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Author Topic: how div . E = 0 in the dielectric of a coaxial line  (Read 2471 times)

Offline malik

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how div . E = 0 in the dielectric of a coaxial line by applying the divergence theorem to a portion of the dielectric.


 

lyner

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how div . E = 0 in the dielectric of a coaxial line
« Reply #1 on: 03/06/2008 09:25:22 »
Doesn't the divergence theorem just mean that the total divergence of the field from a region depends on the charge contained in it.
There is no total charge in a piece of the dielectric; it is just polarised; so the divergence will be zero.
Pictorially, in a section of coax cable dielectric, you have a small area on the inside bit with a high field going in and a large area on the outside with a lower field going out. Net total is zero.
 

Offline malik

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how div . E = 0 in the dielectric of a coaxial line
« Reply #2 on: 03/06/2008 09:32:32 »
can u please write ur answer in mathematical form.
 

lyner

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how div . E = 0 in the dielectric of a coaxial line
« Reply #3 on: 03/06/2008 10:30:33 »
You already did that; I just put it into words.
There is no net charge in the dielectric so divD = 0
Are you asking me to do your homework for you?
 

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how div . E = 0 in the dielectric of a coaxial line
« Reply #3 on: 03/06/2008 10:30:33 »

 

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