A ball (radius $R$) has three layers. For $0<r<a$ it is a conductor with free charge $+Q$. For $a<r<b$ it is a linear dielectric $\epsilon$ with free charge embedded in it with density $\rho_{free}(r) = \left(\frac{\rho_0}{a^2}\right)r^2$. From $b<r<R$ it is a conductor again with some amount of charge on it $q$ which makes the field vanish for $r\lt R$. I'm supposed to:

  1. Find the polarization $\vec P$ in the dielectric
  2. Find bound volume charge density in the dielectric
  3. Find free and bound surface densities at each surface $r=a,b,R$

So I tried using Gauss's law here with a sphere of radius $a<r<b$ as my surface to get $\vec P$:$$\oint \vec D\cdot d\vec A =Q_{free}+\int_a^r \rho_{free} dr$$$$D = {Q+\frac{\rho_0}{3a^2}(r^3-a^3)\over4\pi r^2}$$

Using $\vec D = \frac{\vec E}{\varepsilon}$ and $\vec P = \varepsilon_0\chi_e\vec E$ and $\rho_b=-\nabla\cdot\vec P$ I calculated: $$\vec P = \frac{\varepsilon_0\chi_eQ}{4\pi\varepsilon r^2}+\frac{\varepsilon_0\chi_e\rho_0}{12\pi\varepsilon}({r\over a^2}-{a\over r^2})\hat r$$$$\rho_b = \frac{\varepsilon_0\chi_e}{2\pi\varepsilon}(\frac{Q}{2r^2}-\frac{\rho_0r}{3a^2}+\frac{\rho_0}{6r^2})$$

Two questions:

  1. Is that right?

  2. How do I use that to calculate the surface charge densities $\sigma_b$ and $\sigma_f$ for each surface?


The value of $\rho_b$ is incorrect. Check the formula you're using.

Use $\vec{n}\cdot\ (\vec{D_{out}}-\vec{D_{in}})=\sigma_f$

$\vec{n}\cdot\ (\vec{P_{out}}-\vec{P_{in}})=\sigma_b$

On the respective surfaces.

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  • $\begingroup$ Yes, sorry, $r^1$ was a typo when inputting the question. I'll change it. $\endgroup$ – Art M May 28 '13 at 18:10
  • $\begingroup$ I see. I was using $\nabla\cdot\vec P = \frac{1}{r}\frac{\partial}{\partial r}(r\vec P)$ but it should be $\nabla\cdot\vec P = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2\vec P)$ so really $\rho_b = -\frac{\varepsilon_0 \chi_e \rho_0 r^2}{4\pi\varepsilon a^2}$ $\endgroup$ – Art M May 28 '13 at 18:15
  • $\begingroup$ Yes thats right $\endgroup$ – Man May 28 '13 at 18:16

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