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Potential inside concentric spherical shells with non-uniform charge density

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data
    We are given a two concentric spherical shells with small radius ## a ## and larger radius ## b ##. The inner and outer shells are made of conducting material and there is a volume charge density, ##\rho\left(r\right) ##, that exists between the shells,. The boundary conditions are ##\phi\left(a\right) = 0## and ##\phi\left(b\right) = V_0##.

    Show that the potential for ##a<r<b## is:

    ##\phi\left(r\right) = \dfrac{ab}{b-a} \left[V_0 \left(\dfrac{1}{a}- \dfrac{1}{r}\right) + \int\limits_{a}^{b} \dfrac{r'^2 dr'}{\epsilon_0} \rho\left(r\right) \left(\dfrac{1}{a}-\dfrac{1}{r_<}\right)\left(\dfrac{1}{r_>} - \dfrac{1}{b}\right) \right] ##

    where ##r_< = lesser(r,r')## and ##r_> = greater(r,r')##.


    2. Relevant equations

    ##Q\left(\mathbf{r}\right) = \int\limits_{V}\rho\left(\mathbf{r}\right)dV##

    ##\int\limits_{S}\mathbf{E}\cdot d\mathbf{S} = \dfrac{Q_{enclosed}}{\epsilon_0}##

    ##\phi\left(\mathbf{r}\right) = -\int\limits_{\mathbf{r}_0}^{\mathbf{r}} \mathbf{E}\cdot d\mathbf{l}##

    ##\nabla^2 \phi = -\dfrac{\rho\left(\mathbf{r}\right)}{\epsilon_0}##

    3. The attempt at a solution
    Since the inner surface is kept at zero potential, a charge will be induced on it's surface equal to the total charge of the charge density. And thus, the charge enclosed by a Gaussian sphere of radius ##r## (##a<r<b##) will be:

    ##Q\left(r\right) = -\sigma_{induced}\left(a\right) + \int\limits_{V}\rho\left(r\right)dV = -4\pi \int\limits_{a}^{b} \rho\left(r\right)r^2 dr + 4\pi\int\limits_{a}^{r}\rho\left(s\right)s^2 ds##

    ##E\left(r\right) = \dfrac{1}{4\pi\epsilon_0 r^2}Q\left(r\right) = \dfrac{1}{\epsilon_0 r^2} \left[ -\int\limits_{a}^{b} \rho\left(t\right) t^2 dt + \int\limits_{a}^{r}\rho\left(s\right)s^2 ds \right] ##

    ##\phi\left(r\right) = -\int\limits_{a}^{r} E_{r}dr = \int\limits_{a}^{r} \left\{\dfrac{1}{\epsilon_0 r'^2} \left[ -\int\limits_{a}^{b} \rho\left(t\right) t^2 dt + \int\limits_{a}^{r'}\rho\left(s\right)s^2 ds \right] dr'\right\} = \int\limits_{a}^{r} \left[\dfrac{1}{r'^2}\int\limits_{a}^{b}\dfrac{\rho\left(t\right)}{\epsilon_0}t^2 dt - \dfrac{1}{\epsilon_0 r'^2}\int\limits_{a}^{r'}\rho\left(s\right)s^2 ds \right] dr'##

    Taking the first sub-integral:
    ##\int\limits_{a}^{b}\dfrac{\rho\left(t\right)}{\epsilon_0}t^2 dt = \int\limits_{a}^{b} \left(-\nabla^2 \phi\right) t^2 dt##

    In spherical coordinates, with dependence only on ##r##,
    ##\nabla^2 \phi = \dfrac{1}{r^2}\dfrac{d}{dr}\left(r^2 \dfrac{d\phi}{dr} \right)##

    Hence,
    ##\int\limits_{a}^{b} \left(-\nabla^2 \phi\right) t^2 dt = -\int\limits_{a}^{b}\left[\dfrac{1}{t^2}\dfrac{d}{dt}\left(t^2 \dfrac{d\phi}{dt} \right)\right] t^2 dt = -\int\limits_{a}^{b}\dfrac{d}{dt}\left(t^2 \dfrac{d\phi}{dt} \right) dt = -t^2\dfrac{d\phi}{dt}\bigg|_{t=a}^{b}##

    Plugging into larger integral,
    ##\int\limits_{a}^{r} \left[\dfrac{1}{r'^2}\left(-t^2\dfrac{d\phi}{dt}\bigg|_{t=a}^{b}\right)- \dfrac{1}{\epsilon_0 r'^2}\int\limits_{a}^{r'}\rho\left(s\right)s^2 ds \right] dr' = -\int\limits_{a}^{r}\dfrac{1}{r'^2}\left[b^2\dfrac{d\phi}{dt}\bigg|_{t=b} - a^2\dfrac{d\phi}{dt}\bigg|_{t=a}\right]dr' - \int\limits_{a}^{r}\dfrac{1}{r'^2}u\left(r'\right)dr'##

    where ##u\left(r'\right) = \int\limits_{a}^{r'}\dfrac{\rho\left(s\right)}{\epsilon_0}s^2 ds##

    I feel like this is leading nowhere though... For one, I don't know the value of the potential derivatives anywhere, much less the two surfaces. Second, I don't have an integral of the density from ##a## to ##b##. Finally, I don't have any clue where the ##r_<## and ##r_>## come into play.

    Any suggestion? Am I doing some of this work wrong?
     
  2. jcsd
  3. Oct 21, 2014 #2
    In case anyone was wondering, I did figure this out, after much struggling. I'll leave this here for the next poor soul who needs some direction on a similar problem. The approach is to use Green's function. Consider Poisson's equation:
    $$\nabla^2 \phi = -\dfrac{\rho\left(r\right)}{\epsilon_0}$$
    Carrying out the derivative in only the radial direction (as our charge distribution only varies radially):

    $$\dfrac{1}{r^2}\dfrac{d}{dr}\left(r^2 \dfrac{d\phi}{dr}\right) = \dfrac{1}{r^2}\left(2r\dfrac{d\phi}{dr} + r^2\dfrac{d^2 \phi}{dr^2}\right) = -\dfrac{\rho\left(r\right)}{\epsilon_0}$$
    Our differential equation is:

    $$r^2\dfrac{d^2 \phi}{dr} + 2r \dfrac{d\phi}{dr} = -\dfrac{r^2\rho\left(r\right)}{\epsilon_0}$$
    Following convention in finding the Green’s function for this setup, multiply both sides of the equation by a scalar function [itex]\psi\left(r\right)[/itex] and integrate over the region:

    $$\int\limits_{a}^{b}r^2\psi\left(r\right)\dfrac{d^2 \phi}{dr} dr + \int\limits_{a}^{b}2r\psi\left(r\right) \dfrac{d\phi}{dr} dr = \int\limits_{a}^{b}-\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\psi\left(r\right) dr$$
    Performing integration by parts on each integral:

    $$\begin{align}r^2\psi\left(r\right)\dfrac{d\phi}{dr}\bigg|_{a}^{b} &- \left(2r\psi\left(r\right) + r^2 \dfrac{d\psi}{dr}\right)\dfrac{d\phi}{dr}\bigg|_{a}^{b} + \int\limits_{a}^{b}\phi\left(r\right)\left(2\psi\left(r\right) + 4r\dfrac{d\psi}{dr} + r^2\dfrac{d^2\psi}{dr^2} \right)dr \\
    &+ 2r\psi\left(r\right)\phi\left(r\right)\bigg|_{a}^{b} - 2\int\limits_{a}^{b}\phi\left(r\right)\psi\left(r\right)dr - 2\int\limits_{a}^{b}r\phi\left(r\right)\dfrac{d\psi}{dr}dr = K \end{align}$$

    where [itex]K = \int\limits_{a}^{b}-\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\psi\left(r\right)dr[/itex].
    Canceling like terms:

    $$2\int\limits_{a}^{b} r\phi\left(r\right)\dfrac{d\psi}{dr}dr + \int\limits_{a}^{b} r^2 \phi\left(r\right)\dfrac{d^2\psi}{dr^2} dr + r^2\left[\psi\left(r\right)\dfrac{d\phi}{dr} - \phi\left(r\right)\dfrac{d\psi}{dr}\right]\bigg|_{a}^{b} = K$$
    And we get:

    $$\int\limits_{a}^{b} r^2\left(\nabla^2 \psi\right) \phi\left(r\right) dr + r^2\left[\psi\left(r\right)\dfrac{d\phi}{dr} - \phi\left(r\right)\dfrac{d\psi}{dr}\right]\bigg|_{a}^{b} = K$$
    Letting [itex]\nabla^2\psi\left(r\right) = -4\pi\delta\left(\mathbf{x}-\mathbf{x}'\right) = -\dfrac{4\pi}{r^2}\delta\left(r-r'\right)[/itex] the above equation is:

    $$\int\limits_{a}^{b} r^2 \left( -\dfrac{4\pi}{r^2}\delta\left(r-r'\right)\right)\phi\left(r\right) dr + r^2\left[\psi\left(r\right)\dfrac{d\phi}{dr} - \phi\left(r\right)\dfrac{d\psi}{dr}\right]\bigg|_{a}^{b} = -4\pi\phi\left(r'\right) + r^2\left[\psi\left(r\right)\dfrac{d\phi}{dr} - \phi\left(r\right)\dfrac{d\psi}{dr}\right]\bigg|_{a}^{b} = K$$

    Letting [itex]\psi\left(a\right) = \psi\left(b\right) = 0,[/itex] and noticing that [itex]\phi(a) = 0, \phi(b) = V_0[/itex], the result is:
    $$-4\pi\phi\left(r'\right) + r^2\left[\psi\left(r\right)\dfrac{d\phi}{dr} - \phi\left(r\right)\dfrac{d\psi}{dr}\right]\bigg|_{a}^{b} = -4\pi\phi\left(r'\right) - b^2V_0 \dfrac{d\psi}{dr} = K$$

    Therefore,

    $$\phi\left(r’\right) = \dfrac{1}{4\pi}\left[- b^2V_0 \dfrac{d\psi}{dr} + \int\limits_{a}^{b}\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\psi\left(r\right)dr\right]$$
    Noting that in the region [itex]r<r’, \nabla^2 \psi = 0,[/itex] try the function

    $$\psi\left(r\right) = A + Br^{-1}$$
    This function satisfies the above homogenous differential equation. Applying the boundary condition at [itex]a[/itex],

    $$\psi\left(a\right) = 0 = A + Ba^{-1}, \Longrightarrow B = -Aa$$
    $$\psi_{1}\left(r\right) = A\left(1-ar^{-1}\right)$$
    Similarly, in the region [itex]r>r’, \nabla^2 \psi = 0.[/itex]

    Therefore, assuming a solution and applying a boundary condition at [itex]b[/itex],

    $$\psi\left(b\right) = 0 = C + Db^{-1}, \Longrightarrow D = -Ca$$
    $$\psi_{2}\left(r\right) = C\left(1-br^{-1}\right)$$
    In order to determine the value of the coefficients, one must use the following equations:
    $$\psi_{1}\left(r’\right) = \psi_{2}\left(r’\right)$$
    $$\int\limits_{r’-\varepsilon}^{r’+\varepsilon} \nabla^2 \psi dr = \int\limits_{r’-\varepsilon}^{r’+\varepsilon} -\dfrac{4\pi}{r^2}\delta\left(r-r’\right)dr = -\dfrac{4\pi}{r’^2}$$
    Hence,

    $$A\left(1-ar'^{-1}\right) = C\left(1-br'^{-1}\right) \Longrightarrow A = C\left(\dfrac{r'-b}{r'-a}\right)$$
    and

    $$\int\limits_{r’-\varepsilon}^{r’+\varepsilon} \left[\dfrac{2}{r}\dfrac{d\psi}{dr} + \dfrac{d^2\psi}{dr^2}\right]dr = \dfrac{2}{r}\psi\left(r\right)\bigg|_{r’-\varepsilon}^{r’+\varepsilon} + \int\limits_{r’-\varepsilon}^{r’+\varepsilon}\dfrac{2}{r^2}\psi\left(r\right) + \dfrac{d\psi}{dr}\bigg|_{r’-\varepsilon}^{r’+\varepsilon}$$
    It's clear that the first two terms in will vanish as [itex]\varepsilon \rightarrow 0[/itex]. The third term is:

    $$\dfrac{d\psi}{dr}\bigg|_{r’-\varepsilon}^{r’+\varepsilon} = \dfrac{d\psi_{2}\left(r\right)}{dr}\bigg|_{r = r' +\varepsilon} - \dfrac{d\psi_{1}\left(r\right)}{dr}\bigg|_{r = r' - \varepsilon} = Cb\left(r'+\varepsilon\right)^{-2} - Aa\left(r'-\varepsilon\right)^{-2}$$
    Letting [itex]\varepsilon \rightarrow 0[/itex],

    $$Cbr'^{-2} - Aar'^{-2} = -\dfrac{4\pi}{r'^2}$$
    $$Cb - Aa = -4\pi$$
    $$Cb - Ca\left(\dfrac{r'-b}{r'-a}\right) = -4\pi$$
    $$C = -\dfrac{4\pi}{b-a}\left(\dfrac{r'-a}{r'}\right) \Longrightarrow A = -\dfrac{4\pi}{b-a}\left(\dfrac{r'-b}{r'}\right)$$
    Hence,
    $$\psi\left(r,r'\right) = \begin{cases} -\dfrac{4\pi}{b-a}\left(\dfrac{r'-b}{r'}\right)\left(1-ar^{-1}\right) & r<r' \\
    -\dfrac{4\pi}{b-a}\left(\dfrac{r'-a}{r'}\right)\left(1-br^{-1}\right) & r>r' \end{cases}$$
    $$\psi\left(r,r'\right) = \begin{cases} -\dfrac{4\pi}{b-a}\left(\dfrac{r'-b}{r'}\right)\left(\dfrac{r-a}{r}\right) & r<r' \\
    -\dfrac{4\pi}{b-a}\left(\dfrac{r'-a}{r'}\right)\left(\dfrac{r-b}{r}\right) & r>r' \end{cases}$$
    $$\psi\left(r,r'\right) = \begin{cases} -\dfrac{4\pi ab}{b-a}\left(\dfrac{1}{b} - \dfrac{1}{r'}\right)\left(\dfrac{1}{a} - \dfrac{1}{r}\right) & r<r' \\
    -\dfrac{4\pi ab}{b-a}\left(\dfrac{1}{a} - \dfrac{1}{r'}\right)\left(\dfrac{1}{b} - \dfrac{1}{r}\right) & r>r' \end{cases}$$
    Finally,
    $$\psi\left(r,r'\right) = \dfrac{4\pi ab}{b-a}\left(\dfrac{1}{r_>} - \dfrac{1}{b}\right)\left(\dfrac{1}{a} - \dfrac{1}{r_<}\right)$$
    where [itex]r_< = \min\left(r,r'\right)[/itex] and [itex]r_> = \max\left(r,r'\right)[/itex].

    We also need [itex]d\psi/dr[/itex] at [itex]r=b[/itex]:
    $$\dfrac{d\psi}{dr} \bigg|_{r=b} = -\dfrac{4\pi ab}{b-a}\left(\dfrac{1}{a} - \dfrac{1}{r'}\right)\dfrac{1}{r^2}\bigg|_{r=b} = -\dfrac{4\pi a}{b\left(b-a\right)}\left(\dfrac{1}{a} - \dfrac{1}{r'}\right)$$
    Plugging this into the equation for the potential determined earlier,
    $$\phi\left(r’\right) = \dfrac{1}{4\pi}\left[- b^2V_0 \dfrac{d\psi}{dr} + \int\limits_{a}^{b}\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\psi\left(r\right)dr\right]$$
    $$\phi\left(r’\right) = \dfrac{1}{4\pi}\left[-b^2V_0\left(-\dfrac{4\pi a}{b\left(b-a\right)}\left(\dfrac{1}{a} - \dfrac{1}{r'}\right)\right) + \int\limits_{a}^{b}\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\dfrac{4\pi ab}{b-a}\left(\dfrac{1}{r_>} - \dfrac{1}{b}\right)\left(\dfrac{1}{a} - \dfrac{1}{r_<}\right)dr\right]$$
    $$\phi\left(r’\right) = \dfrac{ab}{b-a}\left[V_0\left(\dfrac{1}{a} - \dfrac{1}{r'}\right) + \int\limits_{a}^{b}\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\left(\dfrac{1}{r_>} - \dfrac{1}{b}\right)\left(\dfrac{1}{a} - \dfrac{1}{r_<}\right)dr\right]$$
     
    Last edited: Oct 21, 2014
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