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Electrodynamics q

  1. Dec 1, 2009 #1
    Electrodynamics q!!!

    Struggling with this module, if someone could get me started in the right direction that'd be great!!!

    An electric field is given by

    E(r,theta,phi,t)=A(sin theta/r)(cos(kr-wt)-(1/kr)sin(kr-wt)) in phi direction

    Show that E obeys all 4 of Maxwells equations in vacuum (no free charges and no free currents)

    for Maxwells 1st eqn I said that there is only a phi component in the electric field so the divergence is d/d(phi) of E. As there is no phi term in the equation this goes to zero. Is this correct? If so how do I continue with Maxwells other equations? Any help wiould be greatly appreciated!!!
     
  2. jcsd
  3. Dec 1, 2009 #2

    CompuChip

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    Re: Electrodynamics q!!!

    You can simply write down the equations using the [itex]\nabla[/itex] operator as you are used to. Then to do the explicit calculation, you can copy the identities from this bookmark-worthy Wikipedia page.
     
  4. Dec 1, 2009 #3
    Re: Electrodynamics q!!!

    ok so for maxwells 1st eqn the div of E is just the phi component and gives 1/(r sin theta)d/d phi(E). Again d/d phi(E) is zero so the whole eqn goes to zero. Is this right? Also I'm not sure how to get maxwells 2nd equation as i have no term for B and when trying to calculate the 3rd eqn i get zero again as there is no phi term in E. Very confused alltogether!!??
     
  5. Dec 1, 2009 #4

    gabbagabbahey

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    Re: Electrodynamics q!!!

    That electric field only obey all 4 Maxwell equations for a particular associated magnetic field....it's up to you to find what that field is :wink:....You shouldn't be getting zero for [itex]\mathbf{\nabla}\times\textbf{E}[/itex]....if you show us your calculation, we should be able to spot your error
     
  6. Dec 2, 2009 #5

    CompuChip

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    Re: Electrodynamics q!!!

    Hint: there is a dB/dt somewhere in the equations.
     
  7. Dec 8, 2009 #6
    Re: Electrodynamics q!!!

    Ok now I am now getting a fairly long formula for the curl of E. I have tried letting that equal to -dB/dt (Maxwells 3rd eqn) and integrating over time to get an expression for -B. However, when i then get the div of B, nearly everthing cancels but not quite everything, so i do not get zero (Maxwells 2nd eqn) (unless, this is, cos^2(theta)=2cos(theta)sin(theta), which i'm pretty sure it doesn't!!). This is quite a long and tedious calculation and I am still not sure I am going down the right path. Is there any more advice or hints anyone can give!!?? Again, any help would be hugely appreciated.
     
  8. Dec 8, 2009 #7
    Re: Electrodynamics q!!!

    I suppose my main problem is finding the particular magnetic field, B, associated with the electric field given by E.
     
  9. Dec 10, 2009 #8

    gabbagabbahey

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    Re: Electrodynamics q!!!

    As long and tedious as it might be, the only way we 'll be able to see where you are going wrong is if you post your calculations....let's start with your result for [itex]\mathbf{\nabla}\times\textbf{E}[/itex]...
     
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