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Consider a thick metallic spherical shell of an inner radius $r=a$ and an outer radius $r=b$. Let an external electric field is applied horizontally ($\theta=\frac{\pi}{2}$ direction) from left to right which breaks the spherical symmetry of the problem.

Due to the applied field, charges will be accumulated in a non-uniform manner on the inner and outer surfaces of the shell. Since the field is assumed to be from left to right, for the outer surface, negative charges will accumulate on the left side and positive charges on the right side (with a gradual variation from left to right). The opposite distribution will happen on the inner surface.

My question is about the electric field inside the empty cavity. Considering a spherical Gaussian surface of radius $<a$, I can use Gauss' theorem to say that the electric flux is zero. But due to the lack of spherical symmetry, I cannot use to Gauss theorem to first say that the electric field (if any) must be along the radial direction and is zero.

Please explain if I am right. Whether the electric field inside the empty cavity is zero or nonzero cannot be ascertained from Gauss law. Is this right to say?

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Due to the applied field, charges will be accumulated in a non-uniform manner on the inner and outer surfaces of the shell

No, the charges will accumulate only on the outer surface of the shell, as the inner surface will not have bear any charges.

Now, remember that the whole conducting shell is at the same potential, which means that the inner surface is an equipotential surface. Since there is no charge in the cavity, the potential within has to satisfy Laplace's equation, $\nabla^2 V = 0$, with boundary conditions given by the (constant value) that $V$ takes on the surface. The only solution to this equation with these boundary conditions is that $V$ takes the same value also within the cavity. Since the electric field is the gradient of the potential, which is constant, we demonstrated that $\vec{E} = 0$ even in the cavity.

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  • $\begingroup$ @Ir1985 Why do you say that the inner surface will not any charges? It is not always true that inner surface is charge-free. Think of a vertical metallic bar on which an horizontal electric field is applied. Now charges will appear on both the left and right side of the bar. Now imagine bending it in the form of the shell. $\endgroup$ – mithusengupta123 Sep 24 at 13:00
  • $\begingroup$ the effect is known as the Faraday cage: the charges induced on the outer surface screen the whole inner conductor, so that the field is zero everywhere in its interior. $\endgroup$ – lr1985 Sep 24 at 14:41
  • $\begingroup$ If you change the shape of a conductor in an electric field, the charges will redistribute to maintain E = 0 in the conductor. $\endgroup$ – R.W. Bird Sep 24 at 18:21
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    $\begingroup$ I like the idea of considering electric potential. We know that in a static situation, charges in a conductor will move until the field inside is zero. That means that if you take a path through the conductor from one point on the inner surface to another, there will be no change in potential. That means that a path through the hole between the same two points will give the same result. For this to be true for any choice of points, there must be no field in the hole (and no charge on the inner surface). $\endgroup$ – R.W. Bird Sep 24 at 18:45
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So, first upon all, answer is no. We can't say only by gauss law whether electric field is zero or non zero in cavity. But resultant electrical field inside cavity is zero. Let's see how? fact that resultant electric field inside conducting surface is zero and this is logical and experimental fact And here you can say by gauss law that charge on inner surface is zero where gaussian spheres radius lie in between a & b as electric field at surface is zero. Here again a logical thing is that as no resultant electric field at inner surface therfore no resultant electric field inside conductor.

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