haruspex said:
For calculating electric field inside the shell due to shell and outside charge ## q_1##, let's consider that there is no charge at the center.
Due to ## q_1## the charge Q is not uniformly distributed on the metallic sphere. Hence, the electric field due to this inside the sphere is nonzero.
Let's take a spherical Gaussian surface. Then, the charge enclosed by this surface is 0. Applying Gauss's law,
$$ \oint (\vec E_{shell} +\vec E_{outside~ charge}) \cdot d\vec A = 0 $$ .....(1)
## \vec E_{shell} \neq 0##
## \vec E_{outside~ charge} \neq 0. ##
Electric field inside (in the meat of) a conductor is 0. Hence, ## \vec E_{shell} +\vec E_{outside~ charge}= 0## in the meat of the shell.
Electric field at the surface of a conductor is perpendicular to its surface as any tangential component gets cancelled by the movement of chargedparticles. Hence, electric field at the inner surface of the shell is radially inward (and the presence of charge q at center doesn't affect this direction). Hence, from (1), ## \vec E_{shell} +\vec E_{outside~ charge}= 0 ## inside the shell.
Thus the force on the central charge is 0. At centre, the electric field due to ## q_1 ## is towards left. Hence, the electric field and force due to the shell is towards right.