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Gauss Law $$\oint \vec E . d \vec A = \frac{Q_{enclosed}}{\epsilon _○}$$ From this question, Gauss Law is more fundamental than Coulomb's Law.
This seems counter intuitive to me when it comes to applying Gauss Law.
In the linked example and in many other cases like finding field due to uniform (sheet distribution of charge, cubical distribution, infinite wire) etc.

In all those cases my teachers as well as the books use Gauss law as $$\vec E.\int d \vec A \implies E= \frac {Q_{enclosed}}{A_{net}\epsilon_○}$$
They take it for granted that the field is equal in magnitude (wherever non-zero) on a Gaussian surface without any explain.

  1. What is the reason behind such an assumption? Please provide a proof
  2. Does this assumption also hold true for any random(irregular shape) non-uniform distribution of charge?

I have been told that we choose Gaussian surface as per convenience so that accounts for the assumptions.

  1. How to prove that the chosen Gaussian surface make the assumption true?
    For example how to prove that a spherical Gaussian surface for a sphere makes the assumption valid.
Qmechanic
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Aurelius
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1 Answers1

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They take it for granted that the field is equal in magnitude (wherever non-zero) on a Gaussian surface without any explain.

They don't "take it for granted." Rather, they are starting you off with a simple pedagogical example.

  1. What is the reason behind such an assumption? Please provide a proof

There is no general proof, since it is not generally true.

Often one tries to find a specific gaussian surface that makes the problem simple to solve. E.g., a spherical surface surrounding a point charge, in which case the field magnitude the same on the whole surface. This is only an example, it is not generally true.

  1. Does this assumption also hold true for any random(irregular shape) non-uniform distribution of charge?

No, the "assumption" of constant field magnitude does not hold for an arbitrary surface. But, Gauss's law does still hold for any closed surface.

  1. How to prove that the chosen Gaussian surface make the assumption true?
    For example how to prove that a spherical Gaussian surface for a sphere makes the assumption valid.

It depends on the specific arrangement of charges and the specific surface of interest. As discussed above your supposed "assumption" is not an assumption, it is just a fact for the specific pedagogical examples you are being taught.

hft
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