# electric field and distribution of induced charge on outer surface of conducting shell enclosing an off-center charge

Halliday and Resnick present a system with a conducting shell enclosing a cavity with a negative charge located somewhere in the cavity, but off-center. The authors claim that the induced negative charge on the outer surface is equally distributed and has the field of a negative point charge because "the shell is spherical and because the skewed distribution of positive charge on the inner wall cannot produce an electric field in the shell to affect the distribution of charge on the outer wall."

My question is whether this logic is solid? I am not questioning whether the description of the outer charge distribution is correct, but rather WHY it is deduced to be this way. (my intuition is leading me to view the outer surface as an equipotential surface, that because it is symmetrical, should have a uniform field). However, I don't see how the author's description of the null field inside the material of the shell implies a foundational principle that charges on one side of a conductor do not contribute to a field on the other side. My understanding of the property of conductors is that the final arrangement of charges must have null field in the conductor, but to me, that doesn't necessitate that the inner induced charge can't contribute a component to the field at the surface.

Am I incorrect in stating that the electric field at some point on the outer surface or just beyond it is the superposition of the field components from ALL of the other charges in the system, including the inner induced charges? So the field contribution from the inner charges shouldn't just be dismissed? In other words, if some of the inner induced charge was to be magically removed, wouldn't this create a new initial condition that would be immediately felt by the outer charges, even before the internal charges of the metal have time to re-equilibrate?

• @lamplamp It's not just that the net charge is zero that implies the field is zero everywhere outside, but also Poisson's equation $\nabla^2\phi=0$. It's pretty easy to show that since the inner shell is an equipotential, the electric field must be spherically symmetric. And the only spherically symmetric solution with zero flux through any surface is the one that is uniformly zero. – Chris Jan 16 '18 at 6:10