Question

Calculate the potential of a uniformly polarized sphere (Ex. 4.2) directly from Eq. 4.9.

Short answer

The value of potential of a uniformly polarized sphere is $\mathrm{V}\left(\overrightarrow{\mathrm{r}}\right)=\overrightarrow{\mathrm{P}}\cdot \frac{1}{4{\mathrm{\pi \epsilon }}_0}{\int }_{}\frac{\hat{\mathrm{l}}}{{\mathrm{l}}^2}{\mathrm{d\tau }}^{\text{'}}$.

The value of polarization vector for the electric field of a homogenous sphere of charge inside the sphere$\rho =1$ is $V(r,\theta )=\frac{\mathrm{Prcos}\theta }{3{\epsilon }_0}$.

The value of polarization vector for the electric field of a homogenous sphere of charge outside the sphere$\rho =1$ is$V(r,\theta )=\frac{\mathrm{P}{R}^3\cos \theta }{3{\epsilon }_0{r}^2}$ .

Step-by-step solution

Write the given data from the question

Reference as Ex. 4.2

Consider $\overrightarrow{P}$will be point along the z-axis.

Consider$I$ will be using as letter.

Consider$l$ as the distance from the source to the point of interest.

Determine the formula of potential of a uniformly polarized sphere and polarization vector for the electric field of a homogenous sphere of charge.

Write the formula of potential of auniformly polarized sphere.

$\mathbf{V}\left(\overrightarrow{\mathbf{r}}\right)=\frac{1}{\mathbf{4}\pi {\mathbf{\epsilon }}_0}{\int }_{\nu }\frac{\overrightarrow{\mathbf{P}}\left({\mathbf{r}}^{\mathbf{\text{'}}}\right)\cdot \widehat{\mathbf{I}}}{{\mathbf{l}}^2}\mathbf{d}{\tau }^{\text{'}}$ …… (1)

Here, $\overrightarrow{\mathbf{p}}$is vector constant in both magnitude and direction,$\stackrel{}{\mathbf{}\mathbf{}\mathbf{r}}$ is inner radius of sphere,$\widehat{\mathbf{I}}$will be using as letter, $\mathit{l}$as the distance from the source to the point of interest and role="math" localid="1657544577909" ${\mathbf{\epsilon }}_0$is relative pemitivity.

Write the formula ofpolarization vector for the electric field of a homogenous sphere inside the sphere.

$\mathbf{V}(\mathbf{r}\mathbf{,}\mathbf{\theta })=\overrightarrow{\mathbf{P}}\cdot \frac{\overrightarrow{r}}{\mathbf{3}{\mathbf{\epsilon }}_0}$ …… (2)

Here, $\overrightarrow{\mathbf{P}}$is vector constant in both magnitude and direction, $\overrightarrow{\mathbf{r}}$is radius of sphere and role="math" localid="1657544779642" ${\mathbf{\epsilon }}_0$is relative pemitivity.

Write the formula of polarization vector for the electric field of a homogenous sphere outside the sphere.

$\mathbf{V}(\mathbf{r}\mathbf{,}\mathbf{\theta })=\overrightarrow{\mathbf{P}}\cdot \frac{{\mathbf{R}}^3}{\mathbf{3}{\mathbf{\epsilon }}_0{\mathbf{r}}^2}\widehat{\mathbf{r}}$ …… (3)

Here, $\overrightarrow{\mathbf{P}}$is vector constant in both magnitude and direction, $\overrightarrow{\mathbf{r}}$is radius of sphere, $\mathbf{R}$is outer radius of sphere and ${\mathbf{\epsilon }}_0$is relative permittivity.

Determine the value of potential of a uniformly polarized sphere and polarization vector for the electric field of a homogenous sphere of charge.

The distance $I$between the source and the place of interest will be represented by the letter $I$due to site limitations.

The sphere has continuous polarization (so $\overrightarrow{P}$is a vector constant in both magnitude and direction). Specify $\overrightarrow{P}$as the z-axis pointer. The potential using (eq. 4.9) is:

Determine the potential of a uniformly polarized sphere.

Substitute$1$for${\overrightarrow{r}}^{\text{'}}$into equation (1).

$V\left(\overrightarrow{r}\right)=\overrightarrow{P}\cdot \frac{1}{4\pi {\epsilon }_0}{\int }_{\nu }\frac{\hat{l}}{l^2}d{\tau }^{\text{'}}$

The electric field of a homogeneous sphere of charge with $\rho =1$may be calculated accurately by multiplying the polarization vector. So, for $r<R$:

Determine thepolarization vector for the electric field of a homogenous sphere of charge inside the sphere.

Substitute$r\cos \theta$for $\overrightarrow{r}$into equation (2).

$V_{ins}(r,\theta )=\frac{\overrightarrow{P}r\cos \theta }{3{\epsilon }_0}$

Therefore, the value of polarization vector for the electric field of a homogenous sphere of charge inside the sphere $\rho =1$is $V(r,\theta )=\frac{\mathrm{Prcos}\theta }{3{\epsilon }_0}$.

Determine thepolarization vector for the electric field of a homogenous sphere of charge outside the sphere.

Substitute$\cos \theta$ for $\hat{r}$into equation (3).

$V_{out}(r,\theta )=\frac{\overrightarrow{P}{R}^3\cos \theta }{3{\epsilon }_0{r}^2}$

Therefore, the value of polarization vector for the electric field of a homogenous sphere of charge outside the sphere $\rho =1$is $V(r,\theta )=\frac{\mathrm{P}{R}^3\cos \theta }{3{\epsilon }_0{r}^2}$.