# Inner and Outer Radius of a Sphere

1. Feb 8, 2009

### jumbogala

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

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

3. 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: Feb 8, 2009
2. Feb 8, 2009

### AEM

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. Feb 8, 2009

### jumbogala

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: Feb 8, 2009
4. Feb 8, 2009

### AEM

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