# Electric field within a solid sphere

1. Feb 13, 2013

### Gee Wiz

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

A nonconducting, solid sphere of radius a is placed at the center of a spherical conducting shell of inner radius b (> a) and outer radius c, as shown in the figure below. A charge +Q is distributed uniformly through the sphere, which thus carries a charge density ρ (C/m3). The outer shell carries a total charge -3Q.

Find the electric field E(r) within the solid sphere, i.e., at a radius r < a

2. Relevant equations
∫EdA=Qenclosed/Eo

3. The attempt at a solution

E*4∏r^2=(ρ*(4/3)*∏*a^3)/Eo

I thought I knew what i was doing, but now i'm not quite so sure. I know what i have above is not the correct answer.

2. Feb 13, 2013

### BruceW

Think about how much charge you are enclosing by your Gaussian surface.

3. Feb 13, 2013

### Gee Wiz

well, i think that i am enclosing something less than Q. Since Q is uniform over the entire sphere. If i am taking only a portion of that sphere, then i think it should be less.

4. Feb 13, 2013

### BruceW

yep. well, technically, the charge density is uniform over the entire sphere. You know that a Gaussian sphere of radius a would enclose Q, so how much charge does a Gaussian sphere of radius r(<a) enclose?

5. Feb 13, 2013

### Gee Wiz

A ratio of what a enclosed. So like, Qr/a?

6. Feb 13, 2013

### Staff: Mentor

It's a ratio, but not that ratio. Suggestion: first find the charge density ρ in terms of Q and a.

7. Feb 13, 2013

### Gee Wiz

That's what i tried to start to do with :E*4∏r^2=(ρ*(4/3)*∏*a^3)/Eo

8. Feb 13, 2013

### Staff: Mentor

You have a sphere of radius a and total charge Q. Forget everything else about the problem for the moment. What is the charge density ρ?

Once you have that, what is the charge on a sphere with the same charge density, but with radius r?

9. Feb 15, 2013

### Gee Wiz

charge density p is charge/volume. That's kinda what i was trying to do. Was i just making it more complicated than i needed to. Should i just find p and then multiply it by the new volume that i am trying to find?

10. Feb 15, 2013

### BruceW

yep. that will give you the charge enclosed by your Gaussian surface (since charge density is constant inside the sphere)

11. Feb 15, 2013

### Gee Wiz

ohhh okay. Well thank you all very much