Linear superposition for electric field

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SUMMARY

The discussion centers on calculating the electric field generated by a non-conducting slab with a charge density of ρ = αx² and a uniformly charged plate located adjacent to it. The problem requires the application of the superposition principle to derive the electric field in four distinct regions: x < -d/2, -d/2 < x < 0, 0 < x < d/2, and x > d/2. Participants confirmed the necessity of breaking down the problem into these regions and discussed the potential to combine regions 2 and 3 into a single expression. Clarifications were sought regarding the definitions of variables d and L, as well as the notation for x².

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AKJ1
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Homework Statement



A rather large non conducting slab of area A and thickness d has a charge density given by ρ = αx2.
The origin is through the center of the slab. That is, it bisects the slab into two equal volumes of d/2 thickness and with an area A, with -L/2 to the left of x=0 and L/2 to the right of x=0.

Suppose there is a large thin plate of UNIFORM charge density sigma on the left side of the aforementioned slab. What is the E field (everywhere) due to this system? Derive an expression for the electric field for the thin plate and then apply the superposition principle. Also give the domains on X for the regions chosen.

Note: You will not have just one answer.

The Attempt at a Solution



I found the E field everywhere due to the slab & due to the thin plate. Now for the E field everywhere, I should break this problem down into four regions and find the superposition of E field in these regions (due to slab and plate):

x<-d/2

-d/2 < x < 0

0<x<d/2

x>d/2

Is this correct?
 
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I would expect that you can combine regions 2 and 3 into a single formula, but splitting them up is not wrong.

d and L are the same thing?

Is x2 meant to be x2?
 

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