Applying Gauss's Law Several Times

Bashyboy

1. The problem statement, all variables and given/known data
I am working on a problem that is similar to this one:

http://www.physicsforums.com/showthread.php?t=598914



2. Relevant equations



3. The attempt at a solution
When r < a, why doesn't the electric fields generated by the concentric conducting, hollow sphere affect the Gaussian surface at that point??

Why is the electric field at a point in the concentric conducting, hollow sphere zero? After all, it doesn't specify that it is in electrostatic equilibrium.

Also, what is the difference between the entire system being in electrostatic equilibrium and just the concentric conducting, hollow sphere being in electrostatic equilibrium? Can an insulator be in electrostatic equilibrium?

rude man

You will probably never have to worry about electrostatic disequilibrium. You were given that the insulating sphere had a uniform charge density. Therefore it's in equilibrium.

Bashyboy

So, you are saying that if I was to generate a Gaussian surface inside of the conductor (say, a sphere), then the flux through this surface is zero, because the electric field through this surface is zero? How can this be? I know that the charge isn't distributed inside of the conductor, the charge resides on the surface, but the Gaussian surface would still enclose the insulator, which has a charge.

EDIT: Something just came to my mind: does the insulator induce the electro-static equilibrium? And if the insulator had a positive charge, it would cause the electrons in the conductor to move to the surface closest to the insulator? Leaving the other surface "less" negative?

Doc Al

Yes.

Charge can rearrange itself on the inner surface of the conducting shell to cancel any charge placed within the hollow of the sphere.

Yes. Any charge placed inside the conducting sphere will induce a charge on the inner surface that cancels its field within the conductor (and for all points beyond).

Bashyboy

So, using the diagram given in the link, if I was to place a Gaussian sphere directly between the insulator and conductor, the electric field lines emanating from the insulator will "hit" the sphere from the inside, and the electric field lines emanating from the conductor "strike" the very surface of the Gaussian sphere; and so, the electric fields combine to cancel out?

And is this the reason why we don't have to consider the electric field of the conductor when generating a Gaussian surface that meets the condition of r < a, because the electric fields lines never even make it that far?

EDIT: Also, is there a way of quantifying the amount of induction that the insulator affects? Would it just be equal and opposite to the charge of the insulator? And the remaining charge would reside on the outside?

Bashyboy

For example, say the insulator has a charge of Q1 = 3.60 C, and the conductor had a total charge of -2.00 C. Would the inner surface of the conductor would have a charge of Q2 = -Q1 ---> Q2 = -3.60 C; and would this then imply that Q3, the outer surface of the conductor, would have a charge of Q2 + Q3 = -2.00 C ----> Q3 = 1.6 C?

Doc Al

That is correct.

Bashyboy

Why is it correct, if you don't mind me asking?

Bashyboy

Is there some law or principle of induction?

Is it possible for someone to answer these questions I have from an earlier post?

SammyS

The reason that we know the charge on the inner surface has equal magnitude and opposite sign of the charge on the inner charged sphere, is that if we use a Gaussian surface that lies within the conducting material, such as a sphere with radius between b & c, the E field is zero everywhere on the Gaussian surface so the net charge within the Gaussian surface is zero.

Bashyboy

So, would electric field at a radius of a < r < b be zero?

EDIT: And would it be zero because the number of electric field lines radiating out of the Gaussian surface equals the number of electric field lines coming in?

rude man

No. Put a Gaussian spherical surface anywhere on a < r < b and what is the charge inside that surface? And so what does Gauss say about the E flux thru that surface?

Doc Al

No. The field there would equal that from the charged insulator. The induced charge on the inner surface of the conducing shell will have no field for r < b.

Bashyboy

So, the flux through a Gaussian only depends on the charge enclosed, further implying that it only depends on the electric field emanating from that enclosed charge?

SammyS

Whoa !

E field lines originate from Positive and terminate on Negative charge. The negative charge on the inner surface of the conductor does not interfere with the electric field at a < r < b .

So the field at a radius of a < r < b is not zero. Gauss's Law says the flux though a sphere of such radius is not zero.

Bashyboy

Why won't it have a field there?

Doc Al

The total flux calculated over the entire closed Gaussian surface depends only on the net charge enclosed within that surface. (That's what Gauss's law says.) Note that the field at any given point on the surface may well depend on other charges.