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Electric flux through cubical box

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data
    If the electric field is constant in direction (horizontal in the x direction) but its magnitude decreases from E1 to E2, determine the flux through a cubical box of side length L if four of the sides of the box are parallel to the field.

    2. Relevant equations
    ∫EdA = flux

    3. The attempt at a solution
    [itex] \int_0^L EdA = EA|_0^L = E_LA-E_0A = A(E_L-E_0) = L^2(E_L-E_0) [/itex]

    is this reasoning correct?
     
  2. jcsd
  3. Jan 31, 2015 #2

    BvU

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    Can't find a flaw in your working. The exercise itself leaves me wondering how one could bring about such a field.
     
  4. Jan 31, 2015 #3
    thanks. yeah it seems like a weird question. Would it be like the field between two parallel plates of opposite charge?
     
  5. Jan 31, 2015 #4

    rude man

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    ε∇⋅E = ρ.
     
  6. Feb 1, 2015 #5
    What does this mean?
     
  7. Feb 1, 2015 #6

    rude man

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    epsilon times the divergence of the E field equals the charge density (aka one of Maxwell's 4 equations). Solving that with your given E field would give you the charge distribution needed to effect the given E field. It was in answer to BvU who wondered how an E field like your given one could be produced. You could solve for ρ(x) if it amused you.


    epsilon = dielectric constant
     
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