Gauss' Law, Spherical Charge Distribution

[Fronzbot]

Before I get into the question I'd just like to state that this is not homework, but questions in my book that I'm going through to prepare myself for the midterm in one week. I got stuck at a few questions, here's the first one. I won't ask the next until I'm done with this and so forth. Normally I'd go to my professor for help but I can literally not understand him when he speaks. It's awful. Luckily he's a lenient grader (at least compared to other EM Fields profs), but yeah. Ok, onto the question then...

Homework Statement

A spherical distribution of charge $$\rho = \rho_{0}[1-(R^{2}/b^{2}]$$ exists in the region $$0\leqR\leqb$$. This charge distribution is concentrically surrounded by a conducting shell with inner radius $$R_{i} (>b)$$ and outer radius $$R_{o}$$. Determine E everywhere.

Homework Equations

$$\int E \cdot ds = Q_{enc} / \epsilon_{0}$$

The Attempt at a Solution

I have the right answers (from the back of the book) but cannot figure out how to get there. Before I write my solution down, here are the correct answers:
$$For 0 \leq R \leq b:$$
$$E_{R1} = \frac{\rho_{0}R}{\epsilon_{0}}(\frac{1}{3}-\frac{R^{2}}{5b^{2}})$$

$$For b \leq R < R_{i}:$$
$$E_{R2} = \frac{2 \rho_{0} b^{3}}{15 \epsilon_{0} R^{2}}$$

$$For R_{i}<R<R_{o}$$
$$E_{R3} = 0$$

$$For R > R_{o}:$$
$$E_{R4} = \frac{2 \rho_{0} b^{3}}{15 \epsilon_{0}R^{2}}$$

Now I didn't get very far, but here's what I have:
$$Q_{enc} = \int \rho dV = \rho_{0}[1 - \frac{R^2}{b^2}]\frac{4\pi R^{3}}{3}$$
$$\int E \cdot ds = E_{R1} 4 \pi R^{2}$$
$$E_{R1} = \frac{\rho [1-\frac{R^2}{b^2}]R}{3\epsilon_{0}}$$
$$E_{R1} = \frac{\rho R}{\epsilon}(\frac{1}{3}-\frac{R^2}{3b^2})$$

I have some work for the next two, but I'd rather go one step at a time here to make sure I completely what's going on. At least this answer is close, but I'm not sure where the book's 5b^2 came from.

 

[KataKoniK]

Dear student,

I understand that you are preparing for your midterm and have encountered some difficulties with a question from your textbook. I appreciate your effort in trying to solve the problem on your own before seeking help.

Firstly, it is important to understand the concept of spherical symmetry in this problem. Since the charge distribution is spherically symmetric, the electric field will also have spherical symmetry. This means that the electric field will only depend on the distance from the center of the distribution, and not on the direction.

Next, let's take a closer look at the charge distribution. The charge density, \rho, is given by \rho_{0}[1-(R^{2}/b^{2}] and is only dependent on the distance from the center, R. This means that the charge enclosed within a radius R is given by Q_{enc} = \rho_{0}[1-(R^{2}/b^{2})] \frac{4\pi R^{3}}{3}. This is the expression you have used in your attempt to find the electric field.

Now, let's move on to the correct answers. For 0 \leq R \leq b, the electric field is given by E_{R1} = \frac{\rho_{0}R}{\epsilon_{0}}(\frac{1}{3}-\frac{R^{2}}{5b^{2}}). To get this answer, we need to use Gauss's Law, which states that \int E \cdot ds = Q_{enc}/\epsilon_{0}. In this case, the charge enclosed is given by Q_{enc} = \rho_{0}[1-(R^{2}/b^{2})] \frac{4\pi R^{3}}{3}. Plugging this into Gauss's Law and solving for E_{R1}, we get the correct answer.

Similarly, for b \leq R < R_{i}, we use Gauss's Law again, but this time the charge enclosed is given by Q_{enc} = \rho_{0}b^{3} \frac{4\pi}{3}. Plugging this into Gauss's Law and solving for E_{R2}, we get the correct answer of E_{R2} = \frac{2\rho_{0}b^{3}}{15\epsilon_{0}R^{2}}.

I hope this explanation helps you understand the problem better and prepares