Electric Field around Conducting Surfaces

[nimadh]

Hey,

Completing AP Problems, I ran into a puzzling answer.

Basically, there was a conducting sphere with charge Q on it and radius R1, and then around it two hollow hemispheres were placed (forming a spherical capacitor) which had no charge, with Radius R2 at the inner edge and R3 at the outer edge.

Basically, the problem asked for the electric field
- a) less than R1
- b) between R1 and R2
- c) between R2 and R3
- d) greater than R3

The answers were given as:
- a) zero
- b) kQ/R^2
- c) zero
- d) zero

I understand a, b, and c, but I could not understand why d is zero. Wouldn't the enclosed charge according to Gauss's law be the charge on the inner sphere? Or would the outer conducting surface somehow neutralize the charge?

Keep in mind that this is an actual Collegeboard Question, so the answers are certainly right (at least using basic physics). The Question is Physics C 1999 #1 for Electromagnetism.

Thanks!

 

[khemist]

The charges (electron, there is no net charge on the outer sphere) on the outer sphere cancel the the charge from the inner sphere. I cannot remember exactly why this happens, but I know it has to do with the ability for charges to move freely within a conducting material.

 

[nimadh]

That's something along what I was thinking, but wouldn't that happen for all cases with a (initially grounded) conductor near a charge? This "neutralizing" of the electric field beyond the conductor?

 

[khemist]

That is similar to the situation of having a hollow conductive sphere, without a charge, and placing a point charge outside of it. The electric field INSIDE of the sphere is zero, because if we used gauss' law around the sphere, the net charge inside is zero!

 

[Meir Achuz]

The answers to c and d are correct if the two shells are grounded.
If they are neutral, then Gauss's law says that the answers to b,c,and d should all be kQ/r^2.

 

[nimadh]

The shells were initially grounded, but then released. But why would that make a difference? Wouldn't grounding make the shells neutral?

 

[fyzxfreak]

Hunh, that's interesting. Could someone please explain how grounding works in the presence of these other charges?
=EDIT=
Wait a second... grounding means some set low potential, right? So in that case there is zero potential difference between the inner and outer capacitor-parts? So there's no field there.
Outside the capacitor... why is the field zero? Has the grounding somehow induced a total negative charge on the inner part of the small capacitor-part?

 

[nimadh]

Fyzxfreak, I think the post before your edit hit it on the head--grounding would get rid of the positive charges and leave a net negative charge on the spheres.

 

[fyzxfreak]

Haha, really? Whoops, I felt like I had no idea what I was talking about.

So what grounding does is hook the inner shell up to a zero potential of sorts... which is why the positive charges would like to leave? ... while the negative charges stay attracted to the +Q? For the outer shell, then, there's nothing inside so... nothing happens. And therefore the field is zero between and outside the capacitor.
IS this handwaving or is this legit?

 

[nimadh]

I think that's a fair way to look at it, except I don't understand what you mean by "For the outer shell, then, there's nothing inside so... nothing happens."

Basically, the way I see it, the outer shell is surrounding a positive charge (positive potential, l like you said), so all the negative charges want to go toward the charge, while the positive charges want to move away. Normally, this would cause a negative charge on the inner edge and positive charge on the outer edge, thus cancelling itself out when you use Gauss's law. But by grounding it, we allow the positive charges to leave, leaving a net charge of -Q on the outer shell and making the net charge of the capacitor zero.

Obviously, of course, there is really only one type of charge, but this aids description.

Is this correct?

 

[fyzxfreak]

Ah, well I think what I mean by nothing happens is that from the point of view of the outer capacitor-shell, there's the inner +Q but also the -Q that formed from grounding the inner capacitor-shell. Thus, there is no net enclosed charge inside the outer capacitor-shell and therefore no field outside of it (by Gauss' Law).

 

[nimadh]

We're on the same page then. Excellent! Thanks! Your random musings brought out my insight :D

 

[fyzxfreak]

Good! Glad to be of help! Hopefully it's correct!

 

[Meir Achuz]

The shells were initially grounded, but then released. But why would that make a difference? Wouldn't grounding make the shells neutral?

Grounding the shells makes the inner one charged with -Q, but the outer one would be neutral.
You can trust the College Board.