Electric flux through cubical box

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SUMMARY

The discussion focuses on calculating the electric flux through a cubical box when the electric field is constant in direction but varies in magnitude from E1 to E2. The correct approach involves using the equation ∫EdA, leading to the expression L²(E_L - E_0) for the flux. Participants also explore the implications of such an electric field configuration, suggesting a relationship to fields between parallel plates of opposite charge. Additionally, the divergence of the electric field relates to charge density, as described by one of Maxwell's equations.

PREREQUISITES
  • Understanding of electric flux and its calculation using integrals
  • Familiarity with Maxwell's equations, particularly the relationship between electric fields and charge density
  • Knowledge of electric fields, including constant and variable fields
  • Basic concepts of electrostatics, including parallel plate capacitors
NEXT STEPS
  • Study the derivation and application of Maxwell's equations in electrostatics
  • Learn about electric flux calculations in varying electric fields
  • Explore the concept of charge distribution in relation to electric fields
  • Investigate the behavior of electric fields between parallel plates and their applications
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone interested in the practical applications of electric fields and flux calculations.

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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