91Ó°ÊÓ

Write the expression for the electric displacement.

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

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

We know that,

$D=\mathrm{Î\mu }E$ …… (2)

Then write the expression for electric field.

$E=\frac{D}{\mathrm{Î\mu }}$ …… (3)

Substitute$\frac{q}{4\mathrm{Ï€}{r}^2}\widehat{r}$ for$D$

$E=\frac{q}{4\mathrm{Ï€}\mathrm{Î\mu }}\frac{\widehat{r}}{r^2}$ …… (4)

Now, write the relation between permittivity$\left(\mathrm{Î\mu }\right)$ and susceptibility of linear dielectric sphere $\left({\mathrm{χ}}_e\right)$.

$\mathrm{Î\mu }={\mathrm{Î\mu }}_0(1+{\mathrm{χ}}_e)$ …… (5)

Here,${\mathrm{Î\mu }}_0$ is the permittivity for free space.

Substitute${\mathrm{Î\mu }}_0(1+{\mathrm{χ}}_e)$ for$\mathrm{Î\mu }$ in equation (4)

$E=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(1+{\mathrm{χ}}_e)}\frac{\widehat{r}}{r^2}$ …… (6)

Therefore, the electric field is$E=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(1+{\mathrm{χ}}_e)}\frac{\widehat{r}}{r^2}$ .

Write the expression for Polarization.

$P={\mathrm{Î\mu }}_0{\mathrm{χ}}_{e}E$ ……. (7)

Substitution$\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(1+{\mathrm{χ}}_e)}\frac{\widehat{r}}{r^2}$ for$E$ in equation (7)

$\begin{array}{c}P={\mathrm{Î\mu }}_0{\mathrm{χ}}_e\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(1+{\mathrm{χ}}_e)}\frac{\widehat{r}}{r^2}\\ P=\frac{q{\mathrm{χ}}_e}{4\mathrm{Ï€}(1+{\mathrm{χ}}_e)}\frac{\widehat{r}}{r^2}\end{array}$ …….. (8)

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

With the expression for volume bound charge density.

${\mathrm{ÒÏ}}_b=−∇â‹\dots P$ ……. (9)

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

$\begin{array}{c}{\mathrm{ÒÏ}}_b=−∇â‹\dots \left(\frac{q{\mathrm{χ}}_e}{4\mathrm{Ï€}(1+{\mathrm{χ}}_e)}\frac{\widehat{r}}{r^2}\right)\\ =\frac{−q{\mathrm{χ}}_e}{4\mathrm{Ï€}(1+{\mathrm{χ}}_e)}\left(∇â‹\dots \frac{\widehat{r}}{r^2}\right)\end{array}$ …... (10)

We know that,

$\left(∇â‹\dots \frac{\widehat{r}}{r^2}\right)=4\mathrm{Ï€}{\mathrm{δ}}^3\left(r\right)$

Substitute$4\mathrm{Ï€}{\mathrm{δ}}^3\left(r\right)$ for$\left(∇â‹\dots \frac{\widehat{r}}{r^2}\right)$ in equation (10)

$\begin{array}{c}{\mathrm{ÒÏ}}_b=\frac{−q{\mathrm{χ}}_e}{4\mathrm{Ï€}(1+{\mathrm{χ}}_e)}\left(∇â‹\dots \frac{\widehat{r}}{r^2}\right)\\ =\frac{−q{\mathrm{χ}}_e}{4\mathrm{Ï€}(1+{\mathrm{χ}}_e)}\left(4\mathrm{Ï€}{\mathrm{δ}}^3\left(r\right)\right)\\ =\frac{−q{\mathrm{χ}}_e}{1+{\mathrm{χ}}_e}{\mathrm{δ}}^3\left(r\right)\end{array}$

Therefore, the volume bound charge density is $\frac{−q{\mathrm{χ}}_e}{1+{\mathrm{χ}}_e}{\mathrm{δ}}^3\left(r\right)$.

Write the expression for surface bound charge density.

${\mathrm{σ}}_b=Pâ‹\dots \widehat{r}$ …… (11)

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

$\begin{array}{c}{\mathrm{σ}}_b=Pâ‹\dots \widehat{r}\\ =\frac{q{\mathrm{χ}}_e}{4\mathrm{Ï€}(1+{\mathrm{χ}}_e)}\frac{\widehat{r}}{r^2}\left(\widehat{r}\right)\\ =\frac{q{\mathrm{χ}}_e}{4\mathrm{Ï€}(1+{\mathrm{χ}}_e)r^2}(\widehat{r}â‹\dots \widehat{r})\\ =\frac{q{\mathrm{χ}}_e}{4\mathrm{Ï€}(1+{\mathrm{χ}}_e)r^2}\end{array}$

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

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

$\begin{array}{c}{\mathrm{σ}}_b=\frac{q{\mathrm{χ}}_e}{4\mathrm{π}(1+{\mathrm{χ}}_e)r^2}\\ =\frac{q{\mathrm{χ}}_e}{4\mathrm{π}(1+{\mathrm{χ}}_e)R^2}\end{array}$

Now, write the expression for total bound surface charge.

$Q_{surface}=\left({\mathrm{σ}}_b\right)4\mathrm{π}{R}^2$ …… (12)

Substitute$\frac{q{\mathrm{χ}}_e}{4\mathrm{π}(1+{\mathrm{χ}}_e)R^2}$for ${\mathrm{σ}}_b$in equation (12)

$\begin{array}{c}{Q}_{surface}=\left({\mathrm{σ}}_b\right)4\mathrm{π}{R}^2\\ =\left(\frac{q{\mathrm{χ}}_e}{4\mathrm{π}(1+{\mathrm{χ}}_e)R^2}\right)4\mathrm{π}{R}^2\\ =\frac{q{\mathrm{χ}}_e}{(1+{\mathrm{χ}}_e)}\end{array}$

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

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.

$Q=∫{\mathrm{ÒÏ}}_{b}d\mathrm{Ï„}$ …… (13)

Substitute $\frac{−q{\mathrm{χ}}_e}{1+{\mathrm{χ}}_e}{\mathrm{δ}}^3\left(r\right)$for ${\mathrm{ÒÏ}}_b$in equation (13)

$\begin{array}{c}Q=∫{\mathrm{ÒÏ}}_{b}d\mathrm{Ï„}\\ =∫\frac{−q{\mathrm{χ}}_e}{1+{\mathrm{χ}}_e}{\mathrm{δ}}^3\left(r\right)d\mathrm{Ï„}\\ =−\frac{q{\mathrm{χ}}_e}{1+{\mathrm{χ}}_e}∫{\mathrm{δ}}^3\left(r\right)d\mathrm{Ï„}\\ =−\frac{q{\mathrm{χ}}_e}{1+{\mathrm{χ}}_e}\end{array}$

Therefore,$Q=−\frac{q{\mathrm{χ}}_e}{1+{\mathrm{χ}}_e}$ .