Baffled by Electric Field of a Plate

[mborn]

Hi,
I have this question:
I am a little confused about the electric field of a very large sheet of something (insulator or conductor)
for a very large sheet, E = sigma_inclosed/2(epsilon_naught), but for a conductor, it is E = sigma_inclosed/epsilon_naught, that is, twice the first case. But I have a question here in my book that says:

A thin conducting plate 50 cm on a length lies in the x-y plane. If it is placed in an external electric field of 8*10^4C directed perpendicular to the plate, find;
1- The charge density of each face of the plate,
2- The total charge on each face.

The answers uses, E = sigma / 2(epsilon_naught) to find sigma,

Shouldn't he used the formula E = sigma_inclosed/epsilon_naught since it is a conducting plate?


M B

 

[learningphysics]

Yes, you're right. It depends on whether they are using Eext, or a different E. Eext=sigma/epsilon_naught.

You can also solve the problem by thinking of superposition (which is where I'm guessing they used sigma/2epsilon_naught). The field produced by the top positive sheet alone is sigma/2epsilon_naught. The field produced by the bottom sheet alone is -sigma/2epsilon_naught. They add together in the center (between the two sheets), and need to cancel the external field. So sigma/2epsilon_naught+sigma/2epsilon_naught=Eext. Then we get the initial result: sigma/epsilon_naught=Eext.

What is the value of sigma they get?

 

[Andrew Mason]

The answers uses, E = sigma / 2(epsilon_naught) to find sigma,

Shouldn't he used the formula E = sigma_inclosed/epsilon_naught since it is a conducting plate?

This is a little tricky. One has to apply Gauss' law to each surface of the conductor sheet. For a conductor, all charge resides on the top and bottom surfaces with 0 field inside.

Using Gauss' law, the flux per unit area through the top surface is: \phi = \sigma_{top} \delta A/\epsilon_0. Since the flux goes in both directions, above the surface and into the interior, the flux from the top surface charge above + the flux from the top surface below (ie toward the inside of the conductor) is the total flux. So
E\delta a_{up} + E\delta a_{interior} = \sigma_{top} \delta a/\epsilon_0
Since the surface charge produces equal flux in the up and down directions:
E\delta a_{up} = E\delta a_{interior}
then:
2E = \sigma_{top}/\epsilon_0

The same applies to the bottom surface of the conductor. You have to work out the charge densities on the top and bottom surfaces such that the field inside the conductor is 0 in the presence of the external field.

AM