$\overrightarrow{{\mathbf{E}}_{\mathbf{2}}}$Find the field inside a sphere of linear dielectric material in an otherwise uniform electric field $\overrightarrow{{\mathbf{E}}_{\mathbf{0}}}$(Ex. 4.7) by the following method of successive approximations: First pretend the field inside is just $\overrightarrow{{\mathbf{E}}_{\mathbf{0}}}$, and use Eq. 4.30 to write down the resulting polarization $\overrightarrow{{\mathbf{P}}_{\mathbf{0}}}$. This polarization generates a field of its own, $\overrightarrow{{\mathbf{E}}_{\mathbf{1}}}$ (Ex. 4.2), which in turn modifies the polarization by an amount $\overrightarrow{{\mathbf{P}}_{\mathbf{1}}}$. which further changes the field by an amount $\overrightarrow{{\mathbf{E}}_{\mathbf{2}}}$, and so on. The resulting field is ${\overrightarrow{\mathbf{E}}}_{\mathbf{0}}\mathbf{+}{\overrightarrow{\mathbf{E}}}_{\mathbf{1}}\mathbf{+}{\overrightarrow{\mathbf{E}}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$ . Sum the series, and compare your answer with Eq. 4.49.

The uniform electric field is ${\overrightarrow{E}}_0$.

The polarization caused in the presence of an electric field $\overrightarrow{E}$is

$\overrightarrow{\mathbf{P}}\mathbf{=}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{X}}_{\mathbf{e}}\overrightarrow{\mathbf{E}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, ${\mathrm{\epsilon }}_0$ is the permittivity of free space and ${\mathcal{X}}_e$is the dielectric constant of the medium.

The electric field generated by a polarization $\overrightarrow{P}$is

$\overrightarrow{\mathbf{E}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}{\mathbf{\epsilon }}_{\mathbf{0}}}\overrightarrow{\mathbf{P}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

The sum of an infinite geometric series is

$\mathbf{S}\mathbf{=}\frac{\mathbf{a}}{\mathbf{1}\mathbf{-}\mathbf{r}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Here, a is the first term and r is the common ratio.

From equation (1), the polarization caused by the uniform electric field ${\overrightarrow{E}}_0$is

role="math" localid="1658484892351" ${\overrightarrow{P}}_1={\epsilon }_0{\mathcal{X}}_e{\overrightarrow{E}}_0$

From equation (2), the corresponding electric field generated by the polarization ${\overrightarrow{P}}_1$is

$$\begin{gathered} {\overrightarrow{E}}_1=-\frac{1}{3{\epsilon }_0}{\overrightarrow{P}}_1 \\ =-\frac{1}{3{\epsilon }_0}{\epsilon }_0{\mathcal{X}}_e{\overrightarrow{E}}_0 \\ =-\frac{1}{3}{\mathcal{X}}_e{\overrightarrow{E}}_0 \end{gathered}$$

This field creates another polarization which again results in another electric field

role="math" localid="1658485109126" $$\begin{gathered} {\overrightarrow{E}}_2=\left(-\frac{1}{3}{\mathcal{X}}_e\right)\left(-\frac{1}{3}{\mathcal{X}}_e\right){\overrightarrow{E}}_0 \\ =\frac{{{\mathcal{X}}^2}_e}{9}{\overrightarrow{E}}_0 \end{gathered}$$

This cycle continues indefinitely. The total electric field is then

$$\begin{gathered} \overrightarrow{\mathit{E}}\mathit{=}{\overrightarrow{\mathit{E}}}_{\mathit{0}}\mathit{+}{\overrightarrow{\mathit{E}}}_{\mathit{1}}\mathit{+}{\overrightarrow{\mathit{E}}}_{\mathit{2}}\mathit{+}\mathit{.}\mathit{.}\mathit{.}\mathit{.} \\ \mathit{}\mathit{}\mathit{}\mathit{=}{\overrightarrow{\mathit{E}}}_{\mathit{0}}\mathit{+}\left(\mathrm{-}\frac{\mathrm{1}}{\mathrm{3}}{\mathcal{X}}_e\right){\overrightarrow{\mathit{E}}}_{\mathit{0}}+\mathit{(}\mathit{-}\frac{\mathit{1}}{\mathit{3}}{\mathit{X}}_{\mathit{e}}\mathit{)}\mathit{(}\mathit{-}\frac{\mathit{1}}{\mathit{3}}{\mathit{X}}_{\mathit{e}}\mathit{)}{\overrightarrow{\mathrm{E}}}_{\mathit{0}}+.... \\ ={\overrightarrow{\mathrm{E}}}_{\mathit{0}}\left[1+\mathit{(}\mathit{-}\frac{\mathit{1}}{\mathit{3}}{\mathit{X}}_{\mathit{e}}\mathit{)}+\mathit{(}\mathit{-}\frac{\mathit{1}}{\mathit{3}}{\mathit{X}}_{\mathit{e}}\mathit{)}\mathit{(}\mathit{-}\frac{\mathit{1}}{\mathit{3}}{\mathit{X}}_{\mathit{e}}\mathit{)}+....\right] \end{gathered}$$

To do the sum of this infinite series, equation (3) is used

$$\begin{gathered} \overrightarrow{E}={\overrightarrow{E}}_0\left[\frac{1}{1-\left(-{\displaystyle \frac{1}{3}}{\mathcal{X}}_e\right)}\right] \\ ={\overrightarrow{E}}_0\left[\frac{1}{1+{\displaystyle \frac{{\mathcal{X}}_e}{3}}}\right] \end{gathered}$$

Thus, the net electric field is ${\overrightarrow{E}}_0\left[\frac{1}{1+{\displaystyle \frac{{\mathcal{X}}_e}{3}}}\right]$