# Electric field within a solid sphere

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

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
Homework Helper
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?

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

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

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

jtbell
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?

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

ohhh okay. Well thank you all very much