Gauss's Law / Electric Field

[jesuslovesu]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

I know how to get the electric field of the aluminum plate.
$$\sigma$$ = Q/A
2EA = $$\sigma$$A/e0
E = $$\sigma$$/2e0

For the glass plate, I'm not so sure... Initially I would think E = $$\sigma$$/e0, but it isn't...

 

[Doc Al]

Realize that the aluminum plate is a conductor.

 

[m2003]

The electric field for the glass plate can be found using Gauss's Law, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. In this case, the enclosed charge is the same for both the aluminum and glass plates (since they have the same total charge Q), but the permittivity of free space is different for each material. Therefore, the electric field for the glass plate will be different from that of the aluminum plate. The exact value can be calculated using the same equation as for the aluminum plate, but with the permittivity of glass (e_glass) substituted in place of e0. This shows that the electric fields for the two plates will be different, despite having the same total charge. This is due to the difference in the permittivity of the two materials, which affects how the electric field is distributed over their surfaces.