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Gauss' Law - find magnitude of the electric field

  1. Sep 17, 2014 #1
    A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x=d and x=−d. The y- and z-dimensions of the slab are very large compared to d and may be treated as essentially infinite. The slab has a uniform positive charge density ρ.

    find magnitude of the electric field due to the slab at the points 0≤x ≤d.

    and magnitude of the electric field due to the slab at the points x≥d


    Equations: Gauss's Law (too hard to type)



    So i was working through this and am stuck about the area and charge. I doesn't give charge and am i just to assume that the height of the cylinder i choose is just arbitrary z?
     
  2. jcsd
  3. Sep 17, 2014 #2

    collinsmark

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    For clarity, Gauss' Law is

    [tex] \oint \vec E \cdot \vec{dA} = \frac{Q_{enc}}{\varepsilon_0} [/tex]

    The charge enclosed within the Gaussian surface is the volume of the Gaussian surface times the charge density, ρ, for the situation where the Gaussian surface is completely within the material. For the situation where one end of the Gaussian surface sticks out of the material, the volume in question is intersection of the Gaussian surface and the material (the charge is the volume of material that happens to be within the Gaussian surface, multiplied by ρ).

    By the way, you could use a cylinder for your Gaussian surface. But you could also use a "box" for your Gaussian surface if you would rather stick with Cartesian coordinates. It's totally up to you. (Either way is fine. It won't affect the final answer).

    Keep in mind the vector dot product in the left hand side of the equation. Only certain sides of the cylinder or box will have a non-zero [itex] \vec E \cdot \vec{dA} [/itex] product. Those sides of the cylinder/box that have [itex] \vec E [/itex] perpendicular to [itex] \vec{dA} [/itex] do not contribute, and you can ignore them. [Edit: you may also ignore any sides of the cylinder/box that have zero [itex] \vec E [/itex] at those particular sides.]

    I think you mean, 'x'. (At least for the first part of the problem, where 0≤x ≤d.)
     
  4. Sep 18, 2014 #3

    rude man

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    Put a rectangular gaussian box with the left side at x < -d and the right at 0 < x < d. You know the E field at x < -d and you know the contained charge. Only E flux is out of the side at 0 < x < d. That gives you E(0<x<d).

    Part b is obvious if you figured out the E field at x < -d.
     
  5. Sep 18, 2014 #4

    collinsmark

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    It's even easier if you put one side of the Gaussian box at x = 0, invoking symmetry to determine the E at x = 0, then let the other side vary between 0 and an arbitrary x. But that way still works too. Either way. :smile:
     
  6. Sep 18, 2014 #5

    rude man

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    Quite so, cm. Quite so! :thumbs:
     
  7. Feb 9, 2016 #6
    Can someone help me to solve this question:
    .Three concentric hollow metallic spherical shells of radii r1,r2 and
    r3, where r1 < r2 < r3, carry charges +2Q,−3Q and +Q, respectively.
    Determine the charge on the inner and the outer surface of each sphere
     
  8. Feb 9, 2016 #7

    SammyS

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    It's best to start a new thread to post your question.

    When doing that, you should use the template that will be provided for you.
     
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