Unsymmetrically Placed Charge in a Hollow Conductor

[modulus]

This doubt cropped up in my mind while going through my old electrostatics notes...

We consider a spherical shell with some thickness (so that two different charge distributions may be possible on the inner and outer surfaces of the shell) made of some conducting material, and electrically neutral (uncharged).
Now, when we place any charge at the center of the shell, then charges are developed on the inner and outer surfaces of the shell. These induced charges are supposed to ensure that there is no net electric field anywhere inside the conductor (between the inner and outer surfaces). We also observe, though, that once the induced charges are formed and have taken their place on the conductor's inner and outer surfaces, the charge at the center of the shell remains at equilibrium.
This makes sense, because the induced charges are supposed to ensure no net electric field in many concentric spherical surfaces (which constitute the volume of the shell) that are centered at the charge itself. So , when the induced charges remove any electric field from the concentric shells, it makes sense that the net field at the center of these shells comes out to be zero by symmetry.

But what if the charge was placed somewhere not so symmetrical within the shell? Would it experience no net force due to the induced charges which ensure no net electric field within the conductor (but not within the cavity, where our charge is asymmetrically placed)? What if the shell had some charge on it prior to the introduction of the charge in the cavity? And will electrostatic shielding take place for this charge, when the shell is placed in some external electric field??

 

[kuruman]

Go through your old electrostatic notes one more time and concentrate on the section of solving boundary value problems using the method of image charges. If you place charge $q$ at distance $d$ from the center of a conducting shell of radius $R$, the potential in the region inside is the same as the potential generated by a point charge $q'=-qR/d$ placed at distance $R^2/d$ from the center of the shell on the axis of symmetry. That is the case when the shell is grounded and it is at zero potential. If the shell is initially at potential $\varphi_0$ because of some charge placed on it, then you need an additional image charge $q''=4\pi\epsilon_0 R \varphi_0$ placed at the center of the shell to adjust its potential to $\varphi_0$. In either case the force on $q$ can be calculated using Coulomb's law. Yes, if you turn on an external electric field, the charge inside the cavity and on the inside surface will not be affected. The surface distribution on the outside may change in response to the external electric field, but this change cannot be communicated to the charges on the inner surface or inside the cavity. That's because the electric field in the space between the outer and the inner surface is and remains zero whether the external electric field is on or off. Charges communicate with other charges via electric fields. No electric field between outside and inside - no communication.