Gauss's Law Sphere Problem

In summary, the problem involves determining the electric field within a spherical cavity of radius a that is removed from a larger sphere of radius 2a made of nonconducting material with a uniform volume charge density. Using Gauss' law and the principle of superposition, it can be shown that the electric field within the cavity is uniform and given by Ex = 0 and Ey = a/30. This is different from the electric field inside a perfect conductor, which would result in a zero electric field within the cavity. The approach to solving this problem includes using spherical symmetry and allowing for a change of coordinates. However, it is important for the OP to first understand the basics of finding the electric field at a given point due to a uniformly charged
  • #1
GreenLantern674
27
0
A sphere of radius 2a is made of nonconducting material that has a uniform volume charge density . (Assume that the material does not affect the electric field.) A spherical cavity of radius a is now removed from the sphere, as shown in Figure P19.62. Show that the electric field within the cavity is uniform and is given by Ex = 0 and Ey = a/30. (Hint: The field within the cavity is the superposition of the field due to the original uncut sphere, plus the field due to a sphere the size of the cavity with a uniform negative charge density -.)
p19-62.gif


I really don't have any idea how to solve this one. I have a vague idea about ratios of volume to charge. I know it must involve something with that because of the measurement of charge density, but some specifics would really be appreciated.
 
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  • #2
Have you done similar problems in finding the gravity at a certain depth inside a mine or some such stuff? It's a very common problem. If you don't have any idea, it becomes difficult for anybody to solve it wholly for you.

The fact you must know is that the field inside a very thin hollow uniform shell with uniform surface charge density is zero.
 
  • #3
Yes, but the electric field in the cavity isn't zero. That's what initially confused me.
 
  • #4
That only happens in a perfect conductor, irrespective of the shape of the cavity. This is different.

EDIT: Temporarily withdrawn rest of post, to think of easier solution.
 
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  • #5
First tell me what is the field at the surface of a uniformly charged sphere of total charge Q?

And the field inside a thick hollow spherical shell with uniform volume charge density?
 
  • #6
I'm not sure why you'd want the hollow shell case...

The way I'd do it is to use Gauss' law to work out the vector E for a uniformly charged sphere (it is a very simple expression)
Then do the superposition of the two spheres at a given point r, remembering to allow for change of coordinates.

And I think there's a typo in your answer.
 
  • #7
mda said:
I'm not sure why you'd want the hollow shell case...

To use spherical symmetry and then use Gauss law. Could you use Gauss law to find the field inside a uniformly charged cube?

The way I'd do it is to use Gauss' law to work out the vector E for a uniformly charged sphere (it is a very simple expression)
Then do the superposition of the two spheres at a given point r, remembering to allow for change of coordinates.

That will be the correct way. However, the intention of the HW forum is not to solve it for the OP but to nudge him in the right direction, and give a big helping hand if need be over the difficult math parts. I am sure there are scores of people in PF who will readily solve the whole problem for him/her.

In this case, if the OP can't answer the simple questions that I've asked, then he/she is not ready for this problem.
And I think there's a typo in your answer.

Why don't you point it out?
 
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1. What is Gauss's Law Sphere Problem?

Gauss's Law Sphere Problem is a mathematical problem that involves the application of Gauss's Law to calculate the electric field of a charged sphere. It is used to determine the electric field at any point outside or inside a uniformly charged sphere.

2. How is Gauss's Law applied to solve the Sphere Problem?

Gauss's Law states that the electric flux through any closed surface is equal to the enclosed charge divided by the permittivity of free space. In the case of the Sphere Problem, the closed surface is a sphere and the enclosed charge is the total charge of the sphere.

3. What are the assumptions made in solving the Gauss's Law Sphere Problem?

The assumptions made in solving the Gauss's Law Sphere Problem include: the charged sphere is uniformly charged, the electric field is spherically symmetric, and the electric field outside the sphere is the same as that of a point charge located at the center of the sphere.

4. Can Gauss's Law be applied to a non-uniformly charged sphere?

No, Gauss's Law can only be applied to a uniformly charged sphere. For non-uniformly charged spheres, other methods such as integration must be used to calculate the electric field.

5. What are the real-world applications of Gauss's Law Sphere Problem?

Gauss's Law Sphere Problem has various applications in physics and engineering, such as determining the electric field inside a charged conducting sphere, calculating the capacitance of spherical capacitors, and analyzing the behavior of charged particles in a uniform electric field.

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