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CopyOfA
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Homework Statement
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')##.
Homework Equations
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##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}##
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?