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Uniform Slab-Finding Electric Field Using Gauss Law

  1. Mar 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Uniform Slab: Consider an infinite slab of charge with thickness 2a. We choose the origin inside the slab at an equal distance from both faces (so that the faces of the slab are at z = +a and z = −a). The charge density ρ inside the slab is uniform (i.e., ρ =const). Consider a point with coordinates (x,y,z). Using Gauss’ law, find the electric field
    (a) when the point is inside the slab (−a < z < +a),
    (b) and when the point is outside the slab (z > a or z < −a).
    (c) Sketch the Ez vs z graph.
    (d) If the density was not constant at its a function of z like ##ρ=Bz^2## then calculate the upper steps again.

    2. Relevant equations
    Gauss Law

    3. The attempt at a solution
    a) I took a cylinder Gaussian surface inside the slab forand from that I found ##E=\frac {ρz} {2ε_0}## .z is the height of the point that we choose from the origin.
    b)I took a cylinder again and from that I found ##E=\frac {ρa} {2ε_0}##
    c)The field will be constant cause ρ and a is constant also ##ε_0## so As z increases it inrease until a.And from that its constant.
    d)Then Electric field will be ##E=\frac {Bz^3} {6ε_0}## for inside , ##E=\frac {Ba^3} {6ε_0}## for outside ?
    Is these true ?
     
    Last edited: Mar 11, 2017
  2. jcsd
  3. Mar 11, 2017 #2

    TSny

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    You have the right approach. But your answer is off by a numerical factor. Can you show in more detail how you got your result? In particular, how was your cylindrical Gaussian surface oriented and positioned?
     
  4. Mar 11, 2017 #3
    I took a cylinder like height is z and radius is x.So ##Eπx^2=Q/ε_0## ##Q=ρπx^2z## But theres two sides so we should multiply this by 2 so ##2Eπx^2=\frac {Q} {ε_0}## and then ##E=\frac {pz} {2ε_0}##
     
  5. Mar 11, 2017 #4
    Oh wait
     
  6. Mar 11, 2017 #5
    Height is ##2z##
     
  7. Mar 11, 2017 #6
    ##E=\frac {ρz} {ε_0}## ?
     
  8. Mar 11, 2017 #7

    TSny

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    Yes.
     
  9. Mar 11, 2017 #8
    Then (b) is ##E=\frac {pa} {ε_0}##
    (c) will be the same (it increase until ##z=a## then its constant)
    for (d) inside ##E=\frac {Bz^3} {3ε_0}## and outside ##E=\frac {Ba^3} {3ε_0}## and the graph again rises at z rises then its constant ??
     
  10. Mar 11, 2017 #9

    TSny

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    Looks good. Make sure your graphs include negative values of z as well as positive values of z. So you will need to think about the sign of Ez for negative z.
     
  11. Mar 11, 2017 #10
    I see thanks
     
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