Physically what is happening is this:
- When you touch the positively charged source to the conductor (the metal sphere), electrons leave the conductor through the point of contact.
- This leaves the point of contact on the conductor with a large deficit of electrons, and thus the point has a positive charge density.
- The positive charge density produces an electric field in the conductor, which immediately pulls on remaining electrons in the conductor.
- The electrons remaining spread out until they have eliminated all of the electric fields in the conductor (if there were remaining fields, the electrons would continue to rearrange).
- The electrons will now be 'more spread out' than the protons; the difference between the new electron surface density and the original tells you the distribution of 'excess positive charge' on the surface.
I hope this helps, let me know if you have an application in mind for this; I often times find it helpful in thinking about problems to temporarily ignore the fact that in practice there is only one charge carrier (the electron) and just think about excess positive charge as positively charged particles spreading out.
You have two questions. One is "what is the charge distribution on a non symmetric conductor" and the other is "how can the inner surface of a conductor be charged?"
Q1:
The charges will distribute themselves in a manner to minimize total energy (whenever you don't know an answer on the exam, say that ;). Since it's a conductor, they will distribute themselves in such a way that the potential anywhere on the surface is the same (otherwise, the charges would have a gradient they could move along).
Q2
Assumptions:
Let's assume electro statics i.e. no time dependence, and perfect conductors.
Also let's assume the tubes are infinitely long. Therefore, this becomes a 2D problem of concentric circles (senior electromagnetis becomes much simpler when you realize they only ever test you on a handful of geometries).
Solution:
Point 1) The electric field inside a perfect conductor is zero (otherwise a current would arise, rearranging the charge until the field is cancelled out). Therefore, within the material of the thin outer tube, the electric field is zero.
Point 2) However, the integral of the electric field along a path (it's a path integral because we're in 2D!) is equal to the enclosed charge (Gauss' law). From symmetry, there is no angular dependence.
Conclusion) The only way for the first point and the second point to be both correct is if you have a charge in the inner surface of the tube. The charge on the inner surface of the tube needs to cancel out the electric field from the rod, so it is equal and opposite to the charge of the rod. The outer surface also has a charge, equal to the net charge of the tube minus the charge on the inner surface of the tube.
Remember:
The fundamental rule of a conductor is that the electric field within it is zero. There's no fundamental rule against charges on the inside surface of a conductor!
Further investigation:
Realistically charges are electrons in thermal equilibrium. How does this affect our mental image of "charges on an infinitely thin surface"?
Best Answer
You have a constraint system. For every surface-element the tangent component of the electrical field strength must be zero (else the surface charge would start to move).
Consider a perpendicular edge of a cylindrical configuration as shown in the following picture.
Assume that the surface charge distribution would be uniform (constant charge per area).
There we have small surface charge elements $\sigma dA_1$ and $\sigma dA_2$ with $dA_1=dA_2$ generating a force on a small surface charge element $\sigma dA_3$ very close to the edge.
Thereby, $\sigma dA_1$ and $\sigma dA_2$ are just samples of the field generating surface charge elements over which we have to integrate. Nevertheless, we can discuss the main effect exemplary with their help. Both $\sigma dA_1$ and $\sigma dA_2$ have the same distance from $\sigma dA_3$ so that the absolute value of their field strength at the position of $\sigma dA_3$ is the same.
The tangent component $\vec F_{2\parallel}$ of the force $\vec F_2$ caused by $\sigma dA_2$ on $\sigma dA_3$ is small since $\vec F_2$ is directed almost normal to the surface.
On the other side the force $\vec F_1$ caused by $\sigma dA_1$ on $\sigma dA_3$ directly has tangent direction ($\vec F_1 = \vec F_{1\parallel}$).
Therefore, the sum of tangent forces $\vec F_{1\parallel}+\vec F_{2\parallel}$ on $\sigma dA_3$ will point towards the edge. In principle his holds also for the other charge elements contributing to the electrical field at the location of $\sigma dA_3$. Consequently, the surface charge $\sigma dA_3$ will move towards the edge until there is so much charge in the edge that for all surface charge elements the tangent component of the field strength is zero. ("The edge repells further charge.")
Eventually, at edges the surface charge density becomes infinite. Instead of a finite surface density you have a finite line charge density in the edge.
The effect at curved surfaces is similar but not so drastic.
The statement of the book that only spheres admit uniform charge distribution is only right if you restrict your considerations to bounded conductors and fast enough decaying fields at infinity.
If you admit infinite conductors then you also have uniform surface charge on a circular cylinder.
If you admitt outer charge distributions then you can adjust these outer charges such that the charge distribution on a considered smooth surface is uniform. Thereby, no restrictions are made on the shape of the surface.