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Electric Flux thru the top face of a cube

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data
    An electric field given by [tex]\vec E = 4.0 \hat i - 3.0(y^2+2) \hat j[/tex] pierces a 2.0 meter by 2.0 meter by 2.0 meter cube. What is the electric flux thru the top face?


    2. Relevant equations
    [tex]\phi = \int \vec E \cdot d \vec A[/tex]


    3. The attempt at a solution
    I'm aware that you can take a shortcut to solving this problem by considering that the electric field at y = 2.0 is a constant [tex]\vec E = 4.0 \hat i - 3.0([2.0]^2+2) \hat j = 4.0 \hat i - 18.0 \hat j[/tex].

    then this leads to, [tex]\phi = \int \vec E \cdot d \vec A = \vec E \int d \vec A = \vec E \cdot \vec A[/tex]

    And since [tex]\vec A[/tex] has no x-component, we only need to consider the y component. The area of the top face is just 2 times 2.

    [tex]\phi = -18.0 * (2^2) = -72 \frac{Nm^2}{C}[/tex]

    ---
    But how do I solve this same problem without taking the shortcut?

    2. Relevant equations???
    Evaluating a flux integral (surface is oriented upward):
    [tex]\int_S \int \vec F \cdot \vec N dS[/tex]

    The unit vector normal to the surface:
    [tex]\hat N dS = \frac{\bigtriangledown g(x,y,z)}{||\bigtriangledown g(x,y,z)||} dS = \bigtriangledown g(x,y,z) dA[/tex]

    4. The attempt at a solution without the shortcut
    The surface is probably given by [tex]z = g(x, y) = y - 2[/tex]
    [tex]\hat N dS= dA \hat j[/tex]
    [tex]\phi = \int_S \int \vec E \cdot \vec N dA = \int_R \int [4.0 \hat i - 3.0(y^2+2) \hat j] \cdot [dA \hat j] = \int_R \int -3.0(y^2+2)dA = -3.0 \int_0^2 \int_0^2 (y^2+2) dxdy = -40.0[/tex]

    :( this is obviously the wrong answer, any clue on what i'm doing wrong?

    Thanks in advance!
     
    Last edited: Jan 26, 2013
  2. jcsd
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