Electric flux through cubical box

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Homework Help Overview

The discussion revolves around calculating the electric flux through a cubical box when the electric field is constant in direction but varies in magnitude. The problem involves understanding the implications of the electric field's behavior and its relation to the geometry of the box.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the original poster's reasoning regarding the calculation of electric flux and question the feasibility of creating such an electric field. There are inquiries about the relationship between the electric field and charge distribution.

Discussion Status

Some participants express agreement with the original poster's calculations, while others raise questions about the nature of the electric field described in the problem. There is an exploration of related concepts, such as the charge distribution necessary to produce the given electric field.

Contextual Notes

Participants note the unusual nature of the problem and the implications of the electric field's characteristics, suggesting that it may relate to fields generated by charged plates. There is also mention of Maxwell's equations in the context of the discussion.

toothpaste666
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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


\int_0^L EdA = EA|_0^L = E_LA-E_0A = A(E_L-E_0) = L^2(E_L-E_0)

is this reasoning correct?
 
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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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thanks. yeah it seems like a weird question. Would it be like the field between two parallel plates of opposite charge?
 
BvU said:
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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rude man said:
ε∇⋅E = ρ.
What does this mean?
 
toothpaste666 said:
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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