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
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]