Why is there no electric field outside a spherical, conducting, non-grounded shell with a dipole in the center? If you place a dipole at the center of a conducting, non-grounded, spherical shell, the electric field outside the shell is zero. However, if there is a point charge in the center, there is an electric field outside.
I understand that the net charge in the system is zero, because of the dipole, and then maybe it's a violation of Gauss's law to have an electric field outside the shell. However, why does dipole by itself (which has an electric field) not violate Gauss's law?
Thanks in advance.
Edit:
This is my source for this question. It's from the book Conquering the Physics GRE by Yoni Kahn and Adam Anderson.


 A: The exterior surface of the shell is an equipotential $V_0$. Assuming that the potential outside the sphere approaches a finite, constant value $A$ as $r \to \infty$, then a solution to Laplace's equation satisfying these boundary conditions is $V(r) = A + (V_0-A)R/r$.  By the uniqueness of the solutions of Laplace's equation, this is the only possible solution for the exterior region given these boundary conditions.  Thus, the exterior potential must be spherically symmetric, and so must the exterior electric field $\vec{E} = (B/r^2) \hat{r}$.  Invoking Gauss's Law and noting that there is no net charge for any surface enclosing the sphere, we can conclude that $B = 0$, and so $\vec{E} = 0$.
It is, of course, possible to consider situations where the potential does not approach a constant value at infinity;  a standard example of this is the problem of a conducting sphere in an external uniform electric field.  But if we demand that the exterior potential approaches a constant as $r \to \infty$, and there is no net charge inside the sphere, then the exterior potential must be constant and the exterior electric field must be zero.
Finally, I suspect that this argument can be adapted to a non-spherical shell as well.  But the precise details elude me at the moment...
