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Flux through a cube with non uniform electric field

  1. Oct 2, 2004 #1
    This is really frusterating me, my book provides horrible examples and i have no idea how to go about this problem.
    There is a cube with sides L= .3m and an electric field = (-5 N/C X m) x i +(3 N/C x m) z k i= i hat k= k hat
    I know that the flux = the integral of the E . dA (dot product), but the form that i have for the electric field is not making sense to me.
    How can i find the electric flux through the faces of a cube when i am given the electric field mentioned above?
     
  2. jcsd
  3. Oct 2, 2004 #2

    pervect

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    I'm guessing a bit, but it sounds like the electric field at a point (x,y,z) is

    -5 x newtons/columb-meter in the [tex]\hat i[/tex] direction, and
    3 z newtons/columb-meter in the [tex] \hat k[/tex] direction.

    Usually [tex] \hat i[/tex] points in the x direction, [tex]\hat j[/tex] points in the y direction, and [tex]\hat k[/tex] points in the z direction. Does this sound right?

    If so, it shouldn't be too hard to integrate the flux through a cube face....
     
  4. Oct 3, 2004 #3
    i'm still lost
     
  5. Oct 3, 2004 #4
    What pervect is trying to say in fewer words than I'm about to say is you have the field given by -5x + 3z. The flux is the integral of E dot dA right? However, since dA does not change on each face throughout the face E dot dA becomes E*A for each face. Sum the flux on each face up. I'll help you out with a little of the math:

    The cube has one face facing the positive and negative directions for x, y and z. For example, the flux on the face pointing in the positive/negative x direction is <-5x, 0, 3z> * (0.3^2). The only difference is the value of x for each face, where <i, j, k> is a vector. Now sum up all 6 faces (you can ignore the ones facing the +y and -y directions)
     
    Last edited by a moderator: Oct 3, 2004
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