Question

A sphere of radius R carries a polarization

P(r)=kr,

Where k is a constant and r is the vector from the center.

(a) Calculate the bound charges ${\mathbf{\sigma }}_{\mathbf{b}}$and ${\mathit{\rho }}_{\mathbf{b}}$.

(b) Find the field inside and outside the sphere.

Short answer

(a) The value of bound charges ${\sigma }_b$is kR and $p_b$is -3k .

(b) The value of field inside and outside the sphere is${\overrightarrow{E}}_{in}=-\frac{kr}{{\epsilon }_0}\stackrel{⏜}{r}and{\overrightarrow{E}}_{total}=0$ .

Step-by-step solution

Write the given data from the question.

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

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

Determine the formula of bound charges σb andρb, field inside and outside the sphere.

Write the formula ofbound surface charges.

${\mathit{\sigma }}_{\mathbf{b}}\mathbf{=}\overrightarrow{\mathbf{P}}\mathbf{.}\overrightarrow{\mathbf{r}}$ …… (1)

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

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

${\mathit{\rho }}_{\mathbf{b}}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathbf{.}\overrightarrow{\mathbf{p}}$ …… (2)

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

Write the formula offield inside the sphere.

${\overrightarrow{\mathbf{E}}}_{\mathbf{i}\mathbf{n}}\mathbf{=}\frac{\mathbf{\rho }\overrightarrow{\mathbf{r}}}{\mathbf{3}{\mathbf{\epsilon }}_{\mathbf{0}}}$ …… (3)

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

Write the formula offield outside the sphere.

${\mathbf{Q}}_{\mathbf{total}}\mathbf{=}\mathbf{4}{\mathbf{\pi R}}^{\mathbf{2}}\mathbf{\sigma }\mathbf{+}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\pi R}}^{\mathbf{3}}\mathbf{\rho }$ …… (4)

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

(a) Determine the value of bound surface charges σb and bound volume charges ρb.

Determine the bound surface charges ${\sigma }_b$.

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

$\sigma =kR$

Determine the bound volume charges ${\rho }_b$.

Substitute $\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^{2}kr\right)$for$\nabla .\overrightarrow{p}$ into equation (2).

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

Therefore, the value of bound charges ${\sigma }_b$is $kR$and${\rho }_b$ is -3k .

(b) Determine the value of field inside and outside the sphere.

Determine the field inside the sphere due to the uniform sphere of charge $\rho$.

Substitute $-3k$for $\rho$into equation (3).

$$\begin{gathered} {\overrightarrow{E}}_{in}=-\frac{3k\overrightarrow{r}}{3{\epsilon }_0} \\ =-\frac{kr}{{\epsilon }_0}\stackrel{⏜}{r} \end{gathered}$$

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.

$$\begin{gathered} Q_{total}=4{\mathrm{\pi R}}^2\mathrm{kR}+\frac{4}{3}{\mathrm{\pi R}}^3\left(-3\mathrm{k}\right) \\ =4{\mathrm{\pi kR}}^3-4{\mathrm{\pi kR}}^3 \\ =0 \end{gathered}$$

Then,

${\overrightarrow{E}}_{out}=0$

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