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Electric field Q

  1. Sep 24, 2005 #1
    Seems quite strange that i cant solve simple questions like this... but on the other hand applying Gauss Law to cylinders spheres and such isnt that hard for me... ANYWAY
    One of the cube's edges is placed at the origin.
    A cube with edges 1.4m is oriented as shown in the diagram. Find the flux and the charge inside the cube when the electric field is
    a) (3 Nm/c) y j
    b) (-4N/C)i + [6N/C + (3Nm/C)y] j

    for a) since there are TWO faces in the X only. THe Y and Z componenets are orthogonal to the field thus are zero. [tex] \Phi = 2 \int E \bullet dA = \frac{q_{enc}}{\epsilon_{0}}[/tex]

    [tex] \Phi = 2 \int 3y [d(y^2)] = 2 \int 6y^2 dy = 4y^3 [/tex]
    since y = 1.4m flux is 10.976 Nm^2/C
    the charge in teh cube as a result is [tex] 10.976=\frac{q_{enc}}{\epsilon_{0}} [/tex]

    for b), since there are two faces for the X and Y sides
    in the case of the X the field is constnat so [itex]\int E dA = EA[/itex]
    [tex] \Phi = 12(1.4)^2 + 2 \int (6+3y)2y dy [/tex]
    blah blah blah leads to
    [tex] \Phi = 12(1.4)^2 + 2 [6(1.4)^2 + 2(1.4)^3] [/tex]
    and plugging that equal to Qenc/epsilon yeilds the induced charge

    Am i right? If i am wrong please do tell! Thank you so much!

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    Last edited: Sep 24, 2005
  2. jcsd
  3. Sep 24, 2005 #2


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    Homework Helper

    The way I read #1(a), the Field is parallel
    to the y-axis (you use the j_vec),
    which would pierce the Y-faces
    except that one of these is at y = 0.

    E_vec must *pierce* the surface area,
    so you want dA = dxdz
    ..(dy is parallel dy, so can't make an area)
    take E dot (dx cross dz)
  4. Sep 25, 2005 #3
    im not quite sure about what you just said...
    but for the E field in 1a isnt the EdA cos0?
    And the other side EA cos 180??
    could you explain more in depth?
  5. Sep 25, 2005 #4


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    yes, the dot product gives EdAcos0,
    at the y = 1.4 [m] face ...
    where A = Dx * Dz (not the Dy^2 you were using).

    For the y = 0 face , well y=0 so E=0.

    The E-field at each face is different, so you can't just multiply the integral by 2.
    And the dA is nothing at all like d(y^2) .

    You make the same 2 mistakes in part (b),
    where you conclude that the uniform Ex-component
    indicates the presence of a lot of charge in the box.
    uniform E means zero charge in the region;
    charges cause E-field lines to diverge.
  6. Sep 26, 2005 #5
    ok so for hte first one then
    [tex] \Phi = \int E \bullet dA = \int E \bullet dx dz [/tex]
    the area vector is the cross product of these two vectors
    Can be a scalar?? So A = 1.4^2
    or should it be in vector form?
  7. Sep 26, 2005 #6
    so dx cross dz will give me dy
    so dx = (1.4,0,0)
    and dz = (0,0,1.4)
    is this correct
    because the cross product of these two would yield (0,1.96,0) is this correct?

    anbd then i dot this with the electric field? SO the integral sign goes away? Since ifound a numerical value for dy?
    so the flux is 3(1.96)(1.4)?? AM i on the right track?
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