91Ó°ÊÓ

Consider thefield inside a large piece of dielectric is $E_0$.

Consider the electric displacement is $D_0={\mathrm{Î\mu }}_0{E}_0+P$.

Write the formula ofbound surface charges.

${\mathit{σ}}_{\mathbf{b}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{P}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{r}}$ …… (1)

Here,$\stackrel{\mathbf{→}}{\mathbf{p}}$is the polarization of sphere and$\stackrel{\mathbf{Áåœ}}{\mathbf{r}}$is the vector from the center.

Write the formula ofbound volume charges ${\mathrm{ÒÏ}}_b$.

${\mathit{ÒÏ}}_{\mathbf{b}}\mathbf{=}\mathbf{-}\mathbf{∇}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{p}}$ …… (2)

Here,$\stackrel{\mathbf{→}}{\mathbf{p}}$is the polarization of sphere.

Write the formula offield inside the sphere.

${\stackrel{\mathbf{→}}{\mathbf{E}}}_{\mathbf{i}\mathbf{n}}\mathbf{=}\frac{\mathbf{ÒÏ}\stackrel{\mathbf{→}}{\mathbf{r}}}{\mathbf{3}{\mathbf{Î\mu }}_{\mathbf{0}}}$ …… (3)

Here, $\mathit{p}$are charges on sphere,$\stackrel{\mathbf{Áåœ}}{\mathbf{r}}$is the vector from the center and${\mathit{Î\mu }}_{\mathbf{0}}$is relative permittivity.

Write the formula offield outside the sphere.

${\mathbf{Q}}_{\mathbf{total}}\mathbf{=}\mathbf{4}{\mathbf{Ï€¸é}}^{\mathbf{2}}\mathbf{σ}\mathbf{+}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{Ï€¸é}}^{\mathbf{3}}\mathbf{ÒÏ}$ …… (4)

Here, $\mathit{R}$is radius of sphere,$\mathit{σ}$ are bound surface charges and $\mathit{ÒÏ}$ are charges on sphere.

Determine the bound surface charges ${\mathrm{σ}}_b$.

Substitute $k$for $\stackrel{→}{p}$and R for$\stackrel{Áåœ}{r}$ into equation (1).

$\mathrm{σ}=kR$

Determine the bound volume charges ${\mathrm{ÒÏ}}_b$.

Substitute $\frac{1}{r^2}\frac{∂}{∂r}\left(r^{2}kr\right)$for$∇.\stackrel{→}{p}$ into equation (2).

$$\begin{gathered} {\mathrm{ÒÏ}}_b=-\frac{1}{2}\frac{∂}{∂r}\left(r^{2}kr\right) \\ =-3k \end{gathered}$$

Therefore, the value of bound charges ${\mathrm{σ}}_b$is $kR$and${\mathrm{ÒÏ}}_b$ is -3k .

Determine the field inside the sphere due to the uniform sphere of charge $\mathrm{ÒÏ}$.

Substitute $-3k$for $\mathrm{ÒÏ}$into equation (3).

$$\begin{gathered} {\stackrel{→}{E}}_{in}=-\frac{3k\stackrel{→}{r}}{3{\mathrm{Î\mu }}_0} \\ =-\frac{kr}{{\mathrm{Î\mu }}_0}\stackrel{Áåœ}{r} \end{gathered}$$

From equation (4), since the entire charge contained within the sphere of radius r>R should be zero, we anticipate that the field will be zero outside.

$$\begin{gathered} Q_{total}=4{\mathrm{Ï€¸é}}^2\mathrm{kR}+\frac{4}{3}{\mathrm{Ï€¸é}}^3\left(-3\mathrm{k}\right) \\ =4{\mathrm{Ï€°ì¸é}}^3-4{\mathrm{Ï€°ì¸é}}^3 \\ =0 \end{gathered}$$

Then,

${\stackrel{→}{E}}_{out}=0$

Therefore, the value of field inside and outside the sphere is ${\stackrel{→}{E}}_{in}=-\frac{kr}{{\mathrm{Î\mu }}_0}\stackrel{Áåœ}{r}\mathrm{and}{\stackrel{→}{E}}_{total}=0$.