Electrodynamics: divergence of E in empty space

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Discussion Overview

The discussion revolves around the physical significance of the equation ∇·E = 0 in the context of electrodynamics, particularly in free space. Participants explore its implications regarding the presence of charges and its relation to Maxwell's equations.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • Some participants assert that ∇·E = 0 indicates the absence of charges in free space.
  • Others reference LaPlace's equation in relation to the discussion, questioning the connection to Maxwell's equations.
  • One participant argues that the conclusion that there are no charges arises from the assumption that free charges cannot exist in a vacuum, suggesting a potential flaw in reasoning.
  • Another participant claims that the converse is also true, stating that the absence of charge density (ρ = 0) implies ∇·E = 0 and vice versa.

Areas of Agreement / Disagreement

Participants express differing views on the implications of ∇·E = 0 and whether the reasoning regarding the absence of charges is valid. The discussion remains unresolved with competing interpretations of the relationship between charge density and the divergence of the electric field.

Contextual Notes

Participants do not fully resolve the implications of the assumptions made regarding charges in free space and the conditions under which ∇·E = 0 holds.

Flying_Dutchman
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What is the physical significance of fundamental law del.E=0 in free space ?
 
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Flying_Dutchman said:
What is the physical significance of fundamental law del.E=0 in free space ?
It means there are no charges in free space.
 
Dale said:
It means there are no charges in free space.
del.E=P/e• which is one of the fundamental Maxwell's equation. We arrived at the conclusion that since there r no charges in free space because vacuum can't have any matter therefore del.E=0 . So isn't it wrong to conversely say that since del.E=0 there r no charges in free space when del.E=0 came from assuming that there can not be any Free charges in vaccum?
 
No, the converse is also true.
##\rho = 0 \implies \nabla \cdot \vec E =0##
and
##\nabla \cdot \vec E = 0 \implies \rho = 0##
 

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