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Electric flux through gaussian cube

  1. Sep 30, 2007 #1
    1. The problem statement, all variables and given/known data

    I have a cube with four faces parallel to the field and two perpendicular.

    The field is non uniform, given by E = 3 + 2x^2 in the +x direction.

    3. The attempt at a solution

    So I get [tex]\phi = \int \vec{E} \cdot d\vec{A} = \int EdA = A \int E = A \int (3+2x^2)[/tex]

    But how do I evaluate this with no dx?
     
    Last edited: Sep 30, 2007
  2. jcsd
  3. Sep 30, 2007 #2

    robphy

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    Slow down... you need to justify each of these steps.
    First, a cube has six faces.
    Second, what precisely do you mean by the expression after the second equals-sign (and how do you justify it?) [Hint: it's wrong.]
    [tex]\phi = \int \vec{E} \cdot d\vec{A} = \int EA[/tex]
    For each integral, what is the region of integration?
     
  4. Sep 30, 2007 #3
    Sorry, I meant to say four parallel faces.

    and I just realized that the perpendicular faces would have constant E over them since all points on both faces are at the same x value.

    I said [tex]\int \vec{E}\cdot d\vec{A} = \int EdA [/tex]because the angle between E and A vectors is 0

    Would this make sense?
    [tex]\int \vec{E}\cdot d\vec{A} = \int EdA = E \int dA = EA [/tex] and I'm given the dimensions of the cube as well as it's position along the x axis.
     
    Last edited: Sep 30, 2007
  5. Sep 30, 2007 #4

    robphy

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    Technically, you have an integral over the closed surface of the cube... and by breaking the surface up conveniently into six faces, you have the sum of six fluxes through each face (an open surface), four of them being zero. So, what you write actually refers to one of the two remaining faces (which have the same form, but not same value). Note that you need to take the directions of the area-vectors (taken from the original cube) into account.
     
  6. Sep 30, 2007 #5
    So I basically get +EA and -EA because of the antiparallel area vectors for the two faces?
     
  7. Sep 30, 2007 #6

    robphy

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    Yes, although each E is "E on that surface used to calculate the flux".
     
  8. Sep 30, 2007 #7
    Alright, thanks a lot.
     
  9. Sep 16, 2011 #8
    Hi, i am having the same problem. But when you do the calculation of the remaining two faces, will they be zero since the area vectors opposite to each other?

    I know this is a non uniform field. And i was thinking perhaps the field can move in any directions even in the negative x,y,z directions. If i were to calculate the net charged enclosed, i should calculate the e-flux first the use EA = q/Epsilon right?
     
  10. Sep 16, 2011 #9
    In addition my cube sits at a distance 'a' from the x-axis. Is there any integration involved? This part confused me the most
     
    Last edited: Sep 16, 2011
  11. Sep 16, 2011 #10
    Since the field is [3(N/C)+4(N/C m^2)x^2]i^, and x is in meters. So the field's magnitude should be 7 N/C right?

    [​IMG]

    a=b=0.612m
    c=0.7965m

    What shall i do next?
     
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