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Homework Help: Finding the electric flux through the right face, confused on integration!

  1. Jul 3, 2005 #1
    I'm having troubles understanding whats going on here, with the integration. Here is the integral through the right face of the cube.
    I don't know how to insert all the fancey symbols, so here is my key:
    S = integral symbol
    Flux = electric flux symbol, omega or somthing, a circle with a cross down the middle.
    i = vector i in x-axis
    j = vector j in y-axis
    . means the dot product.
    Given: A nonuniform electric feild given by E = 3.0xi + 3.0j pierces the gaussian cube. x = 3.0m.

    Flux = S (E).(dA) = S (3.0xi + 4.0j).(dAi)

    = S [(3.0x)(dA)i.i + (4.0)(dA)j.i] //whats goin on here? are they just distrubting the dA? Why are they allowed to sperate the vector i from dA?

    = S (3.0x dA + 0) = 3.0 S x dA //why is i now 0? wouldn't it be cos(0) = 1? or how do u figure out where the electric feild is pointing with the equation: 3.0xi + 4.0j.

    = 3.0 S (3.0)dA = 9.0 S dA.

    How do you insert symbolic symbols so my future posts won't looks this messy? Thanks. Picture is attached.
     

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  3. Jul 3, 2005 #2

    HallsofIvy

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    Science Advisor

    According to your attachment, the "right face" of the cube is the plane x= 3.0 and the (outward) unit normal is i so the dA= dydz i. Therefore
    (3.0xi+ 4.0j). dA= 3.0x dydz= 3.0 x dA where dA= dydz.

    i did not become "0" the dot product of two vectors is a scalar (number).
    (3.0xi).(i)= 3.0x, of course.
     
  4. Jul 4, 2005 #3
    Thanks for the responce but i'm still confused.... how do you go from, dA = dydz i.
    then you said dA = 3.0x
    dydz = 3.0 x dA.....You didn't take the derivative of anything did you?
    ^is this the variable x or meaning multiplcation?
     
  5. Jul 4, 2005 #4

    Pyrrhus

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    Homework Helper

    Halls, simply did the dot product, the result was 3x dA, then if you look at the picture x = 3, so 9*A, should be the solution.
     
  6. Jul 4, 2005 #5

    Doc Al

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    Staff: Mentor

    The only component of the field that contributes to the flux through a side is the component perpendicular to that side. For the right side of the cube, that perpendicular direction is the [itex]\hat i [/itex] direction. The component of the field in that direction is [itex]3.0 x \hat i[/itex]; at x = 3 m, that component equals [itex]9.0 \hat i[/itex] (in units of N/C). Since the field is constant over the area of the right side, no integration is needed, just flux = E times Area.
     
  7. Jul 4, 2005 #6
    ohhh i think i finally get it... so because the y component of the electric feild doesn't matter (4.0j), you can just discard it and only worry about the 3.0xi. and because x = 3, you end up with 9.0i. So really is i just telling the direction of the vector? you can just discard it? I'm still confused on one issue though. [itex] \zeta [(3.0x)(dA)\hat i \bullet \hat i][/itex] You said you took the dot product, if A is pointing to the right, and also the electric feild is point right, wouldn't that be cos(0) = 1? how did they get 0? [itex] \zeta [(3.0x)(dA) + 0][/itex] Sorry i'm really really rusty on vectors! :bugeye: that zeta is suppose to be an integral sign, i can't find the integral on the latex guide.
     
  8. Jul 4, 2005 #7

    Pyrrhus

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    It looks like you don't know this:

    [tex] \vec{i} \cdot \vec{i} = \vec{j} \cdot \vec{j} = \vec{k} \cdot \vec{k} = 1 [/tex]

    [tex] \vec{i} \cdot \vec{j} = \vec{j} \cdot \vec{k} = \vec{i} \cdot \vec{k} = 0 [/tex]

    Ah and the integral is

    [tex] \int [/tex]
     
    Last edited: Jul 4, 2005
  9. Jul 4, 2005 #8
    ahhh! thanks so much, I had no idea that property even existed. Damn luckly i'm not going to be a mechanical engineeer.
     
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