# Electric field inside the cavity of an irregularly shaped shell

Is the electric field inside a cavity of a charged irregular shell (with no charge inside the cavity) zero, even if the shell is not spherical, but has any irregular shape? Explain.

In 2D there is a nifty trick to see that the potential must be constant. In a charge-free region of space the electric potential obeys Laplace's equation $$\nabla^2 V=0$$. Consider points as points in the complex plane, perform a conformal map of the irregular shell to a circle, recognize that solutions of Laplace's equation in one domain are transformed to solutions in the other domain, use the shell theorem to see that the circle case must have constant potential, and hence the potential in the region must be constant and the electric field zero.
The more "proper" and easy approach is to use a Gaussian surface (which is naturally 3D, and doesn't depend on whether one can actually find a conformal map). For a closed surface $$S$$ $$\int\int_S \mathbf{E}\cdot d\mathbf{A} = Q/\epsilon_0$$ where $$Q$$ is the enclosed charge inside the surface. If you make the surface go through the conductive shell there is no field across it at all (since you are inside a conductor), and you can conclude that $$Q$$ must be 0.