External-excitation Van de Graaff polarities

[eigenmax]

My question regards the polarities of the excitation supply and terminal in an external-excitation VDG.
As you all probably know, external excitation VDGs use a voltage supply (usually 5,000-10,000V) to influence the charge on the lower pulley, instead of just relying on frictional contact to remove electrons from the belt.
My question is, if I place the positive of my supply at the lower pulley, will the positive electrode of the supply draw electrons to it from the belt (as a positive object is an electron deficient one), thereby making a electron-deficient belt and a positive top terminal?
If I use the negative electrode will I have a negative terminal?

There is a little diagram of the excitation supply at the top right.

 

[sophiecentaur]

The section of belt passing the 'excitation supply unit' will become charged by Induction. (the opposite charges are taken to ground via the comb) At the top, there is no path to ground so the charges on the belt will get transferred directly to the outer surface of the sphere by mutual repulsion
I'm not sure of the function of the rest of the lower unit, namely the conductive pulley and the apparent common the other side of the charged bar. Perhaps the induction is better with those extra parts. Or perhaps it's for operation without the external excitation supply.

 

[eigenmax]

The article I was reading was in The Amateur Scientist, a book written 1960 as a compilation of all the Amateur Scientist columns from the Scientific American throughout the 50s. It describes the operation of standard (self-exciting) VDGs, but this is all it has on the external excitation. I'm also not sure what the charged bar is for.