91Ó°ÊÓ

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

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.

$E=−∇V$ ……. (2)

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

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

Write the expression of Polarization.

$P={\mathrm{Î\mu }}_0{\mathrm{χ}}_{e}E$ …… (3)

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

Therefore, $\mathrm{P}={\mathrm{Î\mu }}_0{\mathrm{χ}}_{\mathrm{e}}{\mathrm{V}}_0\frac{\mathrm{R}}{{\mathrm{r}}^2}\widehat{\mathrm{r}}$

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

Then the write the expression for surface bound charge.

${\mathrm{σ}}_b=Pâ‹\dots \widehat{n}$ …… (4)

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

Then

$\begin{array}{c}{\mathrm{σ}}_b={\mathrm{Î\mu }}_0{\mathrm{χ}}_e{V}_0\frac{R}{R^2}(\widehat{r}â‹\dots \widehat{n})\\ =\frac{−{\mathrm{Î\mu }}_0{\mathrm{χ}}_{e}V}{R}\end{array}$

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

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

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

Then the charge is,

$\begin{array}{c}{Q}_{total}={\mathrm{σ}}_{for}\left(4\mathrm{Ï€}{R}^2\right)\\ =4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R{V}_0\end{array}$

Then,

$\begin{array}{c}{\mathrm{σ}}_{for}\left(4\mathrm{Ï€}{R}^2\right)=4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}R{V}_0\\ {\mathrm{σ}}_{for}=\frac{{\mathrm{Î\mu }}_0{V}_0}{R}\end{array}$

Then the total free charge in

${\mathrm{σ}}_f=\frac{{\mathrm{Î\mu }}_0{V}_0}{R}$on northern hemisphere.

$\left(\frac{{\mathrm{Î\mu }}_0{V}_0}{R}\right)(1+{\mathrm{χ}}_e)$on southern hemisphere

b)

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

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

${\mathrm{σ}}_f=\frac{{\mathrm{Î\mu }}_0{V}_0}{R}$ on the northern hemisphere.

Then,

${\mathrm{σ}}_{total}=\frac{{\mathrm{Î\mu }}_0{V}_0}{R}$( on northern hemisphere)

Now, write the expression on the southern hemisphere.

$\begin{array}{c}{\mathrm{σ}}_b=\frac{−{\mathrm{Î\mu }}_0{\mathrm{χ}}_e{V}_0}{R}\\ {\mathrm{σ}}_f=\frac{\mathrm{Î\mu }{V}_0}{R}\end{array}$

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

$\begin{array}{c}{V}_0=\frac{Q_{total}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}\\ =\frac{\left({\mathrm{σ}}_{total}\right)\left(4\mathrm{Ï€}{R}^2\right)}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}\\ =\frac{{\mathrm{Î\mu }}_0{V}_0}{R}\frac{R^2}{{\mathrm{Î\mu }}_{0}r}\\ =V_0\frac{R}{r}\end{array}$

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

c)

Here, $V=V_0$at$R=r$ and $V→0$at$r=\mathrm{∞}$ 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.