Inner and Outer Radius of a Sphere

In summary: Now r determines the volume of the shell enclosed by the Gaussian sphere. Don't use the volume formula you started with, but the formula for the volume of a spherical shell of inner radius x and outer radius r. (It's in the same section of the same table in your book as the formula for the volume of a sphere. One more math error: the volume of the shell is (4/3)(pi)(y^3) - (4/3)(pi)(x^3) = (4/3)(pi)(y^3 - x^3). You have to use the volume formula for a shell. So the charge density is 5/(4/3)(pi)(y^3 - x^3)
  • #1
jumbogala
423
4

Homework Statement


A sphere, which is hollow, has an inner radius (denoted x) and an outer radius (denoted y).

The sphere does not conduct.

It has a charge of 5 C distributed uniformly throughout it.

1) What is the charge density by volume in the body of the hollow sphere?

2) How much charge is enclosed in a gaussian sphere of radius x < r < y?


Homework Equations


(4/3)(pi)(r3) is the volume of a sphere with radius r.


The Attempt at a Solution



I know you divide the charge by the volume of the sphere. Since the sphere is hollow, can you assume that the 5 C of charge are distributed in the hollow part too, or only on the edge?

My instinct is to calculate the volume using radius y, but I'm not sure.
 
Last edited:
Physics news on Phys.org
  • #2
jumbogala said:

Homework Statement


A sphere, which is hollow, has an inner radius (denoted x) and an outer radius (denoted y).

The sphere does not conduct.

It has a charge of 5 C distributed uniformly throughout it.

1) What is the charge density by volume in the body of the hollow sphere?

2) How much charge is enclosed in a gaussian sphere of radius x < r < y?


Homework Equations


(4/3)(pi)(r3) is the volume of a sphere with radius r.


The Attempt at a Solution



I know you divide the charge by the volume of the sphere. Since the sphere is hollow, can you assume that the 5 C of charge are distributed in the hollow part too, or only on the edge?

My instinct is to calculate the volume using radius y, but I'm not sure.

The wording of this problem is a trifle confusing. Here is what I think: You have a spherical shell of inner radius x and outer radius y. The sphere is empty for r < x. The charge is uniformly distributed throughout the material that makes up the spherical shell. You can figure the volume of the shell by subtracting the volume of the empty interior portion from the volume of the whole sphere (i.e. use r = y in the volume formula) From that you can get an expression for the charge density (Charge/Volume). I see you don't have to calculate E, but just need the charge enclosed in a Gaussian sphere inside the shell. So calculate the volume of material enclosed in the Gaussian sphere the same way as you calculated the volume of the spherical shell, it's just that now your outer radius is the radius of your Gaussian sphere. Multiply that by the charge density and you're done. Don't be alarmed if the final answer is messy.
 
  • #3
Okay, I think I understand what you mean.

Volume of the empty part of the sphere:
(4/3)(pi)(x^3)

Volume of the entire sphere:
(4/3)(pi)(y^3)

Subract entire sphere from empty part...
(4/3)(pi)(y^3) - (4/3)(pi)(x^3) = volume of the shell only

Divide 5 by volume of shell only, to get charge density.

At this point I get a bit confused. I understand that I need to find the volume of the Gaussian sphere, then multiply by charge density. But the Gaussian sphere has to have a radius that is less than y, so why does y become the radius of the Gaussian sphere?
 
Last edited:
  • #4
jumbogala said:
Okay, I think I understand what you mean.

Volume of the empty part of the sphere:
(4/3)(pi)(x^3)

Volume of the entire sphere:
(4/3)(pi)(y^3)

Subract entire sphere from empty part...
(4/3)(pi)(y^3) - (4/3)(pi)(x^3) = volume of the shell only

Divide 5 by volume of shell only, to get charge density.

At this point I get a bit confused. I understand that I need to find the volume of the Gaussian sphere, then multiply by charge density. But the Gaussian sphere has to have a radius that is less than y, so why does y become the radius of the Gaussian sphere?

Somewhere we mis-communicated. Your are right, the radius of the Gaussian sphere is NOT y. Call it r where x< r < y.
 

Related to Inner and Outer Radius of a Sphere

1. What is the difference between the inner and outer radius of a sphere?

The inner radius of a sphere is the distance from the center of the sphere to its inner surface, while the outer radius is the distance from the center to its outer surface. In simpler terms, the inner radius is the radius of the sphere itself, while the outer radius includes the thickness of the sphere's walls.

2. How do you calculate the inner and outer radius of a sphere?

To calculate the inner and outer radius of a sphere, you can use the formula r = V/4π, where V is the volume of the sphere and r is the radius. The inner radius will be half the diameter of the sphere, while the outer radius will be half the diameter plus the thickness of the walls.

3. Can the inner radius of a sphere be larger than the outer radius?

No, the inner radius of a sphere cannot be larger than the outer radius. This is because the inner radius is the distance from the center to the inner surface, while the outer radius is the distance to the outer surface. Since the inner surface is always closer to the center, the inner radius will always be smaller than the outer radius.

4. How does the inner and outer radius of a sphere affect its volume?

The inner and outer radius of a sphere are directly related to its volume. As the inner and outer radius increase, the volume of the sphere also increases. This is because the volume of a sphere is directly proportional to the cube of its radius. So, a small change in the radius can result in a significant change in the volume.

5. What is the significance of knowing the inner and outer radius of a sphere?

Knowing the inner and outer radius of a sphere is important for various applications, such as in engineering and architecture. It helps in determining the size and shape of objects that need to fit inside or outside of the sphere. Additionally, it is also useful in calculating the volume, surface area, and other properties of the sphere.

Similar threads

  • Introductory Physics Homework Help
Replies
23
Views
861
  • Introductory Physics Homework Help
Replies
2
Views
252
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
4K
  • Introductory Physics Homework Help
2
Replies
36
Views
596
  • Introductory Physics Homework Help
Replies
9
Views
809
  • Introductory Physics Homework Help
Replies
18
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
805
Back
Top