7-6-00

Sections 16.9 - 16.10

A conductor is in electrostatic equilibrium when the charge distribution (the way the charge is distributed over the conductor) is fixed. Basically, when you charge a conductor the charge spreads itself out. At equilibrium, the charge and electric field follow these guidelines:

- the excess charge lies only at the surface of the conductor
- the electric field is zero within the solid part of the conductor
- the electric field at the surface of the conductor is perpendicular to the surface
- charge accumulates, and the field is strongest, on pointy parts of the conductor

Let's see if we can explain these things. Consider a negatively-charged conductor; in other words, a conductor with an excess of electrons. The excess electrons repel each other, so they want to get as far away from each other as possible. To do this they move to the surface of the conductor. They also distribute themselves so the electric field inside the conductor is zero. If the field wasn't zero, any electrons that are free to move would. There are plenty of free electrons inside the conductor (they're the ones that are canceling out the positive charge from all the protons) and they don't move, so the field must be zero.

A similar argument explains why the field at the surface of the conductor is perpendicular to the surface. If it wasn't, there would be a component of the field along the surface. A charge experiencing that field would move along the surface in response to that field, which is inconsistent with the conductor being in equilibrium.

Why does charge pile up at the pointy ends of a conductor? Consider two conductors, one in the shape of a circle and one in the shape of a line. Charges are distributed uniformly along both conductors. With the circular shape, each charge has no net force on it, because there is the same amount of charge on either side of it and it is uniformly distributed. The circular conductor is in equilibrium, as far as its charge distribution is concerned.

With the line, on the other hand, a uniform distribution does not correspond to equilbrium. If you look at the second charge from the left on the line, for example, there is just one charge to its left and several on the right. This charge would experience a force to the left, pushing it down towards the end. For charge distributed along a line, the equilibrium distribution would look more like this:

The charge accumulates at the pointy ends because that balances the forces on each charge.

A clever way to calculate the electric field from a charged conductor is to use Gauss' Law, which is explained in Appendix D in the textbook. Gauss' Law can be tricky to apply, though, so we won't get into that. What we will do is to look at some implications of Gauss' Law. It's also a good time to introduce the concept of flux. This is important for deriving electric fields with Gauss' Law, which you will NOT be responsible for; where it'll really help us out is when we get to magnetism, when we do magnetic flux.

Electric flux is a measure of the number of electric field lines passing through an area. To calculate the flux through a particular surface, multiply the surface area by the component of the electric field perpendicular to the surface. If the electric field is parallel to the surface, no field lines pass through the surface and the flux will be zero. The maximum flux occurs when the field is perpendicular to the surface.

Even though we won't use this for anything, we should at least write down Gauss' law:

Gauss' Law - the sum of the electric flux through a surface is equal to the charge enclosed by a surface divided by a constant , the permittivity of free space.

What is the permittivity of free space? It's a constant related to the constant k that appears in Coulomb's law. The relationship between the two is this:

Gauss' Law is a powerful method of calculating electric fields. If you have a solid conducting sphere (e.g., a metal ball) that has a net charge Q on it, you know all the excess charge lies on the outside of the sphere. Gauss' law tells us that the electric field inside the sphere is zero, and the electric field outside the sphere is the same as the field from a point charge with a net charge of Q. That's a pretty neat result.

The result for the sphere applies whether it's solid or hollow. Let's look at the hollow sphere, and make it more interesting by adding a point charge at the center.

What does the electric field look like around this charge inside the hollow sphere? How is the negative charge distributed on the hollow sphere? To find the answers, keep these things in mind:

- The electric field must be zero inside the solid part of the sphere
- Outside the solid part of the sphere, you can find the net electric field by adding, as vectors, the electric field from the point charge alone and from the sphere alone

We know that the electric field from the point charge is given by kq / r^{2}. Because the charge is positive, the field points away from the charge.

If we took the point charge out of the sphere, the field from the negative charge on the sphere would be zero inside the sphere, and given by kQ / r^{2} outside the sphere.

The net electric field with the point charge and the charged sphere, then, is the sum of the fields from the point charge alone and from the sphere alone (except inside the solid part of the sphere, where the field must be zero). This is shown in the picture:

How is the charge distributed on the sphere? The electrons must distribute themselves so the field is zero in the solid part. This means there must be -5 microcoulombs of charge on the inner surface, to stop all the field lines from the +5 microcoulomb point charge. There must then be +2 microcoulombs of charge on the outer surface of the sphere, to give a net charge of -5+2 = -3 microcoulombs.