How is the $E$-field getting canceled between outer and inner surface of a neutral conducting spherical shell? I am reading Purcell's E&M book and in one of the example questions, it shows that there is no E field between outer and inner surface after a a point charge is located at an arbitrary position inside a neutral conducting shell. My question is the positive charge +q induces -q on the inner surface and +q on the outer surface. There is an E field drawn in RED going from +q(inside the shell) to -q(inner surface) and -q(inner surface) to +q(outer surface). I drew the fields on the left picture.
The author states that the E field inside the conducting material is ZERO. In the absence of an external electric field, how is the electric field from +q(outer surface) to -q(inner surface) getting canceled?

I believe the question was answered Electric field inside a conductor and induced charges But I was still confused after reading the explanation.

 A: 
In the absence of an external electric field, how is the electric
  field from +q(outer surface) to -q(inner surface) getting canceled?

It is getting cancelled by the field from charge q located inside the shell.
An equivalent statement is that the induced charges, +q(outer surface) to -q(inner surface), cancel the field produced by change q inside the conductor, i.e., between the inner and outer surfaces of the shell.
A: There is no field inside a (non-driven) conductor because if there were a field, it would move free electrons around until there is no field.  In this case the induced charges on the inner and outer surfaces act like the charges on the plates of a capacitor to produce a field which cancels that from the enclosed charge.  At the point closest to the enclosed charge, the surface charge densities must be equal and at a maximum. (At the far point a minimum.)  At other points the densities must vary to compensate for the tilt of the field from the inner charge. I don't believe that the charge density on the outer surface can be uniform.  I suspect that the field outside the sphere would be the same if the sphere were removed.
A: The relationship between the Charge inside a surface and the intensity of the field on said surface if given by (Gauss theorem):
$
  \oint \vec{E}\cdot d\vec{A}=
    \frac{q_{\mathrm{enc}}}{\varepsilon_0}
$
If $S$ is a closed surface located inside the shell, then:
$E . A = (+q) + (-q) =  0$,
therefore
$
E = 0
$
(since A is non null)
