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Consider this conductor

enter image description here

By Gauss' Law and using the properties of conductor , I can certainly say that a $-Q$ charge will occur at the inner surface of cavity and a $+Q$ charge will occur at the outer surface of conductor, Intuitively we can say that this happens because the point charge inside cavity attracts the free electrons of the bulk of conductor.

I have been taught that the distribution of these surface charges will be uniform but why is this the case? According to me it should be true only when the the point charge and cavity were located at the center of conductor.

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Its not necessary that charge outside will have uniform distribution , its just in special cases like a spherical conductor or like a infinite cylindrical setup that the outer surface may have uniform charge distribution because these are the setups which have most symmtery from centre , otherwise any other conductor surface may not have same distance from the so called centre in your language . So in your case as the outside is a spherical conductor hence the charge distrubution outside surface will be uniform and the field will be such that of a point charge placed at centre of sphere .

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  • $\begingroup$ but why will it would be uniform in case of spherical conductor in above case. $\endgroup$ Jan 30, 2022 at 13:17
  • $\begingroup$ It is necessary in an Ideal scenario that charge outside will have uniform distribution in a spherical conducting material or hollow shell, If it's not uniformly distributed, then soon the charges will adjust themselves in such a way that they will become uniformly distributed and the field will become zero inside the conductor. That's not true according to your first statement! $\endgroup$ Jan 30, 2022 at 13:42
  • $\begingroup$ @TejasDahake Your conclusion may (or many not) be correct, but your argument to support it is flawed. Charge rearranges so that the total potential energy is minimum. It may or may not be uniformly distributed. $\endgroup$
    – garyp
    Jan 30, 2022 at 14:05
  • $\begingroup$ @garyp I haven't given any reason or conclusion yet and I haven't told anywhere that it is because of this phenomenon I just told it rearranges and the field becomes zero, I'm still working on it to provide a full answer that's why I've commented and haven't answered! $\endgroup$ Jan 30, 2022 at 14:14
  • $\begingroup$ @TejasDahake You said more than "the field becomes zero". You said, "in such a way that they will become uniformly distributed". This is the second time today (see the comments to this post ) that you deny saying something that you actually did say. If the problem is carelessness in writing, then please be careful. Without clear expression, we can't know what you mean to say. $\endgroup$
    – garyp
    Jan 30, 2022 at 14:21
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Here is an attempt to answer your question in words.

The conductor and hence its outer surface is an equipotential.

The electric field lines outside the conductor must be at right angles to the outer surface and hence must be radiation from the centre of a spherical conductor.

Very near the outer surface the electric field cannot change from position to position as that would mean that different amounts of work would be needed to bright a test charge to the surface and that would imply that the surface was not an equipotential.

If the electric field strength near the outer surface is the same then the charge density on the outer surface must also be the same.

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  • $\begingroup$ But electric field , even if different, will be perpendicular to surface of conductor so still conductor is equipotential? $\endgroup$ Jan 30, 2022 at 14:24
  • $\begingroup$ But with a spherical conductor because of the right angle the field must radiate from the centre of the sphere and then must be equal in strength because the surface is an equipotential. $\endgroup$
    – Farcher
    Jan 30, 2022 at 15:52
  • $\begingroup$ Yes field must radiate from center but even if it is unequal in strength the surface will be equipotential because field will be normal to surface, so how can we say that field strength will necessarily be equal? $\endgroup$ Jan 30, 2022 at 15:56

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