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By gauss law we can say that there is no electric field inside the hollow sphere (as $q=0$) , but we say that charges on a charged conductor accumulates in the surface or interface between the metal and air medium , then we can say that the charges accumulate in the inside of the spherical shell , which produces electric fields normal to the surface hence there is an electric field inside the cavity.

enter image description here The first image shows the charges only accumulating on the top part of hollow sphere , whereas second image shows the charges accumulating in the inside the second and first radius of the hollow sphere which creates an electric field inside

These two statements are contradicting each other , and I know gauss law holds well , what would be the flaw in the second argument ?

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  • $\begingroup$ You haven’t asked a question. $\endgroup$
    – Ghoster
    Oct 8, 2023 at 5:39
  • $\begingroup$ thank you for pointing out , I have added what I intended to ask @Ghoster $\endgroup$
    – Naveen V
    Oct 8, 2023 at 5:42
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    $\begingroup$ I don't understand where's the contradiction. Explain better or include a drawing. $\endgroup$
    – Pizzaguy07
    Oct 8, 2023 at 5:49
  • $\begingroup$ @Pizzaguy07 I have added an image $\endgroup$
    – Naveen V
    Oct 8, 2023 at 6:05

1 Answer 1

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Two flaws:

  1. The fact that charges accumulate on the outside of a conductor does not mean that there have to be charge accumulations on every part of the outside of this conductor. The charge distribution can be non-homogeneous. And in this particular case there are no charges accumulated on the "inner outside" of the spherical shell. (Edit/added: The charges are on the outside of a charged sphere (not shell) because then they are furthest apart, thus giving the system the configuration of least energy. This is why they do not move to lower radii. And turning the sphere into a spherical shell, thus restricting radii from below, does not affect a single thing in this argument.)

  2. Even if there were a charge accumulation on the inner surface of the shell (spherical symmetry assumed, of course): It would produce a field that is zero for smaller radii.

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  • $\begingroup$ can you explain how the field is zero for the smaller radii and how no charges are in the inner outside ? $\endgroup$
    – Naveen V
    Oct 8, 2023 at 6:05
  • $\begingroup$ I've made an addition for point no. 1. Point 2 I do not think I need to explain, this is a super-standard textbook exercise and the way you open your question it sounded like you are familiar with this or at least do not doubt this. $\endgroup$
    – kricheli
    Oct 8, 2023 at 6:13
  • $\begingroup$ Perhaps see physics.stackexchange.com/questions/150238/… and links therein. $\endgroup$
    – kricheli
    Oct 8, 2023 at 6:15

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