Charge Density on a Conductive Slab

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The discussion revolves around calculating charge density on a conductive slab influenced by an infinite sheet of charge. The initial approach involved balancing equations based on charge distributions on either side of the slab, but this led to confusion as the calculations did not yield the expected results. The participant recognized the need to consider the electric field within the slab and how it relates to surface charge densities. They are seeking clarity on how to accurately calculate the field and charge densities based on the setup. The conversation emphasizes the importance of understanding electric fields in conductive materials for solving such problems.
MengMei
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Homework Statement
An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -4.7 μC/m2. A thick, infinite conducting slab, also oriented perpendicular to the x-axis, occupies the region between a = 2.8 cm and b = 4.5 cm. The conducting slab has a net charge per unit area of σ2 = 87 μC/m2. (Recall that the surface charge densities σa and σb on the slab surfaces at a and b, respectively, sum to equal the net charge per unit area: σa + σb = σ2.)

What is σb, the charge per unit area on the surface of the slab located at x = 4.5 cm?
Relevant Equations
σa + σb = σ2
Okay, so I tried thinking of this as like a simple balancing of equations. There's an infinite sheet of charge on the left and a conductor on the right with some charge already on it. My thought process was that the side nearer to the charged sheet would have 4.7 more μC/m2 than the far side. Knowing this, I assumed that the near side would have 87-4.7 = 82.3 and the far side would have 87+4.7 = 91.7. That didn't work, so I took at step back, looked at the equation and then thought, "Oh, the charges on each side of the slab have to equal 87."

So I thought to half the total net charge and then find the difference between the two sides due to the infinite sheet with charge. That also didn't work. Now I'm just really confused and don't know what to do anymore.
 
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Instead of guessing, think about the field inside the slab. What do you know about that, and how can you calculate it from the surface charge densities?
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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