- #1
mr_coffee
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I'm looking at how you find E in a Nonconducting sheet. It all makes sense until the last part. Visualize a thin, infinite, nonconducting heet with a uniform positive surface charge density [tex] \delta [/tex]. A sheet of thin plastic wrap, uniformily charged on one side, can serve as a simple model. Let us find the electric field E a distance r in front of the sheet.
So they ended up using a closed cylinder with ened caps of area A, arranged to pierece the sheet perpendicularly. I know that the E field is going to hit the two end caps. So your going to have 2 EA's. But if you find the flux through the opposite end wouldn't it be E cos(180) A = -EA. then the other side would be E cos(0)A = EA, so wouldn't the EA's cancel out? The book shows them both being positive:
Eo(EA + EA) = [tex]\delta[/tex]A.
E = [tex]\delta[/tex]/(2Eo).
Thanks.
So they ended up using a closed cylinder with ened caps of area A, arranged to pierece the sheet perpendicularly. I know that the E field is going to hit the two end caps. So your going to have 2 EA's. But if you find the flux through the opposite end wouldn't it be E cos(180) A = -EA. then the other side would be E cos(0)A = EA, so wouldn't the EA's cancel out? The book shows them both being positive:
Eo(EA + EA) = [tex]\delta[/tex]A.
E = [tex]\delta[/tex]/(2Eo).
Thanks.