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

  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.

    Attached Files:

    • 222.jpg
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  2. jcsd
  3. HallsofIvy

    HallsofIvy 41,034
    Staff Emeritus
    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. 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. Pyrrhus

    Pyrrhus 2,205
    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. Doc Al

    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. 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. Pyrrhus

    Pyrrhus 2,205
    Homework Helper

    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. 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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