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COnducting Slab Question

  1. Sep 13, 2008 #1
    A conducting slab of side length of w carries a net charge of Q. Find the charge density and electric field just outside one face of the slab.

    *So would the charge density = Q/w*w*2 because there are 2 sides?
    *And would the E just be density/2*(epsilon = permissivity constant) because you are only *dealing with one face of the slab?

    Next, the slab is placed next to an infinite sheet of charge Y so that the faces are parallel. What's the field and density on each face?

    *I think on the face farther from the sheet of charge Y, the fields are in the same direction. So you would add the fields together. This would be (density of slab)/2*epsilon + (density of sheet)/2*epsilon.

    *Is it right that on the face closer to the sheet, the direction will be reversed? And the slab surface will have a positive charge (because it's conducting and so the excessive charges are on the surface) but then the sheet will be positive too, and so you need to substrace the two forces. But I don't understand how, if the slab is positive on the surface near the sheet, the field would be positive (because wouldn't the field be negative since it is going against the direction of the x axis?).

    *And I don't see why the density of each of the slab surfaces isn't E (field) * epsilon * 2 (since you are dealing with each side of the slab).

    Thank you so much.
     
  2. jcsd
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