Closed integral in electromagnetics

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SUMMARY

The discussion centers on the application of Gauss's Law in electromagnetics, specifically the equation relating the total charge enclosed by a surface to the closed integral of the electric displacement field (D) over a surface (S). The equation is expressed as total charge enclosed = closed integral of (D . ds) over S, which is further equated to Dr (closed integral of dS over S). Dr represents the radial component of the electric displacement field, confirming its significance in the context of the law.

PREREQUISITES
  • Understanding of Gauss's Law in electromagnetics
  • Familiarity with vector calculus and surface integrals
  • Knowledge of electric displacement field (D) and its components
  • Basic principles of charge distribution in electrostatics
NEXT STEPS
  • Study the derivation and applications of Gauss's Law in electromagnetics
  • Learn about vector calculus, focusing on surface integrals and their physical interpretations
  • Explore the concept of electric displacement field (D) and its role in material media
  • Investigate charge distribution and its effects on electric fields in different geometries
USEFUL FOR

Engineering students, physicists, and anyone studying electromagnetics or vector calculus who seeks a deeper understanding of Gauss's Law and its mathematical implications.

tidesong
Well I'm studying this under engineering but its also maths so I posted this here and hope its in the right forum

The question is: When using Gauss's Law i have this equation in my notes: total charge enclosed by surface = closed integral of ( D .ds) over S

the next line (in my notes) equates this to Dr( closed integral of dS over S)

Is this a mathematical manipulation, or is this due to some law? And is Dr just a constant or does it have some meaning? Thanks.
 
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Ah thanks! If Dr is the radial component then it makes sense! I'm trying to guess what my notes mean too :P
 
Yeah! Our notes are so lousy! Thanks for helping out =)
 

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