Conductors and their charge?

Question

Why does excess positive charge stay on the surface of a conductor?

This is what I understood from: How does positive charge spread out in conductors?

and other resources on the web:

• If there is a electric field inside the conductor they will pull on the electrons
• Therefore there can be no field inside the conductor
• It follows from Gauss's Law that there are no charges inside

My questions:

• If there are positive charges inside the conductor, they will attract the electrons. But the electrons are already being attracted by the nucleus they belong to so why would they move? All electrons have electric fields already acting on them (the electric field of the nucleus) so why would adding new ones make a difference?

• If the positive charges are distributed on the surface, the field would only be zero right at the centre.The fields would cancel out in the centre because of symmetry but the field anywhere other than the centre would be non-zero. So how would the electrons be an equilibrium?

Please see the details on the bounty

Answer 1

The defining property of a conductor is that charge is free to move within it. Hence, if there existed an electric field within the conducting medium, charge would move until the field became zero. It follows that $\vec{E} = 0$ inside of a conductor.

Gauss's law therefore implies: $$ \rho=\epsilon_0\nabla\cdot\vec{E}=0, $$

since $\vec{E} = 0$ within the bulk of the conductor, all of the excess charge must reside on the surface.

To address your two questions specifically;

• In a metal, the electrons flow freely around like a fluid. They are not associated with any particular nucleus.
• The charges will do whatever they need to, in order to make the field zero inside. This defines how the charge acts on the surface. Your assumption that you know the charge distribution and from that you can determine the field is backwards.

Answer 2

Let's start with the definition of a "perfect" conductor:

A body in which electrons are free to move under the action of an electric field.

• Corollary-1:Under electrostatic conditions the field inside a conductor must be zero.

  Suppose on the contrary,that a non-zero field indeed existed inside a conductor.Then the field would act on the electrons moving them until the electrons themselves produce a field to cancel the external field.Since under electrostatic conditions there is no motion of charges,the field in the bulk must be zero.

• Corollary-2:The charge density at any point inside a conductor must be identically zero.This implies any excess charge must be on the surface.

  This is simple.According to Gauss' Law: $$\nabla\cdot \mathbf{E} = \frac{\rho}{\epsilon}$$

  Since $\mathbf{E}$ is zero inside the conductor,the charge density is also zero.

Now returning to the given problem.Suppose I have a perfect conductor.Now by some mechanism I remove some electrons from the conductors thereby giving the conductor a positive charge.

The charge now redistributes as follows:

• If there are positive charges inside the conductor, they will attract the electrons. But the electrons are already being attracted by the nucleus they belong to so why would they move? All electrons have electric fields already acting on them (the electric field of the nucleus) so why would adding new ones make a difference?

Answer Yes,the nucleus attracts the electrons also.But conductors are such substances in which the this force is not so strong compared to external forces due to the external field.As a result the electrons are loosely bound and free to move.

However in insulators,this force of attraction of the nucleus is strong and the electrons are not free to move.

• If all the charges are on the surface,then how is the conductor in equilibrium?

Answer: On a conductor, a surface charge will experience a force in the presence of an electric field. This force is the average of the discontinuous electric field at the surface charge. This average in terms of the field just outside the surface amounts to: $$ P =\frac{\sigma^2 }{2\epsilon} $$ where $\sigma $ is the surface charge density.

This pressure tends to draw the conductor into the field, regardless of the sign of the surface charge.But at the same time the mechanical forces(mechanical stress) of the conductor counters this electrostatic pressure and the conductor stays in equilibrium.

Answer 3

This answer is physically the same as the accepted answer, but with a few more formulas.

The only assumption you need is the local conservation of charge (intuitive) and the Gauss's law. Consider the continuity equation for charge: $$ \frac{\partial \rho}{\partial t}+\nabla.\mathbf J=0$$ $$\mathbf J=\sigma \mathbf E \tag{ definition of conductors}$$

$$\,\,\,\,\rho=\epsilon\nabla.\mathbf{E} \tag{ Gauss's law}$$ so we find that $$\rho(\mathbf r, t)=\rho_0(\mathbf r)e^{-\sigma t/{\epsilon}} $$

Assuming a good conductor ($\sigma\gg 0$), this relation tells us that the charge density leaves the volume very quickly (instantly for a perfect conductor). Because of the global conservation of electric charge, the charge can not just disappear, and it will reside on the surface of the conductor.

Answer 4

Let's address your second question. Suppose there is a neutral conductor in the beginning. Now you transfer some positive charge to it. Lets zoom in to see what happens to a small charge element.

1. It experiences the electric field of the constituent electrons and protons.
2. It experiences electric repulsion from the other "charge elements".

Since the conductor is neutral, we can say that (on an average) the positive charge density (due to the constituent protons) and negative charge density (due to the continuously moving electrons) cancel each other at each and every point on the bulk of the conductor. So the net effect of [1] is zero.

Now lets see how the remaining portion of the excess charge (free charge) affects the charge element in question. Because these are like charges, they induce repulsion. The infinitesimal charge element feels a repulsive force and tries to move away from the initial position. But remember: this is happening with each "charge element". So, it is a nonlinear process: each element moves away to under the repulsive force, this redistributes the remaining charge distribution, which in turn modifies the repulsive force on each individual element, etc. How long does this continue? Until the free charge has redistributed itself (depending upon the shape of the conductor) so as to attain a minimum energy configuration. Once this configuration is reached, we may deduce the following:

1. There is no free charge in the bulk of the conductor.
2. There is no net electric field at ANY point (not just the center) inside the conductor.

If there were any non-constant free charge distribution inside, the potential inside would vary from point to point. And it would be possible to lower the energy of the configuration further by transferring charge from some point at higher potential to a point at a lower potential. Since, energy COULD be lowered, this suggests that there WOULD be a tendency for doing the same: i.e. there would be a non-zero electric field at each point. Hence, this particular configuration would be prone to redistribution. Actually it is really simple, as long as there is an unbalanced force at any point of a particular charge distribution, the charge will rearrange itself. And it will continue to do so until none of the points have a finite electric field: i.e. it reaches the minimum energy configuration.

Answer 5

In a conductor electrons are losely bounded to any single nuclei as they are significantly influenced by multiple nuclei at the same time, this enables the electrons to move from one nuclei to another with ease and this easiness is the reason we call them free electrons.

So what happens when we pull out some electrons from somewhere and give positive charge ?
The positive charges repel each other and attract electrons. As the electrons are not very tightly bound to any particular nuclei they move towards the positive charge. What happens is that if the positive charge was at center, electrons start moving in radially and leave positive charge (absense of electron) from where they come, this in turn attracts the electrons further away from centre and process continues untill charge reaches the surface.

So why field everywhere become 0, when charges are uniformly distributed ? Fist of all an important point is that since inside their is equilibria of nuclei and electron they do not have net field of their own.
Now suppose a spherical shell, if you cut it from middle, you have 2 equally charged hemispheres which make field zero at centre; however if you cut it from any other location you will have two unequally charged parts which cancel out the field at the respective location.

Now to answer you query from bounty.
Lets say you have a neutral conducting sphere, and you introduce positive charge at its center. Electrons move in radially untill the positive charge reaches the surface. Since no net charge is left inside the surface of the sphere we get a zero magnitude electric field inside the sphere in accordance with gauss law.
The zero magnitude electric field can also be understood as if on any particular point some charges are close then at more distance so many more charges exist that electric field becomes zero. Equal charges exist at equal distance with respect to the centre and unequal charges exist at unequal distances for other points.