Electric field within a solid sphere

[Gee Wiz]

Homework Statement

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

Homework Equations

∫EdA=Qenclosed/Eo

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.

 

[BruceW]

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

 

[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.

 

[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?

 

[Gee Wiz]

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

 

[jtbell]

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

 

[Gee Wiz]

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

 

[jtbell]

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?

 

[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?

 

[BruceW]

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

 

[Gee Wiz]

ohhh okay. Well thank you all very much