First think of a neutral wire - a metal with diameter r and conductivity $\sigma$ and is very long. One end of the wire is grounded, and at the other end of the wire there is a switch that can connect to the Positive terminal of a battery. If you want to get into more details you can say the at whole thing is enclosed by a cylinder parallel to the wire very far away, and the ground terminal of the wire and the ground of the battery is connected to the respective ends of the cylinders.
Before you close the switch. Inside the wire there are equal numbers of positive charges and negative charges because the wire is neutral. If we look somewhere near the middle of wire (ignore the end effects) we would say there is no electric field inside the wire or outside the wire. There is no current flow, and no potential drop along the length we choose near the middle of the wire.
We want to look closely as small segment near the middle of the wire. We close the switch and wait a little bit to avoid any time it takes for any electromagnetic waves to propagate and for the charge to redistribute. Then we are in a steady state, equilibrium situation as long a the battery potential doesn't change.
One end is ground, the other end is at +V, there is a current J flowing in the wire, electrons coming from ground at one end and flowing into the positive terminal of the battery at the other. The current $J$ is constant.
$$J=nev$$ where $v$ is the drift velocity of the electrons. $n$ is the number of electrons and $e$ is the charge of the electron.
If I look at the small segment at the middle of the wire, the number of electrons moving from the left into the segment is equal to the number moving out the right end of the segment. If I count all the protons, all the other electrons that are still bonded to their atoms and the electrons that are free to move at the drift velocity I find that the number of positive and negative charges are the same. So in the steady state situation it is like the wire before I through the switch, there is no electric field outside the wire generated since the total charge + and - sums to zero.
{If I zoomed in to the lattice of the metal and was on the length scales of atoms, I would see variations in electric field and dipoles etc, but they all average out. We are using Maxwells equations in a macroscopic sense.}
However, we do have a current $J$ and that current since it is moving charge will produce a B field which we can sense outside the wire and $$J=\sigma E$$
where $\sigma$ is the conductivity. This tells us that there is an electric field inside the wire that is parallel to the current. There is no radial component to the electric field inside the wire. This electric field has to exist otherwise the there would be no motive for the electrons to move with a drift velocity in the wire. We know that the potential drops along the length of the wire from +V to ground at the other end. If the wire is a constant diameter and the same material along the whole length the the potential drop for equal length segments is constant. This also tells us that the electric field is constant and in the direction of current flow since the negative derivative of the potential is the electric field, and if the slope is constant when we take the derivative we get a constant magnitude for E. This is just
$$E=-\nabla V$$
All of that is for inside the wire, but we also need to have the correct potential at the surface of the wire, and in the space outside of the wire. This is where the surface charge comes in.
We said the radial electric field on the inside of the wire was zero. That is still true if we put a ring of surface charge on the wire, since the electric field of that surface charge will point radially out, or radially in depending on the sign of the surface charge, and the radial component pointing perpendicular out on one patch of charge is canceled out inside the wire by the patch on the other side of the ring. The radial electric field outside the wire pointing out from the wire is also perpendicular to the inner surface of the outer surface that is far away.
So we have a boundary value problem. Inside the wire $E_{radial}=0$ outside the wire the electric field is starting at some value and dropping off. At the boundary you use a Gaussian pillbox to find the magnitude of the surface charge.
In terms of potential which is a scaler, $$V_{inside}=-Ez$$ where z is the coordinate along the length of the wire and the potential outside will fall of as the log of the distance from the wire.
Something like $$V_{outside}=Alog(r/B)$$. Where $A$ is related to -Ez and $B$ is related to the radius of the wire. And at the surface of the wire $V_{inside}=V_{outside}$
The when you work out the amount of charge (note that this sigma is different than the sigma I used for conductivity) you find that $$\sigma=A\frac{-Ez}{log(r/B)}$$ where A and B are constants depending on the details and r is the radius of the wire.
If you look at this you have positive surface charge as you increase z and get closer to the +battery terminal. The amount of surface charge density goes down as you make the radius of the wire bigger.
you can choose where your ground is. If you say you have a + battery terminal and a - battery terminal you can put z=0 at the center of the wire, and as you make z negative you would have negative surface charge.
By the way, this wasn't really taught well in textbooks until a few years ago and still isn't necessarily obvious. There are a bunch of papers in American Journal of Physics that go into it in some detail. From an electrical engineering perspective for a lot of problems you only care about V=IR since you can measure currents voltages and resistances, easily and it is harder to measure fields and surface charges directly.
