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Since the electric field inside a conductor is zero that means the potential is constant inside a conductor, which means the "inside" of a conductor is an equal potential region.

Why do books also conclude, that the surface is at the same potential as well?

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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:

Electron bands

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.

Debye length

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.

The idea

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!

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This result can be understood mathematically. Suppose the system has reached equilibrium and all charges have stopped moving so that electrostatics applies. Then the potential is a harmonic function $\Delta \varphi = 0$ in $\mathbb{R}^3$ and the conductor is a closed and bounded region in $\mathbb{R}^3$.

A general property of harmonic functions is the maximum principle. In short, if $\varphi$ is harmonic on an open set $\Omega$ and $B \subset \Omega$ is a closed and bounded region, then $\varphi$ has no local minimum or maximum in the interior of $B$ and the absolute maximum and minimum of $\varphi$ occur on the boundary of $B$. In particular, if $\varphi$ is constant on the interior of $B$ it must be (the same) constant on the boundary of $B$.

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Please explain the last line: "In particular, if φ is constant on the interior of B it must be (the same) constant on the boundary of B." Also can you suggest me a lucid reference on potential theory? Physics books are making me sick. – omephy Apr 18 at 3:44

The change in potential between two points is

$$ \Delta V = \int_a^b \mathrm{d}\vec{\ell} \cdot \vec{E} $$

but inside the conductor $\vec{E} = 0$ so that integral between any two interior points is also zero, accordingly the interior is all at the same potential.

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Yes I know that, that is what I have said already, I am ok with that part. The inside of the conductor is an equipotential region. I am asking about why the surface of the conductor is at the same potential as the interior ? – Revo Mar 24 '12 at 22:14
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. – nibot Mar 24 '12 at 22:19
The surface charge is still free to move about the conductor, so approaching the surface from the inside the above argument still applies. If you want to be pedantic let $a$ lie in the interior and let $b$ approach the surface from the inside and take the limit. – dmckee Mar 24 '12 at 22:22
@dmckee Don't take this personally, but I don't understand your comment, and I think there's just not enough space here to clearly lay out the specifics of what you mean by your limit approach suggestion. – Alan Rominger Mar 25 '12 at 4:24
@AlanSE: This answer just lives in PhysicsLand (tm) where the surface layer is infinitesimal. That is it ignores all the detail that you've written so nicely about. – dmckee Mar 25 '12 at 17:48

Because if the surface is not equipotential then it would mean that there is a tangential component of electric field along the surface.

This component will result in motion of electrons, but since we have static fields this is not possible. Thus by contradiction we can say that surface must be equipotential.

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Hello again! Usually it's better to not answer if you don't have anything different to add-- the other answers cover this in depth.. The entire content of your answer is contained in the quoted section of AlanSE's post, so there's nothing new here. Instead, you can upvote the posts which you like. – Manishearth Apr 5 '12 at 15:04
Sorry for repetition, I should have read other answer too. I sure will keep that in mind next time. – anuragsn7 Apr 5 '12 at 15:08
I'm more curious how the charges create a field tangential to the surface O_O – Alan Rominger Apr 5 '12 at 15:09

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