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Electric-charge density

  1. Sep 15, 2013 #1
    1. The problem statement, all variables and given/known data
    A space has an uniform electric field
    E=(5.00 x 10^3 N/C)[itex]\widehat{x}[/itex] + (6.00 x 10^3 N/C)[itex]\widehat{y}[/itex] + (7.00 x 10^3 N/C)[itex]\widehat{z}[/itex].
    Find the electric-charge density distribution p(r) in this space.


    2. Relevant equations
    u = (1/2)(εo)(E^2) , where εo is the constant called the permittivity of free space and = 8.85x10^-12 C^2/N*m^2

    I used the equation above, but I was unsure if I was using the correct equation. I thought I maybe should have used one of Maxwell's equations instead: div*E=(ρ)/εo


    3. The attempt at a solution
    magnitude of Electric Field = sqrt((5.00 x 10^3 N/C)2 + (6.00 x 10^3 N/C)2 + (7.00 x 10^3 N/C)2)

    magnitude of E = 10488.1 N/C

    u = (1/2)(8.85x10^-12 C^2/N*m^2)(10488.1 N/C)^2
    u = 4.87 x 10^-4 C/m^2

    Is this the correct way to solve this problem?
     
  2. jcsd
  3. Sep 15, 2013 #2

    Dick

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    No, u is an energy density. Use the Maxwell equation. What's the divergence of the constant electric field? You know what a divergence is, right?
     
  4. Sep 15, 2013 #3
    The divergence of a constant electric field is zero, right? I don't understand if the divergence is zero and the divergence is multiplied by the Electric field how you would get a proper answer.
     
  5. Sep 15, 2013 #4

    Dick

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    Yes, the divergence of a constant E field is zero. The divergence of the field isn't multiplied by the E field. Go back and review 'divergence'. Maxwell's equation is telling you an easy thing about the charge density. What is it?
     
    Last edited: Sep 15, 2013
  6. Sep 16, 2013 #5
    So the divergence of an E field is basically taking partial derivatives with respect to x, y, and z of the E field, but since there are no variables in the E field that means the derivative of the constant E field is zero. Then since the constant E field has a divergence of zero, then does that mean that there cannot be a charge density distribution because the E field doesn't act like a source or a sink, and does that also mean that the charge density is also constant? The concept of divergence is really confusing to me.
     
  7. Sep 16, 2013 #6

    Dick

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    The equation is saying that if the divergence of the E field is zero, then the charge density must be zero, not just a constant, isn't it?
     
  8. Sep 16, 2013 #7
    So if the E field is zero and the charge density is also zero, and then does it even matter what the E field has numbers? As long as the E field is constant then the charge density will always be zero.
     
  9. Sep 16, 2013 #8

    Dick

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    No, the numbers don't matter. Any constant field corresponds to zero charge density.
     
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