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The cube box

  1. Sep 21, 2013 #1
    A cubic box of side a = 0.410 m is placed so that its edges are parallel to the coordinate axes, as shown in the figure. There is NO net electric charge inside the box, but the space in and around the box is filled with a nonuniform electric field of the following form: E(x,y,z) = Kz j + Ky k, where K = 3.40 N/(Cm) is a constant. What is the electric flux through the top face of the box? (The top face of the box is the face where z = a. Remember that we define positive flux pointing out of the box.)

    Flux= ∫E*dA
    *= dot product

    I started my integral off with ∫(3.40j + 3.40k)*dxdy and worked it down to:
    ∫∫3.40dxdy where the boundaries of both integrals go from 0-.410. My final answer was .571 N*m^2/C but it was wrong. Where did I mess up along the way? Am I missing a key part of the question?
     
  2. jcsd
  3. Sep 21, 2013 #2

    TSny

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    Note that the electric field E(x,y,z) = Kz j + Ky k has factors of z and y which you seem to have left out when you set up your flux integral.
     
  4. Sep 22, 2013 #3
    Are the factors of z and y the lengths of the sides of the cube? Or are they factors of the electric field in the z and y direction?
     
  5. Sep 22, 2013 #4

    TSny

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    z and y represent the z and y coordinates of the point where you want to evaluate the electric field. For example, if for some reason you wanted to find the field at (x,y,z) = (5, 6, 7), then you would let y = 6 and z = 7 in the formula for E.
     
  6. Sep 22, 2013 #5
    Ohhhhh ok. Thank you!
     
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