Gauss Law: Concentric Spheres

In summary, when the radius of a sphere is less than the charge enclosed by it, the electric field is zero.
  • #1
SarahAlbert
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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?

Homework Equations



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?
 
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  • #2
SarahAlbert said:
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?
 
  • #3
Orodruin said:
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
 
  • #4
SarahAlbert said:
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:
Orodruin said:
Is the charge enclosed by a sphere independent of the radius of the sphere in this case?
 
  • #5
Orodruin said:
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.
 
  • #6
SarahAlbert said:
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##?
 
  • #7
Orodruin said:
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.
 
  • #8
SarahAlbert said:
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##.
 
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  • #9
Orodruin said:
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!
 

1. What is Gauss Law for concentric spheres?

Gauss Law for concentric spheres is a mathematical equation that relates the electric flux through a closed surface to the charge enclosed within the surface. It states that the electric flux through a closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space.

2. How is Gauss Law applied to concentric spheres?

To apply Gauss Law to concentric spheres, we first need to choose a Gaussian surface that is a sphere with the same center as the concentric spheres. Then, we calculate the electric flux through the Gaussian surface and equate it to the total charge enclosed by the surface divided by the permittivity of free space.

3. What is the significance of concentric spheres in Gauss Law?

Concentric spheres play a crucial role in Gauss Law as they help simplify the mathematical calculations. Since the electric field is spherically symmetrical for concentric spheres, the electric flux through the Gaussian surface can be easily calculated using the spherical symmetry of the electric field.

4. Can Gauss Law be applied to non-concentric spheres?

Yes, Gauss Law can be applied to non-concentric spheres as long as the electric field is spherically symmetrical. However, the mathematical calculations may become more complex, and a different Gaussian surface may need to be chosen.

5. What are the applications of Gauss Law for concentric spheres?

Gauss Law for concentric spheres is widely used in various fields, including electromagnetism, electrostatics, and electronics. It helps in understanding and predicting the behavior of electric charges and fields, which is crucial in the design and functioning of various electrical and electronic devices.

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