# Gauss Law: Concentric Spheres

## Homework Statement

Two concentric spheres have radii a and b with b>a. The region between them is filled with charge of constant density. The charge density is zero everywhere else. Find E at all points and express it in terms of the total charge Q. Do your results reduce to the correct values as a->0?

Flux=EA

## The Attempt at a Solution

[/B]
My line of thinking: The flux is independent of the radius because with two concentric circles it only depends on the charge enclosed by the sphere. The flux is then the same for both areas and it is equal to Q/eo. I'm not giving what the density charge is, so isn't my answer simply Q/eo?

Any help would b appreciated. Thank you.

I'm not looking for the solution, I guess I need to know if my line of thinking is correct in any way?

Orodruin
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The flux is independent of the radius because with two concentric circles it only depends on the charge enclosed by the sphere.

Is the charge enclosed by a sphere independent of the radius of the sphere in this case? What if r < a or a < r < b?

Is the charge enclosed by a sphere independent of the radius of the sphere in this case? What if r < a or a < r < b?

a<=r<=b

a (is less than and equal to) r (is less than and equal to) b

Orodruin
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a<=r<=b

a (is less than and equal to) r (is less than and equal to) b
No, you are being asked to find the field at any point. This means that ##r## can be both larger or smaller than either ##a## or ##b##. The question in my previous post still stands:
Is the charge enclosed by a sphere independent of the radius of the sphere in this case?

No, you are being asked to find the field at any point. This means that ##r## can be both larger or smaller than either ##a## or ##b##. The question in my previous post still stands:

See all the examples I've read about say that yes, the charge enclosed is independent of the sphere. However, in those cases the charge was enclosed by the small sphere. In this case the charge is between a and b. So I don't believe the charge is independent of the radius in this case because its located in an area of b-a.

Orodruin
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See all the examples I've read about say that yes, the charge enclosed is independent of the sphere. However, in those cases the charge was enclosed by the small sphere. In this case the charge is between a and b. So I don't believe the charge is independent of the radius in this case because its located in an area of b-a.

So take it step by step and treat one case at a time. You have three different cases: ##r < a##, ##a < r < b##, and ##r > b##.

Start with ##r < a##. What is the enclosed charge of a sphere with a radius less than ##a##?

So take it step by step and treat one case at a time. You have three different cases: ##r < a##, ##a < r < b##, and ##r > b##.

Start with ##r < a##. What is the enclosed charge of a sphere with a radius less than ##a##?

Isn't it zero? According to the problem even though it is constant between b and a, it is zero everywhere else.

Orodruin
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Isn't it zero? According to the problem even though it is constant between b and a, it is zero everywhere else.

Yes, it is zero. What does this imply for the electric field?

When you have answered this question you can go on with the case of ##r > b## and then finally ##a < r < b##.

SarahAlbert
Yes, it is zero. What does this imply for the electric field?

When you have answered this question you can go on with the case of ##r > b## and then finally ##a < r < b##.

Thank you so much for all your help by the way!