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Suppose there is a metallic shell having a uniform charge Q spread over it uniformly. Now we bring a charge Q1 near it. What is the field at the centre of the shell and in general anywhere in the shell?

By Gauss's law we can say that flux through any closed surface in the shell is zero but because of the asymmetry, I don't think we can say anything more.

What would happen if the shell was not made of a conductor?

What would be the field at the centre if the metallic shell was deformed?

If possible, please try to give an answer not involving complicated maths since I have knowledge of only first order differential equations

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If the shell is a perfect conductor, the field inside it at the center -- or anywhere inside, for that matter -- is going to be zero. Even the shape of the shell is not relevant: the same result holds for a conductor of any shape. The reason for this is that, by definition, a perfect conductor is assumed to have infinitely many free-charge carriers that can orient themselves in response to an external field and cancel out said field so that there is no current (hence the name electrostatics). In other words, a conductor is infinitely polarizable.

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  • $\begingroup$ Yes I understand that field on the shell is zero because charge carriers can orient themselves in response to the field. But I don't understand why there won't be any field inside the shell( the volume where there is vacuum and no charge carriers). $\endgroup$
    – Parth Bhardwaj
    Commented Jan 2, 2021 at 15:10
  • $\begingroup$ Simply use Gauss' law: $\nabla \cdot \mathbf{E} = \rho_{free}/\epsilon$. There are no free charge-carriers inside, so $\nabla \cdot \mathbf{E} = \mathbf{0}.$ This holds for any infinitesimally small area inside the conductor, so $\mathbf{E} = \mathbf{0}$. $\endgroup$
    – Yejus
    Commented Jan 2, 2021 at 15:15
  • $\begingroup$ I don't understand how can we say that if flux of electric field is zero then electric field is also zero, since the situation is asymmetric. $\endgroup$
    – Parth Bhardwaj
    Commented Jan 2, 2021 at 15:24
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Consider a conductor of any shape filled inside with a conducting material and having charge Q on it. Now bring a charge Q1 near it. The charges on the outer surface of conductor will align so that field inside is zero everywhere. Now after the electrostatic condition has been established, I scoop out a cavity inside the conductor. Assuming perfect vacuum, notice that the forces on any of the charges on the conductor or on the charge Q1 do not change. This means that the charge distribution is unchanged after scooping out cavity. So even after scooping out cavity, the field inside the cavity is still zero everywhere. Now I can make this cavity large enough so that the remaining part is just a shell and this argument will still be valid. So in your case also, field inside shell is zero everywhere. Now by the reasoning it is clear why this will not necessarily hold in the case of a non conducting material.

This result can be generalized further, although the rigorous proof is through the Laplace equation. Consider a conducting shell( any shape) with a cavity having some charge (which implies there is equal and opposite charge on the inner surface of cavity) and there is also some charge outside the conductor and on the outer surface of the conductor. We find that net field due to charge on outer surface and due to charge outside the shell is zero everywhere inside the cavity and inside the conducting material. Also net field due to charge inside the cavity and the charge on the surface of cavity is zero everywhere outside the cavity. The conductor sort of acts as a shield against the 2 fields. This phenomenon is called electrostatic shielding

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