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Electric charge in a shell

  1. Apr 13, 2008 #1
    1. The problem statement, all variables and given/known data

    (Note: Don't worry about significant digits. I just want to be able to do the question and will worry about significant digits on the exam.)

    In the figure below, a non-conducting spherical shell of inner radius a=2.00cm and outer radius b=2.40cm has (within its thickness) a positive volume density [tex]\rho[/tex]=A/r, where A is a constant and r is the distance from the centre of the shell. In addition, a small ball of charge q=45.0 fC is located at the centre.

    What value should A have if the electric field in the shell (a[tex]\leq[/tex]r[tex]\leq[/tex]b) is to be uniform?

    (See attached file for the picture. sorry it isn't very good. That's a "q" beside the +ve charge in the middle.)

    2. Relevant equations

    Charge density : [tex]\rho= \frac{dq}{dV}[/tex]

    Volume : [tex]V = 4\pi r^3[/tex]

    Electric field : [tex]E = \frac{1}{4\pi \epsilon} \int \frac{dq}{r^2}[/tex]

    3. The attempt at a solution

    E at a distance r anywhere within [a, b] is constant.
    E is caused by the point charge and distribution of charges in the volume.

    [tex]E_{net}[/tex] is constant, therefore


    [tex] \frac{1}{4\pi \epsilon} \int \frac{dq_{charge}}{r_{charge}^2} = - \frac{1}{4\pi \epsilon} \int \frac{dq_{volume}}{r_{volume}^2}[/tex]

    the E for a point charge doesn't need to have an integral, because it works out that way.

    [tex] \frac{q_{charge}}{r^2} = - \int \frac{dq_{volume}}{r_{volume}^2}[/tex]

    Charge density : [tex]\rho= \frac{A}{r} = \frac{dq}{dV}[/tex]

    [tex]dq = \frac{A}{r} dV[/tex]

    [tex] \frac{q_{charge}}{r^2} = - \int \frac{A dV}{r^3}[/tex]

    dV is related to r because [tex]V = \frac{4}{3} \pi (r^3 - a^3)[/tex]

    then [tex]r^3 = \frac{3V + 4 \pi a^3}{4 \pi}[/tex]

    [tex] \frac{q_{charge}}{r^2} = - \int \frac{4 \pi A dV}{3V + 4 \pi a^3}[/tex]

    [tex] \frac{q_{charge}}{r^2} = - 4 \pi A \int \frac{dV}{3V + 4 \pi a^3}[/tex]

    [tex] \frac{q_{charge}}{r^2} = - \frac{4 \pi A}{3} ln (3V + 4 \pi a^3)[/tex]

    I don't know what to do after this, and I don't know what to do. I don't know if I messed up something.

    This may be supposed to be done with Gauss' Law, but I couldn't figure out what to do with that.

    Gauss' Law : [tex]\Phi = \oint E da[/tex] (In this case since E and a are in the same direction, the dot product just became E da)

    Attached Files:

  2. jcsd
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