The "first level" answer was given by nibot in a comment.
The entire conductor must be equipotential. If there were a potential difference from one part of a conductor to another, free electrons would move under the influence of that potential difference to cancel it out.
However, since I have similar curiosity myself I'm going to try to answer in greater depth.
Imagine a conducting sphere with a negative charge on it. There exist a certain number of surplus electrons. A conductor has electrons that are bound in their orbit to a given nucleus and electrons in the conduction band. The entire volume of the sphere beyond a certain distance from the surface is almost perfectly balanced on the scale of a few atoms. That is, the number of electrons balances perfectly with the number of nuclei, and in reality, the "orbit" of the conduction band electrons span many nuclei.
The critical discussion is what happens to the "surplus" electrons. According to the electrostatic potential equations they must all exist exactly on the surface of the sphere. This, of course, is physically absurd. There is, however, nothing absurd about the mathematics of a surface charge. A surface charge gives a well behaved:
- field - which is simply normal to the surface and points toward the surface in the case of electrons. It has the mathematical form of $-|x|/x$ if the surface is at $x=0$.
- potential - which is piecewise linear, having the mathematical form of $|x|$
In the case of the electron-rich conductor I am speaking of, the field and potential can be revised due to the existence of a field from the curved geometry of the sphere. The field is $0$ within the sphere and a negative value just above the surface. The potential is constant within the sphere and linearly increasing (due to negative charge) just above the surface.
The density of surplus charge mathematically follows a Dirac-delta function around the surface. Let's give the surplus charge density the notation of $\rho(x)=-\delta(x)$ (still using negative because these are electrons). How do we resolve this absurdity? Surely it is absurd, because it would imply that the electrons pile up, saturate the conduction band, and would inhabit even higher energy levels. Here is a basic illustration for the conduction band:
I want to call attention to the y-axis, energy. Energy of what? This is the energy of the orbital (for a single electron I believe), which comes from physics that I do not understand myself, including the Pauli exclusion principle and quantum physics. For answering this question, however, I think we need a simple take-away and I will propose such a thing here. Since electrons are not bound to a single nucleus, we will do best to speak of the total charge density at some point. At a given point, the surplus charge has two potentials associated with it, these being the electrostatic potential (I'll call $V_{C}$) and what I will call the conduction band potential (I'll call $V_{B}$) (although I would appreciate someone proposing better terminology).
$$V_C(x) = \int_x^\infty k \rho(x') dx'$$
$$V_B(x) = V_B( \rho(x) )$$
I really don't know what this function for conduction band energy is like. I would imagine that for a conductor in a differential sense it would be linear (see image), and since $\rho$ is a measure of surplus charge it could probably be formalized as $V_B(x) = C \rho(x) $ where $C$ is a constant.
My point is that you can use the combinations of these two potentials to resolve the absurdity of the infinite charge pileup. What fundamental equation should we seek to satisfy? I suggest that every free charge in the conductor will assume the lowest energy state it can. Charges will always move to a lower potential, unless something is holding them back, that "something" is $V_B$, or the conduction band potential, quantum pileup pressure, or whatever we should call it. You can use some math to determine:
$$V_B(x) + V_C(x) = constant = V_C$$
Let $V_C$ by itself represent the potential in the center of the sphere
It's useful to note that $\rho(x)=0$ in the majority of the sphere, so for the majority of the sphere $V_B=0$, and this is why we don't often entertain this concept in basic physics classes (and why it's so hard to you and I to get straight answers about the question). This has an interesting implication that only a small thickness within close proximity to the surface has these interesting dynamics associated with it.
Now, if you combine all of the equations I have presented so far you can obtain a differential equation for the surplus charge density within the vicinity of the surface. I believe this results in a simple 1st order equation that yields a decreasing exponential. I find the boundary conditions a little difficult, because the surplus charge density beyond the surface is $0$, but the function is allowed to be discontinuous there, so I think the needed condition in place of a boundary condition would be $V(-\infty)=V_C$, which is easy to implement.
So my answer is that a conductor is not an equipotential surface if you consider the orbital/quantum effects. In the vicinity of the surface the potential will have the following general form if the surface is at $x=0$ and the conductor is on the -x side.
$$V(x) = \begin{cases}
V_C + c e^{\lambda x}, & \mbox{if} x<0 \\
V_C + c + d x, & \mbox{if} x \ge 0
\end{cases}$$
c and d are constants I don't know the value of
While writing this answer I stumbled upon the concept of the Stern layer. This may be what I'm describing, but I'm not entirely sure.
http://en.wikipedia.org/wiki/Double_layer_%28interfacial%29
This looks an awful lot like what I was describing. I think the Debye length might be useful as well, which seems to be the general scale over which these effects (of charge pileup) matter. So the sound-byte refinement of nibot's answer may just be that the potential of most conductors is very nearly constant because the Debye length is small relative to their total dimensions.
This isn't my area, but the question always drove me nuts personally, and I hope my mega-overkill answer is helpful to someone someday.
Another reference
If one was only interested in a quick qualitative picture (which is our entire readership), they might to better to disregard everything written in this answer and instead stare at this image from a professor at the University of Kiel, Dr. Helmut Föll.
They introduce the helpful terminology of the electrochemical potential, which I've already written about thoroughly up to this point. This potential is comprised on the electrostatic potential added to something else that I still can't find a name for! Nonetheless, the website gives the following formula for the nameless potential.
$$V_B(\rho) = \frac{kT}{e} ln \rho(c) $$
This reference also makes arguments for the linearization of the above function. Why? And what are the limits of this?
We may thus assume within a very good approximation that the carrier density at any point is given by the constant volume density c0 of the field free material, plus a rather small space dependent addition c1(x);
So yes, they did what I thought they did. What about the justification? Here:
As you might know, the Debye length is a crucial material parameter not only in all questions concerning ionic conducitvity (the field of "Ionics"), but whenever the carrier concentration is not extremely large (i.e. comparable to the concenetration of atoms, i.e in metals).
Bam. This is a very strong and well-reasoned argument. When we deal with electrostatics, the electron surplus numbers are extraordinarily small compared to total atom numbers. Even if the surplus is spread among an microscopic Debye length on the surface, it will still be vanishingly small compared to proton density (number per volume). Obviously, these statements break down in special cases, like semiconductors, which are specifically engineered to violate the assumptions laid out here. At that point, you'll head back to the first principles. However, I do have some hesitation with the $ln(\rho)$ form. My intuition is that it would apply specific to a single conduction band. Of course, all energy levels are quantized in the first place, but this is getting far more complicated than what we ever needed.
As a bit of a personal note, I always struggled with chemistry in my early studies. If 10 years ago someone had come up to me and said "hey Alan, chemistry is really about the balancing of electrochemical potential which includes electrostatic potential and other potentials that can be empirically quantified", I think I might be a chemical engineer right now. As it happened, however, I had a fantastic physics teacher and an okay chemistry teacher. It just goes to show the impact of teachers can have and the importance of using non-subjective arguments in the classroom!
Best Answer
Think of potential as like potential energy. If the mobile charges are electrons and they are mostly at rest they will mostly move towards lower potential energy which is higher potential.
So think of it as like a hill with electrons free to roll down hill (higher potential) until they get to a surface of the conductor at which point they are not free to move. So, you can imagine a fence built on the surface of the conductor that keeps things from leaving. And you can imagine some hills inside.
If the hills inside were flat, then indeed you could have some electrons at the edge and everything could sit there in statics. And if the hill inside was actually flat, then things can stay there.
There is a mountain called Mount Saint Helens. It looks like a normal mountain except the top looks like it was just chopped off (it actually exploded in a huge eruption in 1980).
The edge of the top is all at the same height; things on the edge would fall (and move away) if they were free to move. This is exactly what is going on in a conductor.
Now imagine you poured water on the top. The water is free to move if the pile of water gets higher than the region around. Real water has some surface tension but the point is that it can even itself come out because a region of higher water is free to move.
The same thing happens in the conductor. If you had an isolated positive charge, it would create a positive potential about itself which is a negative potential energy around itself. This would attract electrons.
So if electrons moved away from a positive charge leaving a positive charge all by itself, that would not be creating equilibrium because it would attract more electrons to it.
Charges can move. If a charge moved from the surface to another point inside and the charge there moved to a other point inside and so for until eventually a charge near another part of the surface moves to the surface, that is all fine.
So do that. Draw a bunch of surfaces of constant potentiality in red. Then draw a bunch of curves in red that are ways orthogonal to the surfaces, these are the field lines of the electric field. Now there is no force pushing in the direction of the equipotential and the force is all in the direction of the field.
Just because there is a force in a direction doesn't mean that charges go that way; that depends on the initial velocity of the charge. But imagine a region large enough to have many charges in it, the average velocity of the mobile charges could be zero because they bump into the non mobile charges except in the direction of the electric field. There they could have a non-zero average because they give and get from the non mobile charges but they consistently gain momentum in the direction of the electric field.
So now imagine a current everywhere pointing in the direction of the electric field. If you are in electrostatics then the field lines don't make any loops. So they can start at one part of the surface and end on another part of the surface. That is exactly how charge can flow.
This is important. This is saying that if all the charges were placed on the surface in a way that wasn't an equipotential and all the charges were kept at rest and were kept there for some reason so they are there in place long enough for say the speed of light to traverse across the whole object then there would then be electrostatic fields throughout the object.
If you now let go and let all the mobile charges move inside the conductor then the field at that moment is electrostatic and pointed so that current will flow through the body so that charge on the surface flows to other parts of the surface. And it never stops anywhere inside the object.
This is so frustrating. I've basically described how if you add charge to a conductor that has no net charge on the inside then the charge moves around in such a way that it continues to have no net charge on the inside.
I'm saying that your concern about the positive charge never happens in the first place. How rude. But there is some truth. If conductors start with the property of having no net charge in the body of the conductor then they can continue to have that property. That is essential for electrostatics and all statics in general.
I talked about their being no loops that really just means we can keep magnetism out of it (loops of electric fields is associated with changing magnetic fields). But we really had the electric field line go from one part of the surface to the other because there was no non-zero charge density to terminate on.
So what if for some reason you just made a conductor or just started getting to statics. Then there might be charge in the body. Now if you still have the current be proportional to the electric field you get for instance $\vec J=\sigma \vec E$ then you can take the divergence of both sides and get $$\sigma\frac{\rho}{\epsilon_0}=\sigma\vec \nabla \cdot \vec E = \vec \nabla \cdot \vec J$$
Where we used the Maxwell equation $\dfrac{\rho}{\epsilon_0}=\vec \nabla \cdot \vec E$ and we can also take the divergence of $$\vec \nabla \times \vec B=\mu_0\vec J+\mu_0\epsilon_0\frac{\partial \vec E}{\partial t}$$ to get the continuity equation
$$\vec \nabla \cdot \vec J=-\epsilon_0\vec \nabla \cdot\frac{\partial \vec E}{\partial t}=-\frac{\partial \rho}{\partial t}.$$
This means we have $$\frac{\partial \rho}{\partial t}=-\vec \nabla \cdot \vec J=-\frac{\sigma}{\epsilon_0}\rho.$$
Now we have stepped out of electrostatics but that is because we are talking about how it got that way. But now we see the charge density decreases when it is positive and increases when it is negative and in fact it approaches zero exponentially. So a conductor is never perfect you can think of it as something with a super huge $\sigma$ and so it gets super close to no charge in the body in almost no time.
Now, there is another reason I didn't go straight to this. There is no law of physics that says $\vec J=\sigma \vec E$ and in fact even materials that behave approximately like that have the value of $\sigma$ depend on the frequency of the change if it changes or even just depending on the temperature (and current changes the temperature).
If you placed a huge amount of charge deep inside a conductor by say shooting electrons going at almost the speed of light at it. Then yes the conductor might heat and buckle and strain and generally take some time to get that charge to the outside if it was a tremendous amount of charge. No real material is perfect.
But to the degree that the charges follow the electric field then when there is a charge density the field lines will be pointing the direction as to attract a countering charge.
Which is what we started with. If you had an imbalanced positive charge, it will attract mobile electrons to it and reduce the charge imbalance. If you have an excess of electrons in a spot they will push the neighbour mobile charges away from it and thus reduce the density in that region that contained them all.
And the continuous charge density is always for a region that contains many charges it isn't a thing that jumps up at every proton and down at every electron.
So it approaches zero by attracting more mobile charges into it or pushing more mobile charges out of it depending on whether the field diverges there in a positive or negative way and it does that based on whether there is a positive or negative charge density there.
If the equipotential surface wasn't everywhere tangent to the surface then it would dip inside. Sort of like if you divided your room in two and told your sibling not to cross to your half (TV sitcom style) and your brother or sister blew a bubble. If the bubble was tangent to the imaginary surface it wasn't supposed to cross; but if it isn't tangent to that surface then you can follow the surface into your side.
So the equipotential would cross inside the conductor. Which means you can follow it inside to the body of the conductor and from there you can look at the normal to the equipotential and you'll see a field line going there. That field line either terminates on a charge density (where the electric field lines terminate) or goes to the surface of the conductor. So it has two ends either both in the surface, both on densities inside or from a density inside to the surface.
If current moves in the direction of electric fields then current can flow between the two ends. If they are both on the surface then one part of the surface will have less charge at the spot that field line hit it and the the other place the field line hit the surface gains it.
If one of the ends is inside then that charge density will approach the zero exponentially fast and the current will be causing it so an opposite charge will head to the other end of the field line. So two opposite charge densities inside could be weakening each other or a charge imbalance inside can be sending its excess to the surface. And both. A place where field lines terminate can have field lines from multiple directions coming out of it and they could end on different places.
It might help if you drew the fields for two oppositely charged point charges then draw an arbitrarily surface around then so it contains them both. Where those field lines go to each other they are discharging each other. Where those field lines hit the surface you drew is where current is liking up charge on the surface. Eventually all the charge is on the surface. That's a good example to look at.
But the example in the last paragraph doesn't explain what happens when you place charge on the surface. So draw the two charges again. Draw the field lines again. And draw a new surface, but this time draw a new surface but make sure it passes through one of them and contains the other one inside. Again, the current can follow the field lines so the charge that starts out on the surface travels through the body towards the opposite charge and some travels through the conductor from that charge to where the field line terminates on the surface and some actually travels along the surface as a surface current where the surface is tangent to the field. The two concentrations of charge discharge exponentially and you end up with current spread out on the surface.
OK. Now draw a surface where it intersects both charges on its surface. Here is an example where all the charge is in the surface. It moves through the conductor and sometimes along the surface but here is the key. The current doesn't have a nonzero divergence inside the conductor so even though current flows through the body the body never acquires a nonzero charge density. So the charge stays in the surface always even though current flows through the conductor.
Make sure you can see that. Current can flow even if there is no net charge density. So when Feynman says the charge moves around the surface he doesn't mean that there is can't be current inside. But the charge imbalance stays on the outside if there was no change imbalance inside. Charge can flow inside as long as equal amounts flow in and out of every region in the body of the conductor then no imbalance will form.
And as we saw earlier if there originally was a charge imbalance inside then it decays away exponentially. This is how an ohmic material ($\vec J=\sigma \vec E$) behaves. A different material might behaves slightly differently but if it is a good conductor it will behave similarly and you can think of a perfect conductor as an ohmic material with $\sigma=\infty$ so the charge on the surface just flash gets to that final configuration that an ohmic material would go to.
And that's how to think of a perfect conductor. Just imagine a material with $\vec J =\sigma \vec E$ and find out the state it approaches after an infinite amount of time. Then imagine your object gets super close to that in a very short time. So basically imagine that $\vec J =\sigma \vec E$ for a huge $\sigma$ even if your material isn't as simple as having $\vec J =\sigma \vec E.$