We just finished studying Gauss’ law and were puzzled by this thought. If I look at a copper atom and focus on the 29th electron in the 4th shell, according to Gauss’ law, I can draw a Gaussian surface that completely encloses the first three shells plus the nucleus. It is my understanding that the 29th electron will not contribute to the E-field inside this Gaussian because this charge is outside the Gaussian.  However, it does not mean that the 29th electron doesn’t apply an electric force on the inner electrons, especially the 3rd shell electrons. So here is my dilemma: wouldn’t this effect the E-field of the inner shell electrons?

Please, if possible, answer this question without quantum if possible.


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EDIT:

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