# Electric field inside the empty cavity of a thick spherical metallic shell subjected to a horizontal external electric field

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 $$, 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?

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.