Understanding the physics of induced charge over a perfect conductor

1. Mar 2, 2013

fluidistic

Understanding the physics of "induced charge" over a perfect conductor

When we put a perfect conductor, let's say a sphere inside an external electric field, there will be a surface charge distribution different from 0 even though the sphere is electrically neutral.
I don't really understand: I have a sphere totally neutral, so with NO free charges. I put it inside an electric field and then suddenly some "free charges" will move up to the surface and distribute in function of the electric field. How can this be?
I'd understand if I had initially a non neutral conducting sphere with a total net charge Q (evenly distributed over the sphere, in other words $\sigma (\theta , \phi ) = \sigma _0 \neq 0$). But when there's no free charges, how can "free charges" go over the surface of the sphere?
I'd understand if you'd call those charges "charges of polarization" but they are not. They are not "bound charges" (according to Griffith's terminology if I'm not wrong), they are "free charges".
Can somebody explains me what's going on?
Thank you.

2. Mar 2, 2013

Simon Bridge

"electrically neutral" is not the same as "no charge". It means "zero net charge" - which is to say that there are exactly equal numbers of positive and negative charges.

In an electric field, the positive charges are pushed one way and the negative charges are pulled the other way - separating them.

IRL: matter is composed of positive protons and negative electrons. In a conductor, some of the electrons from each atom are only very loosely bound to their atoms so they are easy to shift with an electric field. It's what makes them conductors.

3. Mar 2, 2013

fluidistic

Yes I know. I just find odd that a neutral conductor still has "free charges".
How does this differ from polarization?
Because in the case of the conductor, there is no polarization. Basically E=D or something like that because P=0.

Again, how does this differ from polarization?
Thanks again!

4. Mar 2, 2013

Simon Bridge

And yet you've jest seen that it can have free electrons capable of moving against a background of positive protons. There are a lot of things in physics you will find odd - you get used to it.
In polarization (in dielectrics), the separation is an average displacement within the atom - the electrons are still bound to the atom. In a conductor, we are talking about an electron from one end of a wire being so loosely bound as to be able to travel to the other end of the wire. Each positive ion in the wire, without an applied external electric field, is only paired off with an electron on average. So, IRL, conductors are only electrically neutral on average.

Apart from that, the physics is exactly the same as polarization in a dielectric.

A separation of charges is polarization in general. What you are thinking of as "polarization" is the specific case where it occurs in dielectrics: polarization on a microscopic scale. In a dielectric, the polarization field does not completely cancel the applied electric field. In a conductor it does.

5. Mar 2, 2013

fluidistic

I see, thank you very much Simon for this neat explanation.
So in my case of a neutral spherical conductor that is put inside an external E field, if there's a slight positive free charge on the surface somewhere, there must be a slight negative charge somewhere else though still on the surface so that the total (integral of sigma over the surface of the sphere) charge density remains worth 0, or the net free charge of the conductor in general.

And I believe that if I have a conducting sphere inside both an external field and dielectric(s) medium, there will be both free charges and polarized charges at the surface of the sphere. So in that case $\sigma = \sigma _{\text{pol. or bound}}+ \sigma _{\text{free}}$. Since the surface is at an equipotential, $\sigma$ must be a constant, right?
And in that particular case, the bound charges are from the conductor itself or from the dielectric(s)? I'm guessing the latter. So in reality it's the dielectric that gets polarized right at the surface of the conducting sphere, not the conducting sphere itself?

6. Mar 3, 2013

Simon Bridge

In a cubic centimeter of copper there is about twice avogadros number of free electrons ... is that "slight"? It is balanced by the same number of otherwise unpaired protons in the metal lattice.

A net charge of zero can still have billions of coulombs of both positive and negative charge available to be moved around by electric and magnetic fields. I think this is what you need to grasp.

In the classical model you are using, a perfect, ideal, conductor - by definition - has unlimited positive and negative charges exactly balanced. The effect of an applied field is to change the distribution of charge across the surface of the conductor.

If the conductor is inside a dielectric medium, then the effect of the dielectric's polarization is to reduce the applied electric field experienced by the conductor.

A dielectric inside a closed conducting surface will not be polarized because the field there is zero.

The surface charge distribution of a conductor in an electric field need not be uniform - if it did then the question in post #1 would not come up.

7. Mar 3, 2013

fluidistic

I see. What I wanted to point out is that if sigma is positive in some region on the surface of the neutral sphere, then it must be negative in some other region but still over the surface. All this regardless if sigma is slightly under/above 0 in a region or really "big". But I realize that quantifying sigma here is maybe meaningless.

Oops, you are right. I'm trying to glue my understanding.
I had in mind that the electric field right outside a conductor is normal to the surface, and I think this is true no matter what the distribution of charge sigma is over the surface. However when one starts to get slightly above the surface, E needs not to point away the surface, depending how sigma is.
If I'm lucky enough to have the total sigma worth a constant then the E fields lines are straight lines pointing away/toward the surface of the conductor, exactly where they start/end. In that case and only in that case, if I have a test charge, it would fall over or go away from the surface of the conductor following a straight line.

2 questions:
1)If I put a conductor inside a dielectric and I apply an external E field so that there's a surface charge density over the surface of the conductor. The dielectric will get polarized with bound charges right over the conductor, right? In fact the bound charges of the dielectric are so close to the surface of the conductor that we model both the surface free charges of the conductor and the bound charges of the dielectric to be at the surface of the conductor while in reality only the free charges of the conductor are at the surface. Is this right?
2)I would like to model a perfect conductor as a dielectric. In that case, the susceptibility must be equal to infinity?
Thanks for all so far.

8. Mar 3, 2013

Simon Bridge

You can have positively charged conductors ... however, in a net neutral conductor, a region of positive charge some place does imply a net negative charge elsewhere... which implies there must be an external field.

... depending on the external fields too.
No - IRL materials are not continuous ... in the model classical you are using, the surface of the dielectric is considered infinitesimally close to the surface of the conductor. There is no advantage to including an infinitesimal layer of dielectric as part of the surface of the conductor. We model the bound charges of the dielectric as a polarization field.

... if the susceptability were infinite, then the displacement field would also be infinite.
http://en.wikipedia.org/wiki/Dielectric#Electric_susceptibility
... what is the displacement field inside a conductor?

There are limits to the models you are exploring.
Later you will find out about the band-theory of solids.