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.
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...