91Ó°ÊÓ

Consider a point dipole p is imbedded at the center of a sphere of linear dielectric material (with radius R and dielectric constant ${\mathrm{Î\mu }}_r$).

Write the formula of the electric potential outside the sphere.

${\mathit{V}}_{\mathbf{o}\mathbf{u}\mathbf{t}}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{θ}\mathbf{)}\mathbf{=}\mathbf{-}{\mathit{E}}_{\mathbf{0}}\mathit{r}\mathit{c}\mathit{o}\mathit{s}\mathit{θ}\mathbf{+}\underset{\mathbf{I}\mathbf{=}\mathbf{0}}{\overset{\mathbf{∞}}{\mathbf{∑}}}\frac{{\mathbf{B}}_{\mathbf{I}}}{{\mathbf{r}}^{\mathbf{I}\mathbf{+}\mathbf{1}}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{θ}\mathbf{\right)}$ …… (1)

Here, ${\mathit{Î\mu }}_{\mathbf{0}}$ is linear dielectric material, r is radius, B_I is boundary conditions and P_I is dipole.

Write the formula of the electric potential inside the sphere.

${\mathit{V}}_{\mathbf{i}\mathbf{n}}\left(r,\mathrm{θ}\right)\mathbf{=}\underset{\mathbf{I}\mathbf{=}\mathbf{0}}{\overset{\mathbf{a}}{\mathbf{∑}}}{\mathit{A}}_{\mathbf{I}}{\mathit{r}}^{\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\left(cos\mathrm{θ}\right)$ …… (2)

Here, A_I is boundary condition, r^I is radius and P_I is dipole.

A point dipole of dipole moment p is imbedded at the sphere of linear dielectric material.

Then total dipole moment at the center is expressed as follows:

$\begin{array}{rcl}{p}^{\text{'}}& =& p-\frac{{\mathcal{X}}_e}{1+{\mathcal{X}}_e}p\\ p^{\text{'}}& =& \frac{p}{1+{\mathcal{X}}_e}\\ p^{\text{'}}& =& \frac{p}{{\mathrm{Î\mu }}_r}\end{array}$

Here, ${\mathcal{X}}_e$is the susceptibility.

Determine the potential due to p' as follows:

Now the separation of variables potential outside the sphere is expressed as follows:

$V_{\mathrm{out}}(\mathrm{r},\mathrm{θ})=-E_{0}rcos\mathrm{θ}+\underset{\mathrm{I}=0}{\overset{∞}{∑}}\frac{{\mathrm{B}}_{\mathrm{I}}}{{\mathrm{r}}^{\mathrm{I}+1}}{P}_{\mathrm{I}}\left(\mathrm{³¦´Ç²õθ}\right)$

Now potential inside the sphere is expressed as follows:

$V_{in}(r,\mathrm{θ})=\underset{I=0}{\overset{a}{∑}}{A}_I{r}^I{p}_I\left(\cos \mathrm{θ}\right)$

Since V is continuous across R.

Determine the A_I and B_I by applying the boundary conditions as follows:

$$\begin{gathered} V_{out}{\overline{)(r,\mathrm{θ})}}_r=V_{in}{\overline{)(r,\mathrm{θ})}}_r \\ \frac{B_I}{R^{I+1}}=A_I{R}^I \end{gathered}$$

For $I≠1$,

B = A_IR^2I+1

For $I=1$.

$\begin{array}{rcl}\frac{B_I}{R^2}& =& \frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{p}{R^2}+A_{I}R\\ B_I& =& \frac{p}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{Î\mu }}_r}+A_I{R}^3\end{array}$

The second boundary condition is expressed as follows:

$$\begin{gathered} {\overline{)\frac{∂V}{∂r}}}_{R+}-{\overline{)\frac{∂V}{∂r}}}_{R-}=-∑\left(I+1\right)\frac{BI}{R^{I+2}}{p}_I\left(\cos \mathrm{θ}\right)+\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{2p\mathrm{co}s\mathrm{θ}}{{\mathrm{Î\mu }}_r{R}^3} \\ -∑I{A}_I{R}^{I-1}{p}_I\left(\cos \mathrm{θ}\right)=-\frac{{\mathrm{σ}}_b}{{\mathrm{Î\mu }}_0} \end{gathered}$$

Then,

$\begin{array}{rcl}-\frac{{\mathrm{σ}}_b}{{\mathrm{Î\mu }}_0}& =& -\frac{1}{{\mathrm{Î\mu }}_0}P·\hat{r}\\ & =& -\frac{1}{{\mathrm{Î\mu }}_0}\left({\mathrm{Î\mu }}_0{\mathcal{X}}_{e}E·\hat{r}\right)\\ & =& {\left|{\mathcal{X}}_e\frac{∂V}{∂r}\right|}_{R-}\\ & =& {\mathcal{X}}_e\left\{-\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{2P\cos \mathrm{θ}}{{\mathrm{Î\mu }}_r{R}^3}+∑I{A}_I{R}^{I-1}{P}_I\cos \mathrm{θ}\right\}\end{array}$

Solve further as

$\begin{array}{rcl}-\left(I+1\right)\frac{B_I}{R^{I+2}}-I{A}_I{R}^{I-1}& =& {\mathcal{X}}_{e}I{A}_I{R}^{I-1}\\ -(2I+1)A_I{R}^{I-1}& =& {\mathcal{X}}_{e}I{A}_I{R}^{I-1}\\ A_I& =& 0\end{array}$

For $I=1$:

role="math" localid="1658748165247" $\begin{array}{rcl}-2\frac{B_I}{R^3}-A_I+\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{2p}{{\mathrm{Î\mu }}_r{R}^3}& =& {\mathcal{X}}_e\left(-\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{2p}{{\mathrm{Î\mu }}_r{R}^3}+A_I\right)-B_I+\frac{p}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{Î\mu }}_r}-\frac{A_I{R}^3}{2}\\ \frac{p}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{Î\mu }}_r}-A_I{R}^3+\frac{p}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{Î\mu }}_r}-\frac{A_I{R}^3}{2}& =& -\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{{\mathcal{X}}_{e}P}{{\mathrm{Î\mu }}_r}+\frac{A_I{R}^3}{2}\\ \frac{A_I{R}^3}{2}\left[3+{\mathcal{X}}_e\right]& =& \frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{{\mathcal{X}}_{e}P}{{\mathrm{Î\mu }}_r}\\ A_I& =& \frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{2{\mathcal{X}}_{e}P}{R^3{\mathrm{Î\mu }}_r\left(3{\mathcal{X}}_e\right)}\end{array}$

Solve further as

$\begin{array}{rcl}{A}_I& =& \frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{2{\mathcal{X}}_{e}P}{R^3{\mathrm{Î\mu }}_r\left(3{\mathcal{X}}_e\right)}\end{array}$

For B₁ boundary condition

$B_1=\frac{p}{\begin{array}{rcl}& & 4\mathrm{Ï€}{\mathrm{Î\mu }}_0\mathrm{Î\mu }R\end{array}}\begin{array}{rcl}& & \frac{3{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_r+2}\end{array}$

For $â‰\yen R$, the electric potential outside the sphere is,

role="math" localid="1658751371491" $V_{out}\left(r,\mathrm{θ}\right)=\frac{\mathrm{\pm 賦´Ç²õθ}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\left(\frac{3}{{\mathrm{Î\mu }}_{\mathrm{r}}+2}\right)for\mathrm{r}â‰\yen \mathrm{R}$

For $\mathrm{r}â‰\yen \mathrm{R}$, the electric potential inside the sphere is,

$$\begin{gathered} V_{in}\left(r,\mathrm{θ}\right)=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{\pm 賦´Ç²õθ}}{{\mathrm{Î\mu }}_r{r}^2}+\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{pr\mathrm{³¦´Ç²õθ}}{{\mathrm{R}}^3}\frac{2\left({\mathrm{Î\mu }}_{\mathrm{r}}-1\right)}{{\mathrm{Î\mu }}_r\left({\mathrm{Î\mu }}_{\mathrm{r}}+2\right)} \\ {V}_{in}\left(r,\mathrm{θ}\right)=\frac{\mathrm{\pm 賦´Ç²õθ}}{4{\mathrm{Ï€Î\mu }}_0{r}^2{\mathrm{Î\mu }}_r}\left[1+2\frac{\left({\mathrm{Î\mu }}_{\mathrm{r}}-1\right)}{\left({\mathrm{Î\mu }}_{\mathrm{r}}+1\right)}\frac{{\mathrm{r}}^3}{{\mathrm{R}}^3}\right] \end{gathered}$$

Therefore, the value of the electric potential outside the sphere is $\frac{\mathrm{\pm 賦´Ç²õθ}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\left(\frac{3}{{\mathrm{Î\mu }}_{\mathrm{r}}+2}\right)$and the electric potential inside the sphere is $\frac{\mathrm{\pm 賦´Ç²õθ}}{4{\mathrm{Ï€Î\mu }}_0{r}^2{\mathrm{Î\mu }}_r}\left[1+2\frac{\left({\mathrm{Î\mu }}_{\mathrm{r}}-1\right)}{\left({\mathrm{Î\mu }}_{\mathrm{r}}+1\right)}\frac{{\mathrm{r}}^3}{{\mathrm{R}}^3}\right]$.