Two questions regarding conductors and their properties.

[flyingpig]

Homework Statement

I am so close to this breakthrough I can taste it. I say this because I haven't taken a shower in four days. Anyways go to this link http://physics.bu.edu/~duffy/py106/Electricfield.html

My questions

1)

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

This part. OKay so a conductor is charged, all the excess charges are on the surface canceling with whatever charge that is inside. Meaning all the free electrons inside cannot move because F = qE and F = q(0) = 0.

I don't understand, so what if F = 0? F = 0 could also mean the electrons would move under constant speed right?

Also when they say the surface charges cancel with the inner charges, do they mean vectorially? Because if I have a positive charge inside and some negative charge on the surface, then clearly a field exists between them. If that exists then how can it be 0?

 

[cepheid]

I am so close to this breakthrough I can taste it. I say this because I haven't taken a shower in four days.

Thanks for sharing dude!

This part. OKay so a conductor is charged, all the excess charges are on the surface canceling with whatever charge that is inside.

I'm not sure what you mean by 'cancelling with.' I mean, your quoted statement said there was a NET charge on the sphere. Net means surplus. So there must be some unbalanced charge present. Otherwise the total charge would just be zero, not Q.

Meaning all the free electrons inside cannot move because F = qE and F = q(0) = 0.

Of course free electrons can move in a conductor. That's why we call them "free" electrons. If there is no electric field present, all this means is that there are no forces acting on those electrons. It doesn't mean that they don't move. In any case, I don't see what this has to do with anything or how you arrived at it.

I don't understand, so what if F = 0? F = 0 could also mean the electrons would move under constant speed right?

See above.

Also when they say the surface charges cancel with the inner charges, do they mean vectorially? Because if I have a positive charge inside and some negative charge on the surface, then clearly a field exists between them. If that exists then how can it be 0?

I'm not totally clear on what "they" are saying, but I think the point is that the surface charges are free to arrange themselves in such a way that the field inside the conductor is zero. If it's not zero, charges will be accelerated by the field until they are redistributed in such a way that it becomes zero.

 

[Phrak]

Meaning all the free electrons inside cannot move because F = qE and F = q(0) = 0.

You should really clarify (as should the author you quoted) whether you mean a conducting solid ball or a conducting shell.

If it is a ball, there is no free charge in the interior. Gauss' law is an idealization.Charge doesn't come in units of particles unless you additionally hypothesis it. If we consider all charges to come as particles you could have at most one charged particle in the interior, but we still have to pretend all the other particles on the surface of the ball are smeared out like a fluid.

 

[flyingpig]

I'm not sure what you mean by 'cancelling with.' I mean, your quoted statement said there was a NET charge on the sphere. Net means surplus. So there must be some unbalanced charge present. Otherwise the total charge would just be zero, not Q.

But then what is the point of charging it if the conductor is already going to have a net charge? Like I have a conducting sphere, inside has more electrons than protons and therefore the net charge is negative, if we already know that why are we charging it?

Of course free electrons can move in a conductor. That's why we call them "free" electrons. If there is no electric field present, all this means is that there are no forces acting on those electrons. It doesn't mean that they don't move. In any case, I don't see what this has to do with anything or how you arrived at it.

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.

You say that free electrons move, but this site say they don't. I don't even understand what they mean by distributing themselves so that the E-field is 0. This concept has been bothering me for weeks now. In my textbook, it says this occurs when an external field is applied to a conducting slab to create a field of 0

This is what my textbook did

[PLAIN]http://img651.imageshack.us/img651/9509/92678195.jpg

Now I am assuming that a zero field inside meant that this will happen, where there exist a field inside, there is also an external field outside therefore the net field is 0 and therefore there is no field inside.

[PLAIN]http://img403.imageshack.us/img403/6589/52447869.jpg

Now here is the thing, why can't the following scenario happen?

Scenario A

Since there is a E-field generating, that means a positive charged thing must be producing this E-field, so why can't the E-field terminate when it touches the negative charge? Or does it go further? But that is assuming the positive charge has a greater charge such that it can go pass through the slab

[PLAIN]http://img843.imageshack.us/img843/4670/76394506.jpg

Now my book also said this

As the electrons move, the surface charge densities on the left and right surfaces increase until the magnitude of the internal field equals that of the external field, resulting a net field of zero inside the conductor

What does that mean? What do they mean "increase", what if there is not enough electrons to make that happen?

Scenario B

In all my HW problems where they want you to say that the E-field is 0 for a conductor, they say it like this.

A spherical solid conductor has a net charge of -25uC (an excess of electrons) and has a radius of 10cm. What is the E-field at a) r = 5cm, b)r = 15cm

Okay so this is a conductor, a textbook method of answering would say this is zero because this is inside a conductor, almost just memorizing. But according to the book, this happens only if there is an external E-field to make that happen, but in problems like these it doesn't even say it. Perhaps when a conductor is charged, we immediately assume that it must be in electrostatic equilibrium? But if the charges are "spread out" as we want to believe, that still doesn't mean the E-field is 0 since there is no external E-field to make the inside a zero net field

So perhaps something like this would happen. You got all the negative charges on the surface and some positive charges inside. Clearly there is field and notice the charges have been distributed uniformly (so it was once charged), but once the thing we used to charge is removed (the external field), the field inside wouldn't be zero anymore

[PLAIN]http://img834.imageshack.us/img834/9910/58991724.jpg

 

[flyingpig]

Also, if there are no charges inside a conductor (and hence no field), how come they say that

There are plenty of free electrons inside the conductor

Why!? If there are free electrons to cancel out the protons, then there must be a field.

 

[flyingpig]

Wait can I argue it this way? I'll take a shower and shave tonight if I get this

[PLAIN]http://img703.imageshack.us/img703/7849/85250004.jpg

Alright the dotted sphere is my Guassian Surface and the giant good looking sphere is a conductor. Charges are spread out and therefore it must in electrostatic equilibrium. This concept i didn't understand before because I keep thinking there must be an external field to make that happen - to make the E-field inside a 0.

So here is the math that really explain EVERYTHING

\frac{\sum _{i=1}^{n} Q_{en_{i}}}{\epsilon_0} = \oint \vec{E}\cdot \vec{dA}

Where n (on the summation) is the number of I guess protons and electron charges divided out in coulombs. Since all the excess charges go ON the surface leaving all the remaining charges inside where (and I quote) the charges inside will rearrange themselves such that the E-field is zero meaning inside vectorially they will add up to zero

So we have

\frac{\sum _{i=1}^{n} (-Q) + Q + (- Q)+ Q + (- Q) +...+ Q}{\epsilon_0} = \oint \vec{E}\cdot \vec{dA}

So basically the charges add up and subtract almost like a telescoping series and therefore E = 0 enclosed