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Electrostatics: spherical shell

  1. Oct 1, 2009 #1
    Electrostatics: spherical shell [SOLVED]

    1. The problem statement, all variables and given/known data

    A point charge [tex]Q_1[/tex] is located in the centre of a spherical conducting shell with inner radius [tex]a[/tex] and outer radius [tex]b[/tex]. The shell has total charge [tex]Q_2[/tex]. Determine the electrostatic field [tex]\vec{E}[/tex] and the potential [tex]\phi[/tex] everywhere in space, and determine how much charge is on the inner and outer surfaces of the shell after electrostatic equilibrium has been reached.

    2. Relevant equations

    Gauss law: [tex]\oint_S\vec{E}\cdot d\vec{a} =\frac{ Q_{encl}}{\epsilon_0}[/tex]
    Area of a sphere: [tex]A=4\pi r^2[/tex]
    Surface charge density (homogenous charge distribution): [tex]\sigma = Q/A[/tex]
    Electric field from a point charge: [tex]\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{r}[/tex]
    Electric field from a surface charge: [tex]\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\int_S\frac{\sigma da}{r^2}\hat{r}[/tex]

    3. The attempt at a solution

    I call the charges on the inner and outer surfaces [tex]Q_a[/tex] and [tex]Q_b[/tex] respectively. The electric field inside a conductor is always zero, so Gauss law on a surface [tex]a<r<b[/tex] gives

    [tex]0=\oint_S\vec{E}\cdot d\vec{a} =\frac{ Q_1+Q_a}{\epsilon_0} [/tex]
    [tex]\Rightarrow Q_a=-Q_1[/tex]

    The total charge of the conductor:

    [tex]Q_2=Q_a+Q_b=Q_b-Q_1[/tex]
    [tex]\Rightarrow Q_b=Q_1+Q_2[/tex]

    The total electric field could be calculated by taking the sum of the electric field from each charge.

    [tex]\vec{E}_{tot}(\vec{r})=\vec{E}(\vec{r})_1+\vec{E}(\vec{r})_a+\vec{E}(\vec{r})_b[/tex]

    E-field from the point charge:

    [tex]\vec{E}_1(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{Q_1}{r}\hat{r}[/tex]

    Introduce the surface charge density [tex]\sigma_{a,b}=Q_{a,b}/A_{sphere}[/tex]

    [tex]\begin{align*}\vec{E}_a(\vec{r})=&\frac{1}{4\pi\epsilon_0}\int_S\frac{\sigma da}{r^2}\hat{r} \\
    &=\frac{1}{4\pi\epsilon_0}\int_0^r\frac{Q_a da}{4\pi a^2 r^2}\hat{r} \\
    &= \frac{Q_a}{16\pi^2\epsilon_0a^2}\int_S\frac{rdrd\theta}{r^2}\hat{r} \\
    & =\frac{Q_a}{8\pi\epsilon_0a^2}\int_0^r\frac{1}{r}\hat{r} \\
    & = \frac{Q_a}{8\pi\epsilon_0a^2}\left[ln|r|\right]_0^r
    \end{align*}[/tex]

    which does not make sense. Similar problems arise for E-field from [tex]\sigma_b[/tex]. Can anyone help me out, please?

    Thanks in advance.
     
    Last edited: Oct 1, 2009
  2. jcsd
  3. Oct 1, 2009 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Use the spherical symmetry of the problem: the magnitude of E depends only on r, and it is parallel with the radius at any point. As [tex]d\vec{a}[/tex] is normal to the surface of the sphere and so is [tex]\vec{E}[/tex]

    [tex]
    \oint_S\vec{E} \cdot d\vec{a} =\oint_SEda
    [/tex]

    where E is the magnitude of the electric field at distance r from the centre of the sphere, and it is the same at each point of the surface, therefore

    [tex]
    \oint_S\vec{E} \cdot d\vec{a} =\oint_SEda
    = 4\pi r^2 E=\frac{ Q_{encl}}{\epsilon_0}[/tex]

    You know well that the electric field is zero inside the conducting shell, you know that the enclosed charge is Q1 for r<a and Q1+Q2 for r>b...

    ehild
     
  4. Oct 1, 2009 #3
    Thank you! That was really helpful.
     
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