Continuity of the first Maxwell equation.

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SUMMARY

The discussion focuses on proving the continuity of the first Maxwell equation, specifically the divergence of the electric field, expressed as div(𝐸) = ρ/ε₀. The proof involves applying the divergence theorem, leading to the conclusion that the electric field must remain continuous across surfaces. A key point raised is the need to define the y-direction clearly, as only the parallel component of the electric field is continuous. The proof is contingent upon this clarification.

PREREQUISITES
  • Understanding of Maxwell's equations, particularly the first equation related to electric fields.
  • Familiarity with vector calculus and the divergence theorem.
  • Knowledge of electrostatics, including charge density (ρ) and permittivity (ε₀).
  • Ability to interpret vector components in three-dimensional space.
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  • Study the divergence theorem in detail to understand its applications in electromagnetism.
  • Explore the implications of electric field continuity at boundaries in electrostatics.
  • Review vector calculus, focusing on divergence and its physical interpretations.
  • Investigate the conditions under which electric fields are continuous across different media.
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Students of physics, particularly those studying electromagnetism, electrical engineers, and researchers interested in the mathematical foundations of electric fields.

Bert
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Suppose that we will proof the continuity of the first maxwell equation:

So we have [tex]div(\vec{E})=\frac{1}{\epsilon _0} \rho[/tex] than [tex]\iiint \ div(\vec{E}) = \oint_v \vec{E} d\vec{s}=\iiint \frac{1}{\epsilon _0 } \rho[/tex]
than follewed [tex]E_{y1} l -E_{y2}l=Q[/tex]

Therefore E must continue is this a good proof? Thanks.
 
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You have not defined what "y" direction is. So your y components are ambiguous.

Only the parallel component of the E-field (parallel to the surface) is continuous.

Zz.
 
Thanks if I define my y component as the parallel one. Is than the proof oké?
 

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