Does a point charge inside a conducting shell cause redistribution of charge in the shell? A point charge Q is placed inside a conducting spherical shell at a random place (non-centre).


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*I have read that there is no force on Q from the shell no matter where Q is inside the shell ('there will be a large force from a few electrons pulling the charge one way, and a smaller force but from more electrons pulling the charge the other way').

*But I have read that then the electric field from Q induces a charge (the electrons in the shell rearrange themselves to neutralise the electric field from Q) after which Q will feel an attractive/repelling force from the shell. 
Question: How can there be a force on the electrons in the shell from Q making them redistribute, but no force on Q from the electrons in the shell? Is it true that a charge inside a conducting shell induces a redistribution of charge on the shell? It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?
 A: The force between every electron and the charge is the same according to the Newton's third law. When you add together the forces from all electrons on the charge, the resulting force is zero for the reason in your first bullet. However, the force from the charge on every electron is still there. Thus the answer is yes. 
A: There can't be non-zero electric field in a conductor. No matter how the charge distributed inside or outside of a conductor, electric field inside (in the conducting material) is always zero. So if you place a point charge inside a conducting shell the inner side of the conducting shell will have same amount but opposite charge distributed so that a Gaussian surface in side the conducting material have zero enclosed charge, which is required to make the electric field zero in the conducting material. Now that the total amount of charge has to be conserved the outside of the surface of the conducting shell will have same amount of charge as the point charge inside to compensate the extra charge in the interior side of the shell; so that the net charge from outside of the conducting shell is just the point charge placed inside the shell.  
A: You're confusing two phenomena here.  

I have read that there is no force on Q from the shell no matter where Q is inside the shell ('there will be a large force from a few electrons pulling the charge one way, and a smaller force but from more electrons pulling the charge the other way').

This is (a paraphrase of) the usual argument for the fact that $\vec{E}$ vanishes inside a spherical shell with a uniform charge distribution.  If we were to place a charge $Q$ inside the shell, and held all the charges on the shell fixed (for example, if it was insulating), then $Q$ would feel no force.  
Critically, the usual argument relies on the fact that the charge distribution is uniform over the surface of the shell.  If the charge on the shell isn't evenly distributed, the electric field inside the shell will not be zero.  

But I have read that then the electric field from Q induces a charge (the electrons in the shell rearrange themselves to neutralise the electric field from Q) after which Q will feel an attractive/repelling force from the shell.

If, on the other hand, the sphere is conductive, the charges can move around on it.  Placing the charge $Q$ inside the sphere then causes them to experience a force, and they will redistribute themselves so that they are in equilibrium with $Q$ and with the other charges on the sphere.

It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?

The key difference here is that you're talking about two different distributions of charge on the shell.  Since the shell charges are in different places here, we shouldn't expect the forces on $Q$ to be the same in each case.
As an analogy:  suppose there is a charged bead $q$ that can slide along a rod placed along the $x$-axis.  I then fix a second charge $Q$ in place somewhere along the $y$-axis.  At the moment I put $Q$ in place, it feels some particular force.  But $q$ can slide along the rod, and so it gets pushed away from $Q$, which causes the force on $Q$ to change.
A: The charges induced on the inner surface will apply some force on the charge placed inside the cavity but the charge on the outer surface of conducter won't interact with the charge inside the cavity. It's explanation is given in I.E Irodov Basic laws of Electromagnetism ,page 52-53
A: If the shell is conducting, then placing a charge inside it at a non-center position will certainly cause redistribution of charges on its inner surface. This happens due to the unbalanced forces experienced by the electrons (mobile charge carriers) in the shell.
It is true as you said that the charge placed inside the shell will also experience a force equal and opposite to that of the electrons in the shell, but this force will not displace the charge Q as we are keeping this charge Q fixed by some external (mechanical) means.
So a force acts on Q (equal and opposite to the force that acts on the electrons) which is balanced by the external support, until the electrons redistribute so that every electron in the conductor is in equilibrium under the influence of the forces due to other electrons, ion centers (positive charges of atoms which have given up their electrons into the sea) and the additional charge Q placed inside it. As a result the force on Q after redistribution will also be 0.
The charge developed on the outer surface will be uniform and it will not contribute to the electric field inside due to the fact that uniformly charged sphere has 0 electric field inside it, thereby maintaining every interior point at the same potential with respect to the conductor surface. This can be easily checked by applying Gauss law to a spherical surface that has a radius less than that of the shell, so that the charge enclosed inside it is 0 and that implies that the net flux is 0. Since the sphere surface and charge distribution is radially symmetric, the electric field is the same at every point of the sphere and it is equal to 0.
So hopefully, this answers your third question as well

It doesn't seem to make sense as the force per charge (E-field) inside the shell was zero before we added Q, and then we put a charge of 1Q in there, so Q should feel zero force from the shell?

Earlier before Q was added the shell had balanced charges throughout, but at the instant succeeding it, there is a redistribution that causes equal and opposite charges to appear on the inner and outer surfaces to make the electric field inside the conductor 0. This redistribution causes a force to be exerted on the charge Q. The boundary charges (charges on the inner shell surface) acquire a suitable configuration as described earlier to again neutralize the net electric field developed.
Hope this was helpful. Cheers!
A: The force will no be zero on the charge placed off-center in the conducting sphere because there will be unbalanced forces due to  the fraction of surface near (small distance, high accumulation of opposite charges, hence high force ) and far(Large distance low accumulation of opposite charges, hence low force ) from the charge inside. It can also be justified by Earnshaw theorem which states: a charge particle can't be held in an equilibrium position using electrostatic forces only.
