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Homework Help: Gauss' Law, Spherical Charge Distribution

  1. Sep 26, 2010 #1
    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...

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

    2. Relevant equations
    [tex]
    \int E \cdot ds = Q_{enc} / \epsilon_{0}
    [/tex]


    3. 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:
    [tex]
    For 0 \leq R \leq b:[/tex]
    [tex]
    E_{R1} = \frac{\rho_{0}R}{\epsilon_{0}}(\frac{1}{3}-\frac{R^{2}}{5b^{2}})
    [/tex]

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

    [tex]
    For R_{i}<R<R_{o}
    [/tex]
    [tex]
    E_{R3} = 0 [/tex]

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

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

    [/tex]

    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.
     
  2. jcsd
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