Question

A point charge $\mathit{q}$is imbedded at the center of a sphere of linear dielectric material (with susceptibility${\mathit{\chi }}_{\mathbf{e}}$and radius $\mathit{R}$).Find the electric field, the polarization, and the bound charge densities,${\mathit{\rho }}_{\mathbf{b}}$ and ${\mathit{\sigma }}_{\mathbf{b}}$.What is the total bound charge on the surface? Where is the compensating negative bound charge located?

Short answer

The electric field is$\frac{q}{4\pi {\epsilon }_0(1+{\chi }_e)}\frac{\widehat{r}}{r^2}$ .

The polarization is $\frac{q{\chi }_e}{4\pi (1+{\chi }_e)}\frac{\widehat{r}}{r^2}$.

The volume bound charge density is $\frac{-q{\chi }_e}{1+{\chi }_e}{\delta }^3\left(r\right)$.

The surface bound charge density is $\frac{q{\chi }_e}{4\pi (1+{\chi }_e)r^2}$.

The total bound surface charge is$Q_{surface}=\frac{q{\chi }_e}{(1+{\chi }_e)}$ .

Step-by-step solution

Define function

Write the expression for the electric displacement.

$\mathit{D}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{r}}^{\mathbf{2}}}\widehat{\mathbf{r}}$ …… (1)

Here,$q$ is the charge,$r$ is the radius of dielectric sphere and$\hat{r}$ is the unit vector.

Determine electric field

We know that,

$D=\epsilon E$ …… (2)

Then write the expression for electric field.

$E=\frac{D}{\epsilon }$ …… (3)

Substitute$\frac{q}{4\pi r^2}\widehat{r}$ for$D$

$E=\frac{q}{4\pi \epsilon }\frac{\widehat{r}}{r^2}$ …… (4)

Now, write the relation between permittivity$\left(\epsilon \right)$ and susceptibility of linear dielectric sphere $\left({\chi }_e\right)$.

$\epsilon ={\epsilon }_0(1+{\chi }_e)$ …… (5)

Here,${\epsilon }_0$ is the permittivity for free space.

Substitute${\epsilon }_0(1+{\chi }_e)$ for$\epsilon$ in equation (4)

$E=\frac{q}{4\pi {\epsilon }_0(1+{\chi }_e)}\frac{\widehat{r}}{r^2}$ …… (6)

Therefore, the electric field is$E=\frac{q}{4\pi {\epsilon }_0(1+{\chi }_e)}\frac{\widehat{r}}{r^2}$ .

Determine polarization

Write the expression for Polarization.

$P={\epsilon }_0{\chi }_{e}E$ ……. (7)

Substitution$\frac{q}{4\pi {\epsilon }_0(1+{\chi }_e)}\frac{\widehat{r}}{r^2}$ for$E$ in equation (7)

$\begin{array}{c}P={\epsilon }_0{\chi }_e\frac{q}{4\pi {\epsilon }_0(1+{\chi }_e)}\frac{\widehat{r}}{r^2}\\ P=\frac{q{\chi }_e}{4\pi (1+{\chi }_e)}\frac{\widehat{r}}{r^2}\end{array}$ …….. (8)

Therefore, the polarization is$P=\frac{q{\chi }_e}{4\pi (1+{\chi }_e)}\frac{\widehat{r}}{r^2}$ .

Determine volume bound charge density

With the expression for volume bound charge density.

${\rho }_b=-\nabla \cdot P$ ……. (9)

Substitute$\frac{q{\chi }_e}{4\pi (1+{\chi }_e)}\frac{\widehat{r}}{r^2}$ for$P$ in equation (9).

$\begin{array}{c}{\rho }_b=-\nabla \cdot \left(\frac{q{\chi }_e}{4\pi (1+{\chi }_e)}\frac{\widehat{r}}{r^2}\right)\\ =\frac{-q{\chi }_e}{4\pi (1+{\chi }_e)}\left(\nabla \cdot \frac{\widehat{r}}{r^2}\right)\end{array}$ …... (10)

We know that,

$\left(\nabla \cdot \frac{\widehat{r}}{r^2}\right)=4\pi {\delta }^3\left(r\right)$

Substitute$4\pi {\delta }^3\left(r\right)$ for$\left(\nabla \cdot \frac{\widehat{r}}{r^2}\right)$ in equation (10)

$\begin{array}{c}{\rho }_b=\frac{-q{\chi }_e}{4\pi (1+{\chi }_e)}\left(\nabla \cdot \frac{\widehat{r}}{r^2}\right)\\ =\frac{-q{\chi }_e}{4\pi (1+{\chi }_e)}\left(4\pi {\delta }^3\left(r\right)\right)\\ =\frac{-q{\chi }_e}{1+{\chi }_e}{\delta }^3\left(r\right)\end{array}$

Therefore, the volume bound charge density is $\frac{-q{\chi }_e}{1+{\chi }_e}{\delta }^3\left(r\right)$.

Determine surface density

Write the expression for surface bound charge density.

${\sigma }_b=P\cdot \widehat{r}$ …… (11)

Substitute$\frac{q{\chi }_e}{4\pi (1+{\chi }_e)}\frac{\widehat{r}}{r^2}$for$P$in equation (11).

$\begin{array}{c}{\sigma }_b=P\cdot \widehat{r}\\ =\frac{q{\chi }_e}{4\pi (1+{\chi }_e)}\frac{\widehat{r}}{r^2}\left(\widehat{r}\right)\\ =\frac{q{\chi }_e}{4\pi (1+{\chi }_e)r^2}(\widehat{r}\cdot \widehat{r})\\ =\frac{q{\chi }_e}{4\pi (1+{\chi }_e)r^2}\end{array}$

Therefore, the surface bound charge density is $\frac{q{\chi }_e}{4\pi (1+{\chi }_e)r^2}$.

But the surface charges are distributed only on surface of the sphere, thus substitute $R$for $r$.

$\begin{array}{c}{\sigma }_b=\frac{q{\chi }_e}{4\pi (1+{\chi }_e)r^2}\\ =\frac{q{\chi }_e}{4\pi (1+{\chi }_e)R^2}\end{array}$

Now, write the expression for total bound surface charge.

$Q_{surface}=\left({\sigma }_b\right)4\pi R^2$ …… (12)

Substitute$\frac{q{\chi }_e}{4\pi (1+{\chi }_e)R^2}$for ${\sigma }_b$in equation (12)

$\begin{array}{c}{Q}_{surface}=\left({\sigma }_b\right)4\pi R^2\\ =\left(\frac{q{\chi }_e}{4\pi (1+{\chi }_e)R^2}\right)4\pi R^2\\ =\frac{q{\chi }_e}{(1+{\chi }_e)}\end{array}$

Therefore, the total bound surface charge is $Q_{surface}=\frac{q{\chi }_e}{(1+{\chi }_e)}$.

Because surface charge density and volume charge are independent of angular distribution, the highest negative charge should be at the sphere's center. It means that the charge distribution is symmetric.

$Q=\int {\rho }_{b}d\tau$ …… (13)

Substitute $\frac{-q{\chi }_e}{1+{\chi }_e}{\delta }^3\left(r\right)$for ${\rho }_b$in equation (13)

$\begin{array}{c}Q=\int {\rho }_{b}d\tau \\ =\int \frac{-q{\chi }_e}{1+{\chi }_e}{\delta }^3\left(r\right)d\tau \\ =-\frac{q{\chi }_e}{1+{\chi }_e}\int {\delta }^3\left(r\right)d\tau \\ =-\frac{q{\chi }_e}{1+{\chi }_e}\end{array}$

Therefore,$Q=-\frac{q{\chi }_e}{1+{\chi }_e}$ .