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Difficult Charge Problem

  1. Jul 18, 2007 #1
    1. The problem statement, all variables and given/known data

    19. A nonconducting spherical shell, with an inner radius of 4.0 cm and an outer radius of 6.0 cm, has charge spread nonuniformly through its volume between its inner and outer surfaces. The volume charge density [itex]\rho[/itex] is the charge per unit volume, with the unit coulomb per cubic meter. For this shell [itex]\rho = b/r[/itex], where r is the distance in meters from the center of the shell and b = 3.0 [itex]\mu C/m^2[/itex]. What is the net charge in the shell.

    2. Relevant equations

    [tex]
    \vec{F}_{12} = \frac{k_{e}q_{1}q_{2}}{{r_{12}}^2}\hat{r}_{21}
    [/tex]

    [tex]
    |\vec{F}_{12}| = \frac{k_{e}|q_{1}||q_{2}|}{{r_{12}}^2}
    [/tex]

    3. The attempt at a solution

    This problem is unlike most problems I have solved before so I am not sure if my approach is correct.

    From what was given in the problem I conclude the following equations.

    V = Volume
    A = Area

    [tex]
    \rho = \frac{q}{V}
    [/tex]

    [tex]
    \rho = \frac{b}{r}
    [/tex]

    [tex]
    b = \frac{q}{A}
    [/tex]

    Though, the problem mentions that the charge has spread nonuniformly through its volume between its inner and outer surfaces of the spheres.

    I am unsure as to how I am supposed to use that to solve the problem.

    I assume I can treat the inner and outer spheres: [itex]sphere_{1}[/itex] and [itex]sphere_{2}[/itex]; separately and use the previous equations to come up with,

    [tex]
    \rho_{1} = \frac{q_{1}}{V_{1}}
    [/tex]

    [tex]
    \rho_{2} = \frac{q_{2}}{V_{2}}
    [/tex]

    Volume of a sphere is,

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

    --------------------------------------------------------------------------

    [tex]
    q_{1} = \left(\frac{4}{3} \pi r_{1}^3\right)\left(\frac{b}{r_{1}}\right)
    [/tex]

    [tex]
    sig. fig. \equiv 2
    [/tex]

    [tex]
    q_{1} = 20 nC
    [/tex]

    [tex]
    q_{2} = \left(\frac{4}{3} \pi r_{2}^3\right)\left(\frac{b}{r_{2}}\right)
    [/tex]

    [tex]
    sig. fig. \equiv 2
    [/tex]

    [tex]
    q_{2} = 45 nC
    [/tex]

    Although, the answers, [itex]q_{1} = 20 nC[/itex] [itex]q_{2} = 45 nC[/itex]. Are somewhat close to the book answer, they’re wrong. And because the problem asks for the net charge in the shell, I thought maybe I could take the average of these two figures and get the right answer, however (though it was close) it was still the wrong answer for the net charge in the shell.

    Yea, so I am stuck on how to find the net charge in the shell.

    Any help would be appreciated.

    -PFStudent
     
    Last edited: Jul 18, 2007
  2. jcsd
  3. Jul 18, 2007 #2

    siddharth

    User Avatar
    Homework Helper
    Gold Member



    No. When you say

    [tex] \rho = \frac{q}{V} [/tex]

    notice that [itex]\rho[/tex] changes as a function of r ([itex]\rho=\frac{b}{r}[/itex]). So, you can't just find the total charge as [tex]\rho_{r_2} V_{r_2}[/tex], because then you're assuming that [tex]\rho[/tex] is constant for the whole volume inside, which is not true.

    However, for an infinitesimally (ie, infinitely small) thin spherical shell of thickness dr at a distance of r from the center, what's the charge? Since the shell is infinitesimally thin, it's ok to assume that the charge density remains constant in this extremely small region. Now, add up the charges of all such infinitesimal shells from the inner to the outer radius.

    Are you familiar with integral calculus?
     
    Last edited: Jul 18, 2007
  4. Jul 18, 2007 #3
    Hey,

    Thanks for the information.

    Ok, so then,

    [tex]
    q + C= \int dq
    [/tex]

    And for this problem,

    [tex]
    q = \int_{r_{1}}^{r_{2}} dq
    [/tex]

    [tex]
    \rho = \frac{q}{V}
    [/tex]

    [tex]
    \rho = \frac{b}{r}
    [/tex]

    Which is the same thing as,

    [tex]
    \rho(r) = \frac{b}{r}
    [/tex]

    For now assume [itex]\rho[/itex] is constant, and so therefore,

    [tex]
    \overline{\rho} = \frac{\Delta q}{\Delta V}
    [/tex]

    S.A. = Surface Area of Sphere

    [tex]
    \overline{\rho} = \frac{\Delta q}{S.A.\Delta r}
    [/tex]

    [tex]
    \overline{\rho}(r) = \frac{\Delta q}{S.A.(r)\Delta r}
    [/tex]

    [tex]
    (4\pi r^2)\overline{\rho}(r) = \frac{\Delta q}{\Delta r}
    [/tex]

    [tex]
    \lim_{\Delta r\rightarrow 0} (4\pi r^2)\overline{\rho}(r) = \lim_{\Delta r\rightarrow 0} \frac{\Delta q}{\Delta r}
    [/tex]

    [tex]
    4\pi r^2{\rho}(r) = \frac{dq}{dr}
    [/tex]

    [tex]
    4\pi r^2 \left(\frac{b}{r}\right)dr = dq
    [/tex]

    [tex]
    4\pi rbdr = dq
    [/tex]

    [tex]
    dq = 4\pi brdr
    [/tex]

    [tex]
    \int_{r_{1}}^{r_{2}} dq = 4\pi b\int_{r_{1}}^{r_{2}}rdr
    [/tex]

    [tex]
    q = 2\pi b \left[r^2\right]_{r_{1}}^{r_{2}}
    [/tex]

    [tex]
    q = 2\pi b \left(r_{2}^2-r_{1}^2\right)
    [/tex]

    [tex]
    sig. fig. \equiv 2
    [/tex]

    [tex]
    q = 3.8{\textcolor[rgb]{1.00,1.00,1.00}{.}}x{\textcolor[rgb]{1.00,1.00,1.00}{.}}10^{-8}{\textcolor[rgb]{1.00,1.00,1.00}{.}}C
    [/tex]

    [tex]
    q = 38{\textcolor[rgb]{1.00,1.00,1.00}{.}}nC
    [/tex]

    Ok, I got the right answer however, I want to know if I derived the answer mathematically correct.

    For example is, [itex]S.A.(r)\Delta r = \Delta V[/itex]?

    This reason, I ask is because if you let [itex]\Delta r[/itex] = r

    Then you have,

    [tex]
    S.A.(r)r = V
    [/tex]

    [tex]
    (4 \pi r^2)r \neq \frac{4}{3}\pi r^3
    [/tex]

    [tex]
    4 \pi r^3 \neq \frac{4}{3}\pi r^3
    [/tex]

    So how does that work?

    -PFStudent
     
    Last edited: Jul 18, 2007
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