Does Gauss' Law Hold for Infinite Gaussian Surfaces?

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SUMMARY

Gauss' Law applies primarily to finite surfaces, as demonstrated through the example of a universe filled with a constant non-zero charge distribution. In this scenario, the electric field is zero everywhere, yet any closed surface drawn encloses a non-zero charge, resulting in a surface integral of zero. This indicates that Gauss' Law depends on the assumption that the vector fields vanish at infinity. Therefore, infinite Gaussian surfaces are valid only when the charge distribution is finite in extent.

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  • Understanding of Gauss' Law in electrostatics
  • Familiarity with electric field concepts
  • Knowledge of surface integrals in vector calculus
  • Basic principles of charge distribution
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  • Study the implications of Gauss' Law in different charge distributions
  • Explore vector fields and their behavior at infinity
  • Learn about surface integrals and their applications in electromagnetism
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Students of physics, particularly those focusing on electromagnetism, educators teaching electrostatics, and researchers exploring the implications of Gauss' Law in various contexts.

Ahmes
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Hi,
From what I understand the proof of Gauss' law applies only to finite surfaces.
Can anyone give an example of a charge distribution and an infinite Gaussian surface, where the total flux on it is not proportional to the enclosed charge?

Thanks!
 
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This is similar to what you asked for.
An example I heard was this:
Imagine the universe filled with a constant non-zero charge distribution.

By symmetry, we know the electric field is zero everywhere.

But any closed surface we draw, the enclosed charge is non-zero, but the surface integral is zero.

[my memory is rusty on the conclusion, someone please correct this if I am wrong]
Therefore yes, Gauss' law does depend on an assumption at infinity... it only applies to vector fields that vanish at infinity.


I believe what you asked for though "a charge distribution and an infinite Gaussian surface", will always work if the charge distribution is finite in extent so that the vector field vanishes at infinity. So with that one caveat, I believe infinite Gaussian surfaces are no problem.
 

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