a solution to Maxwell's derived equation:
    ((d^2)E/dx^2) = e(epsilon nought)u(permitivity of free space) x (d^2)E/dx^2. Thanks.
    What I first did was to substitute k = 2pi/lambda, w = 2pi(f). Then I set 2pi(f) as 2pi/lamda x c. Taking 2pi/lambda out and times cos equals 1/lambda. Now I'm stuck.
    Mat
     
    Last edited: Jan 16, 2006
  2. jcsd
  3. Jan 16, 2006 #2

    Galileo

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    To verify the function satisfied the wave equation, you simply put in there. Or look what the left side of the wave equation is for the given E, then calculate the right side and see if they are equal.
     
  4. Jan 16, 2006 #3
    I'm still confused in terms of how to get the sides to equal each other. How do I eliminate the minus and the cos?
     
  5. Jan 16, 2006 #4

    Galileo

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    The only relation you need to use is [itex]\omega/k=c=\frac{1}{\sqrt{\epsilon_0 \mu_0}}[/itex].

    So you have [itex]E(x,t) = E_{max}\cos(kx-wt)[/itex].

    Now what's [itex]\frac{\partial^2}{\partial x^2}E(x,t)[/itex] and what's [itex]\frac{\partial^2}{\partial t^2}E(x,t)[/itex]?

    So does E(x,t) verify the equation?
     
  6. Jan 16, 2006 #5
    I'm sorry I'm a bit lost here. So you do partial differentiation for E(x,t) = Emax(cos(kx-wt)) for x and t? Thank you for your patience.
     
  7. Jan 17, 2006 #6

    Galileo

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    Well, isn't that what the wave-equation says?

    [tex]\frac{\partial^2}{\partial x^2}E(x,t)=\epsilon_0 \mu_0 \frac{\partial^2}{\partial t^2}E(x,t)[/tex]

    What exactly is it you don't understand? Is it the question itself or the way to go about solving the problem? My guess is that you don't understand the question.
     
  8. Jan 17, 2006 #7
    Thanks, I think I got it. So you derivitive for B and E and solve both sides. I couldn't get the question, it was worded kinda weird. Thanks for your help.
     
  9. Jan 18, 2006 #8

    Galileo

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    What answer did you get?
     
  10. Jan 18, 2006 #9

    HallsofIvy

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    Actually, I would think you were the one with the "weird" wording. I have no idea what "solve both sides" means! I know how to solve equations and I even know how to solve problems in general but I don't know what is meant by solving a "side" of an equation!


    Suppose you had a problem that asked to show that x= 3 is a solution to x2- 5x= -6. What would you do? (I hope you would not say "solve the equation.")
     
  11. Jan 18, 2006 #10
    I just did a dervative twice for E = Emaxcos(kx-wt) and for B = Bmaxcos(kx-wt) in terms of x and t. Then I equaled the double derivative of E = Emaxcos(kx-wt) in terms of x with terms of t and solve to get light = light. Same thing for B equation too.
     
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