We always study in our textbooks that when a conductor object is charged in a electrostatic equilibrium state, the charges move out to the surface and are stationed there. But which part of the surface are they exactly at? Is it right beneath the surface or on the outer side? And why?
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1 Answer
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Clasically speaking, you can imagine the charge as being "just inside" the surface (mathematically think of a surface charge as a boundary condition). If you want to be very specific, quantum mechanically speaking the electrons (let's assume a net negative charge) are more likely to be found where their energy is lower, and less likely to be found where their energy is higher.
Being "outside" the surface (i.e. just outside or slightly above the atomic surface) requires a lot of energy because you have to overcome the work function of the material, so the probability there is very low. Being closer toward the middle of the object (farther away form the surface) also requires more energy because of the electrostatic repulsing from the other excess electrons (the other negative charge), and so that is also less likely. Therefore, the wave function (in the position basis) will be sharply peaked just under the surface, and very rapidly decay as you move away from the surface and away from the object, and also decay (at a rate depending on the amount of charge) as you move away from the surface and toward the center of the object.
Also remember that for irregularly shaped objects (i.e. not spheres), the charge will not be uniformly distributed across the surface either. The charge will distribute itself such that all electric field lines intersect the surface orthogonally without any tangential component (because such a tangential component of an E-field would then push surface charges along the surface causing them to redistribute until there were no tangential E-field components). This causes excess charge to tend to "bunch up" at sharp features on the surface such as corners.
