Why Does the Outer Surface of a Conductor Reach 3/2 Q in Gaussian Law Problems?

[IvanSaurus]

Hi there, I have following problem at hand:
(any help would be appreaciated, I have a test tomorrow)

1. Homework Statement

Homework Equations

The Attempt at a Solution

Above is obviously the teachers answer to the problem, what I don't get though; is, where does the 3/2 come from, at r > R2? And how did he get to the conclusion that the electric force is exactly 3/2 away?
Thanks in advance!
- on a sidenote, why does almost every ElectroDynamics question not involve any real cases/numbers /frustration out

 

[Buzz Bloom]

Hi Ivan:

This is not my field, but here is my guess regarding 3/2.

The inner charge +Q/2 attracts electrons within the conductor to its inner surface from its outer surface. So the inner surface gets the charge -Q/2, and the outer surface gets the charge +(3/2) Q.

One more thought: Any spherical surface with charge Q having a uniform distribution over the surface produces zero interior field, and acts externally as it it were a point charge Q at the center. Therefore, within the shell, the center charge +Q/2 together with the inner surface charge -Q/2 nets to a zero charge equivalent. Outside the shell, the three surfaces net to a equivalent center charge of (3/2) Q.

Hopes this helps.

Regards,
Buzz

 

[Doc Al]

what I don't get though; is, where does the 3/2 come from, at r > R2?

What's the total charge contained within a gaussian sphere at that radius?

 

[IvanSaurus]

What's the total charge contained within a gaussian sphere at that radius?

The problem just states that the center of gaussian sphere has charge enclosed, Q/2.. no real values are given.

 

[IvanSaurus]

Hi Ivan:

This is not my field, but here is my guess regarding 3/2.

The inner charge +Q/2 attracts electrons within the conductor to its inner surface from its outer surface. So the inner surface gets the charge -Q/2, and the outer surface gets the charge +(3/2) Q.

One more thought: Any spherical surface with charge Q having a uniform distribution over the surface produces zero interior field, and acts externally as it it were a point charge Q at the center. Therefore, within the shell, the center charge +Q/2 together with the inner surface charge -Q/2 nets to a zero charge equivalent. Outside the shell, the three surfaces net to a equivalent center charge of (3/2) Q.

Hopes this helps.

Regards,
Buzz

Ill try to wrap my head around it some more later, thanks for the answer though :)

 

[Doc Al]

The problem just states that the center of gaussian sphere has charge enclosed, Q/2.. no real values are given.

Give your answer in terms of Q. What's the total charge contained within a gaussian sphere with radius > R2?

 

[IvanSaurus]

Give your answer in terms of Q. What's the total charge contained within a gaussian sphere with radius > R2?

Well that exactly what I can't seem to figure out. I must have overseen some rule regarding Gaussian Surfaces- since appearantly the charge triples outside the surface? from Q/2 to 2Q/3.

The Q/2 is the enclosed charge..

 

[Doc Al]

Well that exactly what I can't seem to figure out. I must have overseen some rule regarding Gaussian Surfaces- since appearantly the charge triples outside the surface? from Q/2 to 2Q/3.

I think you mean that the charge on the outer surface of the conductor is 3/2 Q. That happens to be true, but that is not needed to answer the question.

The Q/2 is the enclosed charge..

That is incorrect. What's the total charge inside the Gaussian surface? Just add 'em up.

 

[IvanSaurus]

I think you mean that the charge on the outer surface of the conductor is 3/2 Q. That happens to be true, but that is not needed to answer the question.That is incorrect. What's the total charge inside the Gaussian surface? Just add 'em up.

Now I am totally lost :) a guess would be "0"..

If the Gaussian surface is the purple ring, and we add up Q/2 - Q/2.
It still confuses me that outer surface of the conductor reaches 3/2 Q

 

[vela]

When r > R2, you have three regions to worry about: r < R1, R1 ≤ r ≤ R2, and r > R2. Try rereading the problem statement and answer the following questions:

1. What's the total charge inside the hollow part of the conductor? This is the charge in region iii.
2. What's the total charge on the conductor? This is the charge in region ii. (Note that you don't care about the distribution of charge here.)
3. What's the total charge outside the conductor? This is the charge in region i.

Then, as Doc Al said, just add them up. That's the total charge inside the Gaussian surface.

 

[Doc Al]

Now I am totally lost :) a guess would be "0"..

No. Remember I'm talking about a Gaussian surface at some point r > R2. (This is called region i in the problem statement.) So everything is included within it.

If the Gaussian surface is the purple ring, and we add up Q/2 - Q/2.

If by the "purple ring" you mean a surface with radius R1 ≤ r ≤ R2, then you would be correct. Since that surface would be within the conducting material, the net charge within it must be zero.

It still confuses me that outer surface of the conductor reaches 3/2 Q

Well, what's the total charge on all surfaces?