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Finding electric flux through a surface, simple problem, whats wrong?

  1. Sep 18, 2005 #1
    At each point on the surface of the cube shown in Figure 23-27, the electric field is parallel to the z axis. The length of each edge of the cube is 4.0 m. On the top face of the cube E = -40 k N/C, and on the bottom face of the cube E = +29 k N/C. Determine the net charge contained within the cube.
    Image: here

    Flux = (29E3 N/C)(4.0)^2;
    Flux = 464000 k going up
    Flux = (-40E3 N/C)(4.0)^2;
    Flux = -640000 k going down
    Net Flux = -640000 + 464000 = -176000

    I submitted it and was inncorrect,
    So i tried added them together and i got 1.104E6, thats the flux, now I found an equation that says, Flux = Qtotal/Eo, so I took Flux*Eo = Qtotal and got 9.7704E-6 and it is still wrong!
     
    Last edited: Sep 18, 2005
  2. jcsd
  3. Sep 18, 2005 #2

    Doc Al

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    Staff: Mentor

    When finding the flux through a surface, think in terms of inward (-) and outward (+).
    That should be negative, since it points inward.
    This is also inward, so negative.
    Redo this, since both are negative.

    Signs matter! The flux is negative.
     
  4. Sep 18, 2005 #3
    So i got -1.104E-6 for the Flux. So Q = -1.104E-6*8.85E-12 = -9.7704E-6? I have only one more chance so i have to make sure i'm doing this one right, thank you!
     
  5. Sep 18, 2005 #4

    Doc Al

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    Well... does that follow from what I said? And do you think what I said is accurate? :wink:
     
  6. Sep 18, 2005 #5
    ahh it was wrong, what did i mess up?
     
  7. Sep 18, 2005 #6

    Doc Al

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    Staff: Mentor

    Nothing that I can see. But those on-line systems can be picky. (Units, significant figures, etc.)
     
  8. Sep 18, 2005 #7
    yeah i hate this system, i'll ask the professor tomarrow whats up, thanks!
     
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