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Electric flux

  1. Mar 9, 2009 #1
    given a cube,each side with a lenth of A, placed at with its back left hand bottom corner at (0,0,0).
    find the electric flux passing through the sides if the electric field is
    a) E=(-3x)i + (2y)j + (4z)k
    b) E=(sin(pi*x/A)i + cos(pi*y/2A)j

    for a) since the area of each side is A2, flux=A2*E, since the cube is at the origin, anything on -x,-y,-z will have o value (for x,y,z) and therfore 0 E and so 0 flux


    flux from
    right= 2*A*A2 =2A3
    left= 2*0*A2 =0
    top= 4*A*A2 = 4A3
    bottom= 4*0*A2 =0
    front= -3*A*A2=-3A3
    back= -3*A*A2 =0

    for b) i cant see how they got the answers, the answers are all 0 other than ffor the left which they say is

    but as far as i can tell, the left flux is [cos(pi*y/2A)j * A2 ] cos(pi/2)=0 so i dont see how they come to this??

    hope i am doing this right, really not sure, any help appreciated
  2. jcsd
  3. Mar 9, 2009 #2


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    Not so fast. The E field is a vector field and you need to consider the flux passing through the surface.

    Φ = ∫ E*dA = ∫ ECosθdA = EA*Cosθ

    Now on each of the three faces from the origin the flux will lie in the plane and hence will have 0 contribution, since cos90 =0.
    (θ is angle taken with the ⊥ )

    But at each of the 3 faces at x,y,z from the origin what will the flux be over the area?
  4. Mar 9, 2009 #3
    dont reallyunderstand, could you pls try pput it into simpler terms, i dont study in english and dont understand what you mean by flux over the area?? where have i gone wrong? i realise that E is a vector
  5. Mar 9, 2009 #4
    but this E has x y and z coordinates
  6. Mar 9, 2009 #5


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    Maybe you have the solution, and I have missed your statement?

    If your answer is the sum of the fluxes that you found through the 3 outer faces 2A3, 4A3, -3A3, then that looks correct. I didn't understand from your solution if this is what you were saying at first.
    Last edited: Mar 9, 2009
  7. Mar 9, 2009 #6


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    Evaluating b) you have 0 through x-y faces because there is no z component.

    In the z-y, the x components evaluated at x=0 and x=A yield again 0 since sin(π ) = 0

    However for the z-x planes evaluated at y=0 and y=A, yields 0 for the Y=A (since cos(π/2) = 0), but cos(0) is 1 so that face will yield a flux over the area A2.
  8. Mar 10, 2009 #7
    i dont understand your workings? could you try step by step please.
    for b) why do you evaluate at y=0 and y=A, why is y=0 an option? is this the left hand side of the cube, which is at the origin?? also why is the flux -A^2? when cos0=1 and not -1
  9. Mar 10, 2009 #8
    i think i got it, since the vector of the field goes like j, (in direction of y+), the flux on the left side is -A^2, (since it goes into the left side and not away from it).
    since the normal to the side is going to y- and E is going to y+ so cos180=-1,

    is this right??
  10. Mar 10, 2009 #9
  11. Mar 10, 2009 #10


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    Yes. It is the left handed side along the origin, and since it is negative directed (into the surface) the flux is -A2
    Last edited: Mar 10, 2009
  12. Mar 10, 2009 #11
    but howcome for the answer to a), i got the minus from the equation, but here i had to use my logic, surely there must be some rule to follow?
  13. Mar 10, 2009 #12


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    In a) you weren't dealing with the complicating factor that you had a negative facing surface with the vector inwardly directed as in the fact that the field vector was positively directed, but inwardly as far as the Gaussian surface was concerned. Those faces had 0 flux crossing the face.

    The rule is flux out is positive. Hence for a point charge and a sphere you have outward vectors (positive charge) over all the surface in all directions, and the integral over the surface area accumulates all those positive field vectors to determine the positive charge inside.

    But in 2) here you had a surface that, while it had a positive field vector across the face, was not positive with respect to its direction crossing the face.
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