A thick spherical shell (inner radius a, outer radius b) is made of dielectric material with a "frozen-in" polarization

$\mathbf{P}\left(\mathbf{r}\right)=\frac{k}{r}\widehat{\mathbf{r}}$

Where a constant and is the distance from the center (Fig. 4.18). (There is no free charge in the problem.) Find the electric field in all three regions by two different methods:

Figure 4.18

(a) Locate all the bound charge, and use Gauss's law (Eq. 2.13) to calculate the field it produces.

(b) Use Eq. 4.23 to find $\mathbf{D}$, and then get$\mathbf{E}$ from Eq. 4.21. [Notice that the second method is much faster, and it avoids any explicit reference to the bound charges.]

Consider a thick spherical shell (inner radius a, outer radius b) is made of dielectric material

Write the formula ofelectrical field produces in all the bound charge.

$\overrightarrow{\mathbf{E}}=\frac{{\mathbf{Q}}_{\mathrm{inc}}}{\mathbf{4}\pi {\mathbf{r}}^2{\mathbf{\epsilon }}_0}$ …… (1)

Here,${\mathbf{Q}}_{\mathrm{inc}}$ is charge inside of the sphere,$\mathbf{r}$ is radius of the sphere and${\mathbf{\epsilon }}_0$ is relative pemitivity.

Write the formula ofelectrical field produces there no free charge anywhere.

$\overrightarrow{\mathbf{D}}={\mathbf{\epsilon }}_0\overrightarrow{\mathbf{E}}+\overrightarrow{\mathbf{P}}$ …… (2)

Here,${\mathbf{\epsilon }}_0$ is relative pemitivity,$\mathbf{E}$ is electric field and$\overrightarrow{\mathbf{P}}$ is polarization.

The bound surface and volume charge are

${\sigma }_b=\overrightarrow{P}\cdot \widehat{n}=\left\{\begin{array}{l}-k/a,r=a\\ k/b,r=b\end{array}\right\}$

$\begin{array}{c}{\rho }_b=-\nabla \cdot \overrightarrow{P}\\ =-\frac{1}{r^2}\frac{\partial }{\partial r}\left(kr\right)\\ =-\frac{k}{r^2}\end{array}$

Inside of the sphere $Q_{inc}=0$so the electric field is obviously zero. Now, in the middle region.

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Determine the electric fieldproduces in all the bound charge.

Substitute $4\pi kr$for $Q_{inc}$into equation (1).

$\begin{array}{c}\overrightarrow{E}=-\frac{4\pi kr}{4\pi r^2{\epsilon }_0}\\ =-\frac{k}{r{\epsilon }_0}\widehat{r}\end{array}$

Therefore, the value of electrical field produces in all the bound charge is $\overrightarrow{E}=-\frac{k}{\gamma {\epsilon }_0}\widehat{r}$.

Determine the value of $D$.

$\begin{array}{c}{\oint }_S\overrightarrow{D}\cdot d\overrightarrow{S}=Q_{f,inc}\\ =0\\ \overrightarrow{D}=0\end{array}$

Since there is no free charge anywhere.

Now, determine the electrical field produces there no free charge anywhere.

Substitute$0$ for$\overrightarrow{D}$ into equation (2).

$\begin{array}{c}0={\epsilon }_0\overrightarrow{E}+\overrightarrow{P}\\ \overrightarrow{E}=-\frac{\overrightarrow{P}}{{\epsilon }_0}\end{array}$

This shows that the electric field is zero between outside and inside ( $r>b$and $r<a$, respectively) and between:

$\overrightarrow{E}=-\frac{k}{{\epsilon }_{0}r}\widehat{r}$

Therefore, the value of electrical field produces there no free charge anywhere is$\overrightarrow{E}=-\frac{k}{{\epsilon }_0\gamma }\widehat{r}$ .