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I was learning about conductors — mainly solid spheres and hollow spheres — and my teacher told me that in the case of solid spheres with a spherical cavity and a charge at the centre of the cavity the surface charge density on the inside surface of the conductor will be uniform; even if the cavity is not at the centre, it could be anywhere.

What I don't get here is that, if the cavity is unsymmetrical with respect to sphere, then the surface charge density should be non-uniform to make net electric field zero for outer charges. Then why is it to be uniform?

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Outer charges are completely screened by the charge on the outer surface of the conductor - their field does not penetrate beyond the surface and certainly not into the cavity. Thus, they have no effect.

Remark
My understanding is that OP is not yet familiar with Gauss theorem or the electrostatics of ideal conductors.

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    $\begingroup$ Than what if the charge in the cavity is not at the centre of the cavity ,the surface charge density on inner surface will still be uniform? $\endgroup$
    – Kushagra Mishra
    Commented May 30 at 9:03
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    $\begingroup$ No, it will not be uniform - see, e.g., figure in this answer $\endgroup$
    – Roger V.
    Commented May 30 at 9:13
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Given the origin of the question, I assume you might not be very familiar with how Gauss's Law works and other more advanced issues, so I'll try to explain it in a more qualitative way.

Let's consider a conductor of arbitrary shape and with a spherical cavity inside, with a charge $q$ located at its center. Since it is a conductor, charges will be induced on its inner surface in such a way that the electric field inside the conductor is zero. Think of a region very close to the cavity (but inside the conductor), meaning the influence of the electric field generated by the induced charges on the inner surface is much more relevant than the influence of the electric field generated by the charges that will be induced on the outer surface. For all purposes of calculating the electric field, only these induced charges matter, and therefore by symmetry, it's necessary for the charge distribution to be uniform, and for the total induced charge to be $−q$, so that the field is zero. The shape of the conductor doesn't matter. Even if it has the shape of a potato, what will make the electric field zero inside it in the regions farther from the cavity is precisely the density of charges on its outer surface, which might not necessarily be uniform, but close to the cavity the only importance is on the charge and the inner surface.

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