Does Gauss' Law Hold for Infinite Gaussian Surfaces?

AI Thread Summary
Gauss' Law is typically proven for finite surfaces, and its application to infinite Gaussian surfaces raises questions about total flux and enclosed charge. An example presented involves a universe filled with a constant charge distribution, where the electric field is zero everywhere, yet any closed surface drawn encloses non-zero charge while the surface integral remains zero. This illustrates that Gauss' Law depends on the assumption that the vector field vanishes at infinity. However, if the charge distribution is finite, infinite Gaussian surfaces can still be valid. Overall, the discussion highlights the limitations of Gauss' Law in certain scenarios involving infinite surfaces.
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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