Confusion about Gauss' law differential form

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SUMMARY

The discussion centers on Gauss' law in its differential form, specifically addressing the divergence of the electric field, represented as ∇·E = ρ/ε₀. Participants clarify that the divergence is non-zero only at charge locations, necessitating the inclusion of a Dirac delta function to accurately represent point charges. The charge density for a single charge is expressed as ρ_single charge(x) = q δ³(x - x₀), highlighting the need for precise mathematical representation in electromagnetic theory.

PREREQUISITES
  • Understanding of vector calculus, particularly divergence
  • Familiarity with electromagnetic theory and Gauss' law
  • Knowledge of Dirac delta functions and their applications
  • Proficiency in SI units and charge density concepts
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  • Study the mathematical implications of Dirac delta functions in physics
  • Explore advanced topics in electromagnetic theory, focusing on charge distributions
  • Learn about the applications of Gauss' law in different coordinate systems
  • Investigate the relationship between point charges and continuous charge distributions
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Students of physics, particularly those studying electromagnetism, educators teaching advanced calculus, and researchers focusing on theoretical physics and charge distributions.

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http://einstein1.byu.edu/~masong/emsite/S1Q50/EQMakerSL1.gif

Hi guys, I have a little confusion about the Gauss' law in differential form over here, obviously, many textbook wrote it in the above form, but actually, the only place at which divE is not zero is at the locations where the charges are present. so I read over here: http://farside.ph.utexas.edu/teaching/em/lectures/node30.html#e343 (equation 208) that there should be a Dirac delta function over there, is it correct? then why does people keep omitting that Dirac delta function and just quoted it as above, where no information regarding the position of the point where we're calculating the divergence is given? Thank you.
 
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The fundamental law, using SI units, is indeed

\vec{\nabla} \cdot \vec{E}=\frac{\rho}{\epsilon_0}.

Here, \rho is the charge density of the charged matter, producing the electromagnetic field.

If the matter consists of a single charge at rest, this density is a \delta distribution

\rho_{\text{single charge}}(\vec{x})=q \delta^{(3)}(\vec{x}-\vec{x}_0).

If you have condensed matter, you can coarse grain the sum of the single-charge \deltadistributions to a continuous charge distribution.
 
Oh I got it now, thank you very much.
 

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