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We can think the charges go to the outer surface of a conductor to minimize the electrostatic potential energy of the system. We can check this using a simple calculation using a charged sphere.

A uniformly charged sphere would have $20\,$% more energy than that of a very thin spherical shell with the same radius and the same charge as in the former case. As the potential energy would be less if we distribute charges uniformly over the surface rather than distributing the charges uniformly throughout, we can say, the conductor prefers having charges on the outer surface. Here's my question then:

Suppose I have a conducting sphere on which a charge $Q$ is to be distributed. Now, suppose I divide the charge into two equal parts, $Q/2$ each, and place them at two diametrically opposite ends of the sphere. Now the potential energy of the configuration becomes $75\,$% less than that for charges distributed uniformly over the surface. Then why doesn't the conductor prefer this kind of distribution which can minimize the potential energy further?

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    $\begingroup$ Because there is a potential energy associated with forcing a bunch of electrons close together to form the two "point" charges: I presume you left that out of the calculation. $\endgroup$ – NickD Aug 25 '17 at 14:49
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    $\begingroup$ Like charges repel, so the charges on the sphere get as far away from each other as they can. This leads directly to a uniform charge distribution on the outer surface of a spherical conductor. $\endgroup$ – David White Aug 25 '17 at 14:52
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If there is any net charge inside the bulk of the conductor Gauss's law guarantees that there is some electric field lines with directions towards the outside of that region.

Any such electric field inside moves charges around inside, so it will only stop when there is no way the charges can move any. That guarantees zero electric field inside, and charges on the surface distributed so the tangential electric fields must be zero. Only fields perpendicular to the surface can survive.

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If you put two equal point charges, say two nuclei with $+|e|Z$, at the opposite points of the conducting sphere, they will stay there. The free electrons of the conductor will start to move due to attraction to these charges until the friction losses (=resistance) dissipate the extra energy into heat. Otherwise there will be current oscillations dissipating due to radiative losses.

In case of negatively charged nuclei the situation will be similar - the electrostatic forces will move free charges until the minimum of the potential energy is reached, the extra energy being lost to heating or to something else.

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I was trying to figure out why charges accumulate perfectly at the surface rather than just mostly at the surface (in my mind it seemed way too crowded at the surface with charge for any more to push its way in...), and I came up with this proof.

First recall that in a perfectly conducting object with excess charge, the charges will move until the electric potential in the object is the same everywhere.

Second recall that the electric potential is the integral of the electric field with respect to distance. This means the electric field in any direction is the derivative of the electric potential in that direction.

Let's call the electric potential field at a point x,y,z in the conductor $V(x,y,z)$ Now, $$E_x(x,y,z) = \frac{d}{dx}V(x,y,z)$$ $$E_y(x,y,z) = \frac{d}{dy}V(x,y,z)$$ $$E_z(x,y,z) = \frac{d}{dz}V(x,y,z)$$

Since the potential is the same everywhere inside the object, all these derivatives are 0, and hence, the electric field is zero everywhere inside the object.

Now, Gauss's law (differential form) says $\nabla \cdot E = \frac{\rho}{\epsilon_0}$ The divergence of the E field is the sum of all three of its spatial derivatives. Since we've just established that the electric field is zero everywhere inside, this means all it's derivatives are zero as well. This means the $\frac{\rho}{\epsilon_0} = 0$ and hence the charge density is zero everywhere inside the object. Note this formulation does not work at the surface, where the charge density will be infinite.

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  • $\begingroup$ 3D charge density is not 0 on the surface, but diverges to infinity: for positive charge, $\frac{\sigma a^2}{a^3}\to + \infty$. $\endgroup$ – Ján Lalinský Jan 22 at 20:40
  • $\begingroup$ @JánLalinský Thanks. Updated the answer accordingly. $\endgroup$ – user234683 Jan 25 at 2:16

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