Question about cross and dot product

  • Thread starter leonne
  • Start date
  • #1
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Homework Statement



Note: [tex]\nu[/tex] is del could not find it...

I need to prove the [tex]\nu[/tex]*E=0
and [tex]\nu[/tex] x E=-dB/dt


Homework Equations


E(s[tex]\phi[/tex]zt)=(Acos(Kz=wt)/s)s^



The Attempt at a Solution


so for the first one [tex]\nu[/tex]*E=0 I thought it would be
d/ds(E) but what they did was 1/s d/ds(s E) no idea why they did it. Is it because its S^? I thought i remember doing something like this.
for the cross product
d/ds d/d[tex]\phi[/tex] d/dz
E 0 0

....I got the right answer but they got 0+dE/dZ [tex]\phi[/tex]^-1/s dE/d[tex]\phi[/tex] z^
no idea how they got the z^ i thought it was 0, well it does =0 but where did the -1/s come from?
thanks for the help[STRIKE][STRIKE][/STRIKE][/STRIKE]
 
Last edited:

Answers and Replies

  • #2
lanedance
Homework Helper
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2
try writing all youyr code in a single tex banner & to write del (\nabla), so you want to show:
[tex]\nabla \bullet E = 0 [/tex]

how about directly applying the product? I can;t really understand teh form of your electric field - mayeb you can re-write it?
 
  • #3
191
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hmm cant edit my post
so need to show that
[tex]\nabla[/tex] [tex]\bullet[/tex] E=0
[tex]\nabla[/tex] [tex]\times[/tex] E= -dB/dT
Well its a wave function
And i need to show that E satisfy Maxwell equation and the boundary conditions.
boundary condition E parallel =0 and B perpendicular=0
lets say E was x^ for the dot product
(d/dx+d/dy+d/dz)(E x^+0+0) so i would have dE/dx am i wrong?
 

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