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Homework Help: Finding flux through surface where E field is function of x

  1. Sep 17, 2015 #1
    1. The problem statement, all variables and given/known data
    In a region of space there is an electric field E(vector E) that is in the x-direction and that has magnitude E=(663 N/(C*m))x

    Find flux for this field through a square in xy plane at z=0 and with side length 0.330m . One side of the square is along the +x-axis and another side is along the +y-axis.

    I understand setting up the integral but I do not understand why there is a y in the answer

    2. Relevant equations

    3. The attempt at a solution

    ΦE= ∫Ecos∅dA , where phi is the angle between the normal area vector and the E field. In this case E-field and A are parallel so cos∅=1 . Thus integral reduces to ΦE=∫EdA. dA of a small piece of the square is determined by dy*dx.

    Then, integral becomes ∫663x*dydx (from 0 to 0.330)

    Final answer should be

    (y*663x^2)/2 (from 0 to 0.330) so final answer is 11.91.

    I understand anti derivative of x is x^2/2 but I am not understanding/seeing where the y factor comes from.

  2. jcsd
  3. Sep 17, 2015 #2
    You have to integrate with respect to [itex]y[/itex] first. Since the term [itex]663x[/itex] does not depend on [itex]y[/itex], you can pull it out, and you have to deal with [itex]\int \ dy[/itex], which is...?
  4. Sep 17, 2015 #3

    ok so bc A=dxdy and that's part of the integral and like you said 663x does not depend on y you have 2 integrals ∫dy=y and ∫663x= 663*(x^2)/2 (from 0 to 0.330). Ok got it. Thank you for the clarification
  5. Sep 17, 2015 #4

    rude man

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    You sure the problem doesn't say the E field is in the z direction?
  6. Sep 17, 2015 #5
    Hello, yes I am sorry the problem statement is in z-direction and I solved it with the z-direction in mind but I failed to revise my problem statement submission here. thanks for pointing that out
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