A conducting sphere at potential ${\mathit{V}}_{\mathbf{0}}$ is half embedded in linear dielectric material of susceptibility ${\mathit{\chi }}_{\mathbf{e}}$, which occupies the region$\mathbf{z}\mathbf{<}\mathbf{0}$ (Fig. 4.35).

Claim:the potential everywhere is exactly the same as it would have been in the

absence of the dielectric! Check this claim, as follows:

1. Write down the formula for the proposed potentialrole="math" localid="1657604498573" $\mathit{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$,in terms of${\mathit{V}}_{\mathbf{0}}$,$\mathbf{R}$,and$\mathit{r}$.Use it to determine the field, the polarization, the bound charge, and the free charge distribution on the sphere.
2. Show that the resulting charge configuration would indeed produce the potential$\mathit{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$.
3. Appeal to the uniqueness theorem in Prob. 4.38 to complete the argument.
4. Could you solve the configurations in Fig. 4.36 with the same potential? If not, explain why.

Given that, the conducting sphere at potential$V_0$ is half embedded in linear dielectric material of susceptibility${\chi }_e$ .

a)

Write the expression for proposed potential in terms of $V_0,R$and $r$.

$\mathrm{V}\left(\mathrm{r}\right)={\mathrm{V}}_0\frac{\mathrm{R}}{\mathrm{r}}$ …… (1)

Write the expression for Electrified.

$E=-\nabla V$ ……. (2)

Substitute the value of$V$is equation (2)

$\begin{array}{l}E=-\nabla \left(V_0\frac{R}{r}\right)\\ E=+V_0\frac{R}{r^2}\widehat{r}\end{array}$

Write the expression of Polarization.

$P={\epsilon }_0{\chi }_{e}E$ …… (3)

Now, put the value of$E$in equation (3)

Therefore, $\mathrm{P}={\mathrm{\epsilon }}_0{\mathrm{\chi }}_{\mathrm{e}}{\mathrm{V}}_0\frac{\mathrm{R}}{{\mathrm{r}}^2}\widehat{\mathrm{r}}$

In the region $Z<0$ and in the region $Z>0$, $P=0$.

Then the write the expression for surface bound charge.

${\sigma }_b=P\cdot \widehat{n}$ …… (4)

Here, $\widehat{n}$points out of dielectric then $\widehat{n}=-\widehat{r}$

Then

$\begin{array}{c}{\sigma }_b={\epsilon }_0{\chi }_e{V}_0\frac{R}{R^2}(\widehat{r}\cdot \widehat{n})\\ =\frac{-{\epsilon }_0{\chi }_{e}V}{R}\end{array}$

This ${\sigma }_b$is on the surface at $r=R$.

The flat surface$Z=0$carries no bound charge since$\widehat{n}=\widehat{z}\perp \widehat{r}$then the volume bound is change.

If $V$is to have spherical symmetry then the net charge must be uniform.

Then the charge is,

$\begin{array}{c}{Q}_{total}={\sigma }_{for}\left(4\pi R^2\right)\\ =4\pi {\epsilon }_{0}R{V}_0\end{array}$

Then,

$\begin{array}{c}{\sigma }_{for}\left(4\pi R^2\right)=4\pi {\epsilon }_{0}R{V}_0\\ {\sigma }_{for}=\frac{{\epsilon }_0{V}_0}{R}\end{array}$

Then the total free charge in

${\sigma }_f=\frac{{\epsilon }_0{V}_0}{R}$on northern hemisphere.

$\left(\frac{{\epsilon }_0{V}_0}{R}\right)(1+{\chi }_e)$on southern hemisphere

b)

Write the total charge configuration is${\sigma }_{total}$uniform on the northern hemisphere.

${\sigma }_b=0$on the northern hemisphere.

${\sigma }_f=\frac{{\epsilon }_0{V}_0}{R}$ on the northern hemisphere.

Then,

${\sigma }_{total}=\frac{{\epsilon }_0{V}_0}{R}$( on northern hemisphere)

Now, write the expression on the southern hemisphere.

$\begin{array}{c}{\sigma }_b=\frac{-{\epsilon }_0{\chi }_e{V}_0}{R}\\ {\sigma }_f=\frac{\epsilon V_0}{R}\end{array}$

Then, write the expression for potential of uniformly changed sphere.

$\begin{array}{c}{V}_0=\frac{Q_{total}}{4\pi {\epsilon }_{0}r}\\ =\frac{\left({\sigma }_{total}\right)\left(4\pi R^2\right)}{4\pi {\epsilon }_{0}r}\\ =\frac{{\epsilon }_0{V}_0}{R}\frac{R^2}{{\epsilon }_{0}r}\\ =V_0\frac{R}{r}\end{array}$

Therefore, the potential of uniformly changed sphere is $V_0\frac{R}{r}$.

c)

Here, $V=V_0$at$R=r$ and $V\to 0$at$r=\infty$ then the boundary conditions are satisfied and everything is consistent then problem 4.35 concludes that this is solution.

d)

The b figure 'a' on the flat surface does not operate with the e potential, as shown in the figures. P is not perpendicular to in this case. As a result, the band charge appears on this surface.

The figure b, on the other hand, has the same potential.