Electromagnetics field question

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SUMMARY

The discussion focuses on calculating electric field strengths (E1 to E5) and surface charge densities (na to nd) between two parallel metal slabs with total charge Q1 = Q. The electric field strengths are determined as follows: E1 = Q/E0, E4 = E2 = 0, and E5 = Q/(2E0). The uniform electric field near a charged sheet is emphasized, and the charge distribution can be analyzed by considering the internal field required to counteract the opposing plate's field.

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Homework Statement


28.P50.jpg

The figure shows two very large slabs of metal that are parallel and distance l apart. Each slab has a total surface area (top + bottom) A. The thickness of each slab is so small in comparison to its lateral dimenstions that the surface area around the sides is negligible. Metal 1 has total charge = Q1 = Q. Assume Q is positive. In terms of Q and A, determine

a. The electric field strengths E1 to E5 in regions 1 to 5.
b. The surface charge densities na to nd on the four surfaces a to d.

2. The attempt at a solution

E1 = Q/E0
because the area vector is opposite to the bottom plate
E4 = E2 = 0 (in a conductor, E is 0)
E5 = Q/(2E0)
because the area vector is opposite to the top plate

I do not know how to determine E3, and I have no clue how to do part b
 
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The field near a uniformly charged sheet is uniform. It only falls off slightly until distance from the sheet is commensurate with its width. Hence the proximate field is not much affected by how the charge is distributed between the two faces of the plates. Also, the affect of each plate in regions 1 and 5 is little altered by relative proximity. This allows the fields to be written down straight away simply by adding or subtracting the fields from the plates as appropriate.

The charge distribution of a plate can be found by thinking in terms of the internal field that needs to be generated to counter the field from the other plate.
 

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