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A question from electrostatics

  1. Aug 19, 2011 #1
    The displacement vector in a cube is defined by the relation D =(4x*xy + z)i+(4xy)j-(z)k c/m2 . Then charge enclosed within the cube of side 1m is …………..
    I know that the displacement vector holds the relation
    ∇ . D = ρ
    Where ρ is the charge density. I found ∇ . D and calculated its volume integral with x,y and z varying between 0 and 1. What I got was 3 coloumb but the correct answer is 4 colomb. Can someone help me with this?
     
  2. jcsd
  3. Aug 19, 2011 #2

    vanhees71

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    2016 Award

    It's not exactly clear to me, how your [itex]\vec{D}[/itex] reads. Please use the LaTeX formulas. I read it as follows

    [tex]\vec{D}=\begin{pmatrix}
    4 x^2 y +z\\
    4 x y \\
    z
    \end{pmatrix}
    [/tex]

    Then

    [tex]\vec{\nabla} \cdot \vec{D}=4xy+4x+1.[/tex]

    The next question is, where your cube is located. I used Mathematica to calculate for your cube, which you give as [itex][0,1] \times [0,1] \times [0,1][/itex]. Then I get 5 (whatever units you use, since these are also inconsistent).
     
  4. Aug 19, 2011 #3
    sir, actually D is having -z as k component. ∇ . D = 8xy + 4x -1 then.(u hav written it as 4xy+4x+1). nothing is mentioned in the question about the location of the cube. i didn't understand your [0 X 1] technique sir. can you please explain?
     
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