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Finding the Scalar Potential of a Sphere with Non-Uniform Charge Density

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A sphere of radius a has a charge density which varies with distance r from the center according to http://img14.imageshack.us/img14/9577/wangsnessproblem594.gif [Broken] where A is a constant and http://img35.imageshack.us/img35/555/wangsnessproblem593.gif [Broken].[/URL] Find the scalar potential http://img35.imageshack.us/img35/9326/phi2.gif [Broken] at all points inside and outside the sphere by using the following formula:

    http://img17.imageshack.us/img17/4628/wangsnesseqn572.gif [Broken]

    Express your results in terms of the total charge Q of the sphere.


    2. Relevant equations

    Volume of the Sphere: http://img10.imageshack.us/img10/8776/spherevolume.gif [Broken]

    Vector Capital R: (From Source to Point) http://img21.imageshack.us/img21/2586/captialr.gif [Broken]


    3. The attempt at a solution

    I'm not sure if I'm over-simplifying this but I figured since the charge density is directly related and varies according to the volume, that is, that I can think of the sphere as containing a series of spheres layered on each other that are infinitely thin, I could take the integral and relate Q to the charge density and the volume. So:

    http://img197.imageshack.us/img197/5717/59work1.gif [Broken]

    Then:

    http://img197.imageshack.us/img197/3756/59work2.gif [Broken]

    Finding that:

    http://img197.imageshack.us/img197/5041/59work3.gif [Broken]

    Plugging that into the main formula above, I get:

    http://img197.imageshack.us/img197/9933/59work4.gif [Broken]

    And putting that in terms of Q:

    http://img180.imageshack.us/img180/7451/59work5.gif [Broken]


    So if I am reading the question correctly, this is the final formula it asks for. I am really suspicious that I missed something and am forgetting a concept that will make this problem more complex than I think it is. I might also be over-thinking things as well. Any insight would be appreciated. Thank you for your time in advance and I apologize about the pictures, as I am still learning MathType.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 22, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Hi MarcZero, welcome to PF!:smile:

    [/URL]

    You are definitely trying to oversimplify things. There is nothing wrong with visualizing the solid sphere as a bunch of spherical shells (which is what I assume you meant) layered on top of each other. But the charge density on each shell will be different, since each shell is a different distance from the center, and the charge density depends on the distance from the center.

    In general, [itex]Q=\int_{\mathcal{V}}\rho(r')d\tau'\neq\rho V[/itex]...it is only when the charge density is uniform throughout the volume that you can make this conclusion. In any case, calculating the total charge on the sphere does not really help you to calculate the scalar potential.

    [/URL]

    No, the formula

    [tex]\Phi(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int_{\mathcal{V}}\frac{\rho(\textbf{r}')d\tau'}{R}[/tex]

    tells you to treat each infinitesimal piece of charge [itex]dq'=\rho(\textbf{r}')d\tau'[/itex] as a point charge, located at [itex]\textbf{r'}[/itex], and integrate (or add up) the contribution to the potential of each piece

    [tex]d\Phi(\textbf{r})=\frac{1}{4\pi\epsilon_0}\frac{dq'}{R}=\frac{1}{4\pi\epsilon_0}\frac{\rho(\textbf{r}')d\tau'}{R}[/tex]

    Each piece of charge will be at a different distance [itex]r'[/itex] from the center, and hence will have a different charge density and will also be a different distance [itex]R=|\textbf{r}-\textbf{r}'|[/itex] from the field point and therefor have a different contribution to the potential.

    This means that [tex]\Phi(\textbf{r})\neq\frac{1}{4\pi\epsilon_0}\frac{Q}{r}[/tex]...You will actually need to integrate in order to find out what the potential really is!
     
    Last edited by a moderator: May 4, 2017
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