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The dielectric shaped as on the figure has dielectric constant $\varepsilon=\varepsilon\left(r\right)$ and free charge density $\rho=\rho\left(r\right)$. What is the electric field and electrostatic potential as a function of $r$? The above is 3D object. We have $R_{1}$, $R_{2}$ and $R$.

I really have no idea how to approach this kind problems. Does electric field in $r<R_{1}$ equal zero? Is the object a Gaussian surface? Does electric field in $R_{1} < r < R_{2}$ depend on dielectric constant?

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Do you know that the problem that you are supposed to solve is $\nabla \cdot \mathbf{E} = 4\pi \rho/\epsilon$? –  Fabian Apr 16 '11 at 17:56
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closed as off-topic by tpg2114, Qmechanic Nov 3 '13 at 2:27

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1 Answer

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You are looking for a solution of Gauss's law $$\nabla \cdot \mathbf{E}(r) = \rho(r)/\epsilon_0\epsilon(r).$$ Because the system is static you can find a potential $\Phi(r)$ such that the electric field is given by $$\mathbf{E}(r) = - \nabla \Phi(r).$$ Thus, the equation becomes $$\Delta \Phi(r) = \frac{1}{r^2} \partial_r [ r^2 \partial_r \Phi(r)] = - \rho(r)/\epsilon_0\epsilon(r);$$ in the last step we used the isotropy of the system. Solving this ordinary differential equation, you obtain the electric potential $\Phi(r)$. In order to solve this equation, we use the substitution $\Phi(r) =f(r)/r$, after which the differential equation reads $$f''(r)=- r\rho(r)/\epsilon_0\epsilon(r).$$ Integrating twice yields $$f(r) = -\int_0^r dr' \int_0^{r'} dr'' \frac{r'' \rho(r'')}{\epsilon_0\epsilon(r'')},$$ where we used the boundary conditions $f(0)=0$, $f'(0)=0$ (which requires that $\rho(r)$ goes to 0 for $r\to0$ sufficiently fast). In principal, you can add $c r$ to $f$ with $c$ an arbitrary constant in order to fulfill "any" boundary condition. Putting everything together, we have $$\Phi(r) = -\int_0^r dr' \int_0^{r'} dr'' \frac{r'' \rho(r'')}{r \epsilon_0\epsilon(r'')}$$ and $$E_r(r)= -\partial_r \Phi(r) = \int_0^{r} dr' \frac{r' \rho(r')}{r \epsilon_0\epsilon(r')} - \int_0^r dr' \int_0^{r'} dr'' \frac{r'' \rho(r'')}{r^2 \epsilon_0\epsilon(r'')}.$$ For more concrete results, you need to provide the explicit form of $\rho(r)$ and $\epsilon(r)$.

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A trivial correction: in SI units, which is almost certainly what the questioner is using, there's no $4\pi$ in Gauss's law. –  Ted Bunn Apr 16 '11 at 18:20
    
@Ted Bunn: thanks I corrected it. –  Fabian Apr 16 '11 at 23:51
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