91Ó°ÊÓ

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

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

$\stackrel{\mathbf{→}}{\mathbf{P}}\mathbf{=}{\mathbf{Î\mu }}_{\mathbf{0}}{\mathbf{X}}_{\mathbf{e}}\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

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

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

$\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{3}{\mathbf{Î\mu }}_{\mathbf{0}}}\stackrel{\mathbf{→}}{\mathbf{P}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

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 ${\stackrel{→}{E}}_0$is

role="math" localid="1658484892351" ${\stackrel{→}{P}}_1={\mathrm{Î\mu }}_0{\mathcal{X}}_e{\stackrel{→}{E}}_0$

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

$$\begin{gathered} {\stackrel{→}{E}}_1=-\frac{1}{3{\mathrm{Î\mu }}_0}{\stackrel{→}{P}}_1 \\ =-\frac{1}{3{\mathrm{Î\mu }}_0}{\mathrm{Î\mu }}_0{\mathcal{X}}_e{\stackrel{→}{E}}_0 \\ =-\frac{1}{3}{\mathcal{X}}_e{\stackrel{→}{E}}_0 \end{gathered}$$

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

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

This cycle continues indefinitely. The total electric field is then

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

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

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

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