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Gauss' Law

  1. Sep 26, 2005 #1
    I know i'm reposting this... but no one seems interested in my thread.. maybe becuase it already has replies (which doesn't mean its been solved!) but whatever

    cube is placed with its edge at (0,0,0) as in the diagram. The sides of the cube are1 .4m. Find the electric flux and the charge inside the cube if the electric field is given by
    a)3yN/C j
    b) -4N/C i + [6N/C + (3Nm/C)y]j

    for a) [tex] \Phi = \int E \bullet dA \hat{j} [/tex]
    [tex] \Phi = \int (3y)(dA) \hat{j} \bullet \hat{j} [/tex]
    [tex] \Phi = 3 \int ydA [/tex]
    but y is constant and A is constant soo
    [tex] \Phi = 3 (1.4)^3 [/tex]
    taht gives it for ONE face. But what about hte other face of the cube? I dont want to proceed with the other question since it is similar to this. ANYWAY., if you could help me it would be really awesome

    thank you for your help and advice!
     

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    Last edited: Sep 26, 2005
  2. jcsd
  3. Sep 26, 2005 #2
    No worries, you're "fluxation" worries will be taken care of. Just as an inital though, if a field vector is constant for the entire domain, chances are pretty good the the net flux is zero throughout the gaussian surface. IN order to investigate though, you need to examine the flux at each face. since you have the dimensions of the cube and the orientation of the electric vector field, then evaluate the flux at each surface. If you're numbers show a NET FLUX of zero, then your in good shape with the general analysis. But if not and/or you're not sure about your numbers, post up what you have done and either myself or some other brave sould will take a look at it.
     
  4. Sep 26, 2005 #3
    but in ths case the electric field is not consatnt its 3y, poiting in the Y direction... Are you sure that the electrifc field would be zero then?
     
  5. Sep 26, 2005 #4
    good call, I was wondering that as well. PLease clarifyand leave nothing out. That value of Y (constant or not) could affect the whole problem.
     
  6. Sep 26, 2005 #5
    i posted the question as it appeared in teh book.

    Is the flux really zero through this cube? The field is non uniform...
     
  7. Sep 27, 2005 #6
    soc na nayone help me with the qeustion?

    the field is not zero.. since the field is not uniform, or doesi t work out to be zero? Please advise!
     
  8. Sep 27, 2005 #7

    Doc Al

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    You'd find the flux through the other relevant side in exactly the same way. (Of course, it's trivial since y = 0 for the other side.) Since the field is only in the [itex]\hat{j}[/itex] direction, only the two sides perpendicular to the y-axis could have any flux (the sides at y = 1.4m, and y = 0).

    What makes you think the total flux is zero?
     
  9. Sep 27, 2005 #8
    so would the electric flux of the other side be the negative value since the angle between E and dA on that side is pi?
    THIS is whats making me think the flux is zero... but i cant believe that since the field is non uniform...

    for the other side
    [tex] \Phi = \int E \bullet dA = \int 3\hat{j} dA\hat{j} cos 180 [/tex]
    i dont think this is right...
     
  10. Sep 27, 2005 #9

    Doc Al

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    Before you worry about the angle that the field makes with dA, first tell me the value of the field on the side y = 0.
     
  11. Sep 27, 2005 #10
    the value is the same as the other side because the same calculation from #1 follows, does it not? that is 3 (1.4)^3?
     
  12. Sep 27, 2005 #11

    Doc Al

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    No. The field is given as 3yN/C j; it depends on y! (The same method is used to calculate the flux for each side, but you have to use the appropriate value of y--which is different for each side.)
     
  13. Sep 27, 2005 #12
    on the left the y value is zero
    on the right side the value is 1.4
    so what happens to the dA however...?
     
  14. Sep 27, 2005 #13

    Doc Al

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    Who cares what happens to dA? If the y value is zero, what is the value of the electric field?
     
  15. Sep 27, 2005 #14
    the value of the elctric field is then zero if y is zero
    right?
     
  16. Sep 27, 2005 #15

    Doc Al

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    Right! So the flux through that side is zero.
     
  17. Sep 27, 2005 #16
    awesome

    now for the other side... is what i have correct?
     
  18. Sep 27, 2005 #17

    Doc Al

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    If you are asking about the flux you caculated through the side with y = 1.4m (that you posted in #1): yes, that looks good.
     
  19. Sep 27, 2005 #18
    so the total flux for the first question that is part a is 3(1.4)^3 nm^2/C and thats that? To find the induced charge... i simply use this [tex] \Phi = \frac{q}{epsilon_{0}} [/tex]

    THank you for the help so far!

    I will solve part b and post my solution up in a little while.

    Once again.. thank you for the help!
     
  20. Sep 27, 2005 #19

    Doc Al

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    That's all there is to it.
     
  21. Sep 27, 2005 #20
    ok for the second part where [itex] E = (-4N/C)\hat{i} + [6N/C + (3N/Cm)y]\hat{j} [/itex]
    ok since the field is uniform for the -4N/C i part and the 6N/C j part... the flux is zero through the x=0 and x=1.4 sides and the y sides which are affected by the 6N/C field

    however for the 3N/cm y j field - for the y=0 the flux is zero, but the y=1.4m the flux is
    [tex] \Phi = \int 3y dA = 3 \int y dA [/tex] since y is constnat and A is also constnat and y = 1.4 and A = 1.4^2
    [tex] \Phi = 3(1.4)^3 [/tex]
    Am i good now? Did i get it right?
    So the only flux is 3(1.4)^3 and the other components of the field have no effect on the flux in the cube?
     
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