Electric flux through cubical box

  • #1
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Homework Statement


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.

Homework Equations


∫EdA = flux

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?
 

Answers and Replies

  • #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.
 
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  • #3
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thanks. yeah it seems like a weird question. Would it be like the field between two parallel plates of opposite charge?
 
  • #4
rude man
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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.
ε∇⋅E = ρ.
 
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  • #6
rude man
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What does this mean?
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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