Solving Gauss's Law Problem for Electric Field

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Homework Statement


A uniformly charged ball of radius a and a total charge -Q is at the center of a hollow metal shell with inner radius b and outer radius c. The hollow sphere has a net charge +2Q. Find the magnitude of the electric field in the regions: r_1 < a,a < r_2 < b,b < r_3 < c, and r_4 > c.

Homework Equations



V = \frac{4}{3} \pi R^3
S = 4 \pi R^2
\oint E(x)dA = \frac{q_{in}}{\epsilon_o}

The Attempt at a Solution



For E(r1 < a):
\rho = \frac{Q_{tot}}{\epsilon_o}
Q_{in,tot} = \rho*\frac{4}{3} \pi r_1^3
\oint_0^rE(x)dA = \frac{q_{in}}{\epsilon_o}

E(r_1) = \frac{\rho\frac{4}{3} \pi r_1^3}{\epsilon*4 \pi r_1^2}

E(r_1) = \frac{\rho*r_1}{3\epsilon_o}

This is actually where I am stuck, I got everything else. Am I supposed to get rid of that volume charge density, \rho?
 
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Yes; in textbook problems, they expect you to express the answer in terms of the variables given in the statement of the problem.

The density is uniform and you know the total charge in the little ball. So just divide that by its volume to find its density.
 
The following may help.

Regards,

Nacer.

http://islam.moved.in/tmp/c.jpg
 
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There is actually a second and third part to this question:

b) Find potentials at points in the regions: r_1 &lt; a, a &lt; r_2 &lt; b, b &lt; r_3 &lt; c, and r_4 &gt;c

For r1 < a,
I use the formula

V_o - V_{r_1} = \int_0^r E(r_1)dr

E(r_1) = \frac{\rho*r_1}{3\epsilon_o}

V_o - V_{r_1} = \frac{\rho}{3\epsilon_o} \int_0^r r dr

V_o - V_{r_1} = \frac{\rho*r_1^2}{6\epsilon_o}

Solving the same way, I got::

V_o - V_{r_2} = \frac{Q}{4 \pi \epsilon_o r_2}

V_o - V_{r_3} = 0

V_o - V_{r_4} = \frac{-Q}{4 \pi \epsilon_o r_4}

Did anyone get the same answer as I did?

Also, I am supposed to sketch the graphs of how Ex depends on a distance(r) from the center of the sphere, and how V depends on a distance(r) from the center of the sphere.

I drew my graphs, but I don't know how to post it on. I don't have a scanner handy, so if anyone can help, please let me know what your graphs looks like.
Thank you very much, you are all so helpful.
 
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