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Electric fields 2

  1. Jul 4, 2009 #1
    1. The problem statement, all variables and given/known data
    A point charge +Q is placed at one of the corners of a cube as shown in the figure.The electric flux through the surface ABCD of the cube due to the charge is
    (1) Q (or Q/[tex]\epsilon_{0}[/tex])
    (2) Q/4 (or Q/4[tex]\epsilon_{0}[/tex])
    (3) Q/6 (or Q/6[tex]\epsilon_{0}[/tex])
    (4) Q/24 (or Q/24[tex]\epsilon_{0}[/tex])
    (5) Q/36 (or Q/36[tex]\epsilon_{0}[/tex])

    2. Relevant equations

    Electric field intensity E=(1/4[tex]\pi[/tex][tex]\epsilon_{0})[/tex]Q/x2

    3. The attempt at a solution

    I really don't know how to do this.I only know the above equation,but I don't know how to apply it here.
    I think a -Q charge would be induced at A,a +Q charge at B,a - Q charge at C and a +Q charge at D,but where do I go from here?

    Thanks in advance.
     

    Attached Files:

  2. jcsd
  3. Jul 4, 2009 #2

    LowlyPion

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    I can't see the picture, but if the charge is right at the corner, then the total flux of the E-Field will be cut into what ... 8 equal pieces? (Think of stacking 8 cubes with the point charge at the common vertex.)

    Now since from Gauss Law you know the total charge of an enclosed surface is Q/εo, then you know that since the cubes are all the same, that 1/8 of the entire field is passing through each cube.

    But think about it for a moment and you will see that each cube has 3 external faces (faces opposite the vertex with the charge) and since each face is identical then the flux through any one of them is ...
     
  4. Jul 4, 2009 #3
    I don't know.
    Could you please tell me how you derived this?
    I know that,according to Gauss' theorem, the total normal electric flux intersecting a closed surface,of any shape,enclosing a charge Q=Q,
    i.e,
    A[tex]\epsilon_{0}E[/tex] = Q
    A is the area enclosed by a charge Q placed at the centre.

    I don't undrestand this part,how does the total electric flux divide into 8?

    This is the picture,
    http://img12.imageshack.us/img12/2296/chargen.png [Broken]

    THANK YOU.
     
    Last edited by a moderator: May 4, 2017
  5. Jul 4, 2009 #4

    LowlyPion

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    Imagine 7 other identical cubes arranged all together in a 2 X 2 X 2.

    The center vertex will have the charge. (The central vertex is the only spot that all 8 cubes would have in common.) This is where the charge is on your one cube. At a corner.

    Since by Gauss Law you know that all the flux through the surface of the 2x2x2 arrangement is Q/εo, then the total of the flux escaping the outward faces of the one cube must be 1/8th of the total.

    But they want the flux through just 1 face. A cube has 6 faces. But in the 2x2x2 arrangement you should satisfy yourself that only 3 of the faces are on the outside. Hence the flux through any one face must be 1/3 of what passes out of that cube.

    Here's a link for Gauss Law:
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c1
     
  6. Jul 4, 2009 #5
    Thank you so much for the detailed explanation.
    so the the electric flux through the face ABCD,would be,
    [1/(8*3)] Q/[tex]\epsilon_{0}[/tex],which would be answer no. (4).
     
  7. Jul 4, 2009 #6

    LowlyPion

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    Yes. That looks like the answer.

    But the understanding is worth much more than just the right answer.

    Good Luck.
     
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