Is this a valid explanation for why charges in a conductor move to the outer surface?

This is a drawing of a spherical conductor with a hollow center. The black dots represent +ive charges, and the arrows represent their electric fields. I am assuming the black dots start on the inner surface. Imagining each position of the inner surface as a plane or sheet, the +ive charge should exert fields in all directions.

My question: Is the reason that the charges relocate to the outer surface because inside the hollow space, the fields pointed center-wise cancel out? This leaves only the field lines pointing outside the sphere, and the +ive charges will continue along the field lines until they can travel no longer. Thus charges in a conductor don't just move to any surface, they must move to the outermost surface.

I know it is common for people to draw conducting +ive spheres with fields just pointing normally outwards from the outer surface. I am wondering if this is because although the surface is much like a sheet with fields pointing both out and into the sphere, we just assume the fields pointing inwards cancel?

By this logic, a conducting sphere of radius R should (rightly) have the same field as a conducting thin spherical shell of radius R.

• Just to make things very clear: Mobile charges move to the surface. Fixed charges don't. – The Photon Apr 6 '19 at 2:33
• I now realize my usage of +ve charges was more abstract than reality. @ThePhoton could I ask you, though, assuming this is a conductor and the electrons are free to move (thus there will be relative +ve charge movement), can you answer my main question. Thank you. – ebehr Apr 6 '19 at 2:56
• Like charges repel, so they try to get as far away from each other as possible. In a conductor, that position is on the outside surface. – David White Apr 6 '19 at 3:48
• What is "+ve" or "+ive"? – Jasper Apr 6 '19 at 9:41
• “Positive” and “negative”. Common shorthand. – Chemomechanics Apr 6 '19 at 16:30