Gauss Law: Concentric Spheres

1. Sep 29, 2015

SarahAlbert

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

Flux=EA

3. The attempt at a solution

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?

2. Sep 29, 2015

Orodruin

Staff Emeritus
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?

3. Sep 29, 2015

SarahAlbert

a<=r<=b

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

4. Sep 29, 2015

Orodruin

Staff Emeritus
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:

5. Sep 29, 2015

SarahAlbert

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.

6. Sep 29, 2015

Orodruin

Staff Emeritus
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 $a$?

7. Sep 29, 2015

SarahAlbert

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

8. Sep 29, 2015

Orodruin

Staff Emeritus
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 $a < r < b$.

9. Sep 29, 2015

SarahAlbert

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