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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.

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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.

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  • $\begingroup$ One thing for sure I want to comment, the OP has not yet defined whether the charge distribution is uniform or defined by some other law. Neither has the OP defined whether the shell is conducting or charges are immobile. $\endgroup$
    – SteelCubes
    Jan 26, 2021 at 11:24
  • $\begingroup$ @SteelCubes - Yes, I assumed that since in the general case there will not be a zero field. The key thing is that the potential along the shell is constant (that also allows using the mean-value theorem for harmonic functions for another kind of proof). $\endgroup$ Jan 26, 2021 at 12:49

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