How Does Gauss' Law Apply to Charged Metal Sheets?

[forty]

Two large flat metal sheets are a distance L apart. The separation L is small compared
to the lateral dimensions of the sheets. Each sheet has a total surface area A, which includes both top and bottom surfaces of the sheet. The thickness of each sheet is very small compared to its lateral dimensions. Metal one has total charge +Q while metal 2 has total charge +2Q. In terms of Q and A and using the Gaussian surfaces shown, determine:

(a) The electric field strengths, E1 to E5, in regions 1 to 5.

(b) The surface charge densities, a to d, on the four surfaces a to d.

Note: The charge distribution will be reasonably symmetric and the magnitude of the field
strength at all points in regions 1 and 5 will be equal. That is the key to working it all out!

http://students.informatics.unimelb.edu.au/serve/cmcleod/stuff/gauss.JPGThe electric fields inside the metal (regions 2 and 4) are zero (if that's wrong god help me).

So i can use gauss' law on the 3 Gaussian surfaces.

Gauss' law => E.dA = q/e

So for gauss 1: E.dA = 3Q/e ?

I have a feeling like usual that this is wrong. the charge Q and 2Q are spread over the entire plate so only the amount inside the the Gaussian surface is enclosed, but i don't know how to work that out from the information given.

Any help would be greatly appreciated.

 

[entorm]

I think that you should calculate the charge surface density for each plate and then get the enclosed charge from that: q=s dA, where s=Q/A (or 2Q/A)

 

[forty]

is it ok to think about this as though the charge in region 4 is Q+ and the charge in region 2 is 0?

 

[forty]

I tried using a superposition approach...

1,2,3,4,5 - 3Q/e, 0 , Q/e , 0 , 3Q/e

does this look even faintly relevant?

(although i still have to work it out using gauss' law)