Archived Electromagnetics field question

AI Thread Summary
The discussion focuses on calculating electric field strengths and surface charge densities between two parallel metal slabs with a total charge Q on one slab. The electric field strengths in regions 1, 4, and 5 are determined, with E1 equating to Q/E0, E4 being zero due to the properties of conductors, and E5 calculated as Q/(2E0). The challenge lies in finding E3 and the surface charge densities on the plates. The uniform electric field near a charged sheet is emphasized, indicating that the charge distribution affects the internal field necessary to balance the opposing fields from the plates. Understanding these principles is crucial for solving the problem effectively.
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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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