Finding the E-Field between parallel plates

[thursdaytbs]

I have a positive and a negative charged plate running parallel to one another. To find the Electric Field to the left of the plates, I say:

E dot dA = Qenclosed / permittivity
Then E times 2pi(r) = Qenclosed / Permittivity
then E = Qenclosed / 2pi(r) x Permittivity

and since Qenclosed to the left and right of the parallel plates is zero, E = zero?

And for in between the two plates I can say the charge enclosed is 2pi(r)^2 times the charge enclosed by that area? divided by 2pi(r) x Permittivity?

I'm a little bit lost as to how to explain in detail how to solve for the Electric Field in between oppositely charged parallel plates. Any help? Thanks.

 

[quasar987]

I have a positive and a negative charged plate running parallel to one another. To find the Electric Field to the left of the plates, I say:

E dot dA = Qenclosed / permittivity
Then E times 2pi(r) = Qenclosed / Permittivity
then E = Qenclosed / 2pi(r) x Permittivity

and since Qenclosed to the left and right of the parallel plates is zero, E = zero?

I assume your Gauss' surface is a cylinder running perpendicular to the plates and passing through both plates? If so, you're correct, but you must use symetry arguments to justify why the integral on the side (as opposed to caps) of the cylinder is 0. This said, I suspect that you've seen in class what is the field of an infinite uniformly charged plate. Is that correct? If so, how can you use that result, together with the principle of superposition of the electric field to find the answer to all 3 questions in 3 microseconds?

 

[Astronuc]

The electric field lines point (by convention) away from positive and toward negative, so they wouldn't cancel if the integral over the surface is done correctly.

See also - http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html