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Gauss's Law / Electric Field

by jesuslovesu
Tags: electric, field, gauss
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jesuslovesu
#1
May16-07, 08:50 AM
P: 201
1. The problem statement, all variables and given/known data
A very large flat aluminum plate of area A has a total charge Q over its surface. The same charge is placed over the upper surface of a glass plate. Compare the electric fields.


2. Relevant equations



3. The attempt at a solution
I know how to get the electric field of the aluminum plate.
[tex]\sigma[/tex] = Q/A
2EA = [tex]\sigma[/tex]A/e0
E = [tex]\sigma[/tex]/2e0

For the glass plate, I'm not so sure... Initially I would think E = [tex]\sigma[/tex]/e0, but it isn't...
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Doc Al
#2
May16-07, 09:25 AM
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Realize that the aluminum plate is a conductor.


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