Question

A point dipole p is imbedded at the center of a sphere of linear dielectric material (with radius R and dielectric constant ${\mathit{\epsilon }}_{\mathbf{r}}$). Find the electric potential inside and outside the sphere.

role="math" localid="1658748385913" $[Aanswer:\frac{pcos\theta }{4{\mathrm{\pi \epsilon r}}^2}\left(1+2\frac{r^3}{R^3}\frac{\left({\epsilon }_r-1\right)}{\left({\epsilon }_r+2\right)}\right),\left(r\le R\right):\frac{pcos\theta }{4\pi {\epsilon }_0{r}^2}\left(\frac{3}{{\epsilon }_r+2}\right),\left(r\ge R\right)]$

Short answer

The value of the electric potential outside the sphere is $\frac{\mathrm{pcos\theta }}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left(\frac{3}{{\mathrm{\epsilon }}_{\mathrm{r}}+2}\right)$and the electric potential inside the sphere is$\frac{\mathrm{pcos\theta }}{4{\mathrm{\pi \epsilon }}_0{r}^2{\epsilon }_r}\left[1+2\frac{\left({\mathrm{\epsilon }}_{\mathrm{r}}-1\right)}{\left({\mathrm{\epsilon }}_{\mathrm{r}}+1\right)}\frac{{\mathrm{r}}^3}{{\mathrm{R}}^3}\right]$.

Step-by-step solution

Write the given data from the question.

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

Determine the formula of value of the electric potential outside the sphere and the electric potential inside the sphere.

Write the formula of the electric potential outside the sphere.

${\mathit{V}}_{\mathbf{o}\mathbf{u}\mathbf{t}}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{-}{\mathit{E}}_{\mathbf{0}}\mathit{r}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{+}\mathbf{\sum }_{\mathbf{I}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{B}}_{\mathbf{I}}}{{\mathbf{r}}^{\mathbf{I}\mathbf{+}\mathbf{1}}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{\right)}$ …… (1)

Here, ${\mathit{\epsilon }}_{\mathbf{0}}$ 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.

${\mathit{V}}_{\mathbf{i}\mathbf{n}}\left(r,\theta \right)\mathbf{=}\mathbf{\sum }_{\mathbf{I}\mathbf{=}\mathbf{0}}^{\mathbf{a}}{\mathit{A}}_{\mathbf{I}}{\mathit{r}}^{\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\left(cos\theta \right)$ …… (2)

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

Determine the value of the electric potential outside the sphere and the electric potential inside the sphere.

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:

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

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:

$V_{\mathrm{out}}(\mathrm{r},\mathrm{\theta })=-E_{0}rcos\theta +\sum _{\mathrm{I}=0}^{\infty }\frac{{\mathrm{B}}_{\mathrm{I}}}{{\mathrm{r}}^{\mathrm{I}+1}}{P}_{\mathrm{I}}\left(\mathrm{cos\theta }\right)$

Now potential inside the sphere is expressed as follows:

$V_{in}(r,\theta )=\sum _{I=0}^a{A}_I{r}^I{p}_I\left(\cos \theta \right)$

Since V is continuous across R.

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

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

For $I\ne 1$,

B = A_IR^2I+1

For $I=1$.

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

The second boundary condition is expressed as follows:

$$\begin{gathered} {\overline{)\frac{\partial V}{\partial r}}}_{R+}-{\overline{)\frac{\partial V}{\partial r}}}_{R-}=-\sum \left(I+1\right)\frac{BI}{R^{I+2}}{p}_I\left(\cos \theta \right)+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2p\mathrm{co}s\theta }{{\epsilon }_r{R}^3} \\ -\sum I{A}_I{R}^{I-1}{p}_I\left(\cos \theta \right)=-\frac{{\sigma }_b}{{\epsilon }_0} \end{gathered}$$

Then,

$\begin{array}{rcl}-\frac{{\sigma }_b}{{\epsilon }_0}& =& -\frac{1}{{\epsilon }_0}P·\hat{r}\\ & =& -\frac{1}{{\epsilon }_0}\left({\epsilon }_0{\mathcal{X}}_{e}E·\hat{r}\right)\\ & =& {\left|{\mathcal{X}}_e\frac{\partial V}{\partial r}\right|}_{R-}\\ & =& {\mathcal{X}}_e\left\{-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2P\cos \theta }{{\epsilon }_r{R}^3}+\sum I{A}_I{R}^{I-1}{P}_I\cos \theta \right\}\end{array}$

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" $\begin{array}{rcl}-2\frac{B_I}{R^3}-A_I+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2p}{{\epsilon }_r{R}^3}& =& {\mathcal{X}}_e\left(-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2p}{{\epsilon }_r{R}^3}+A_I\right)-B_I+\frac{p}{4{\mathrm{\pi \epsilon }}_0{\epsilon }_r}-\frac{A_I{R}^3}{2}\\ \frac{p}{4{\mathrm{\pi \epsilon }}_0{\epsilon }_r}-A_I{R}^3+\frac{p}{4{\mathrm{\pi \epsilon }}_0{\epsilon }_r}-\frac{A_I{R}^3}{2}& =& -\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathcal{X}}_{e}P}{{\epsilon }_r}+\frac{A_I{R}^3}{2}\\ \frac{A_I{R}^3}{2}\left[3+{\mathcal{X}}_e\right]& =& \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathcal{X}}_{e}P}{{\epsilon }_r}\\ A_I& =& \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2{\mathcal{X}}_{e}P}{R^3{\epsilon }_r\left(3{\mathcal{X}}_e\right)}\end{array}$

Solve further as

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

For B₁ boundary condition

$B_1=\frac{p}{\begin{array}{rcl}& & 4\pi {\epsilon }_0\epsilon R\end{array}}\begin{array}{rcl}& & \frac{3{\epsilon }_r}{{\epsilon }_r+2}\end{array}$

For $\ge R$, the electric potential outside the sphere is,

role="math" localid="1658751371491" $V_{out}\left(r,\theta \right)=\frac{\mathrm{pcos\theta }}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left(\frac{3}{{\mathrm{\epsilon }}_{\mathrm{r}}+2}\right)for\mathrm{r}\ge \mathrm{R}$

For $\mathrm{r}\ge \mathrm{R}$, the electric potential inside the sphere is,

$$\begin{gathered} V_{in}\left(r,\theta \right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{pcos\theta }}{{\epsilon }_r{r}^2}+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{pr\mathrm{cos\theta }}{{\mathrm{R}}^3}\frac{2\left({\mathrm{\epsilon }}_{\mathrm{r}}-1\right)}{{\epsilon }_r\left({\mathrm{\epsilon }}_{\mathrm{r}}+2\right)} \\ {V}_{in}\left(r,\theta \right)=\frac{\mathrm{pcos\theta }}{4{\mathrm{\pi \epsilon }}_0{r}^2{\epsilon }_r}\left[1+2\frac{\left({\mathrm{\epsilon }}_{\mathrm{r}}-1\right)}{\left({\mathrm{\epsilon }}_{\mathrm{r}}+1\right)}\frac{{\mathrm{r}}^3}{{\mathrm{R}}^3}\right] \end{gathered}$$

Therefore, the value of the electric potential outside the sphere is $\frac{\mathrm{pcos\theta }}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left(\frac{3}{{\mathrm{\epsilon }}_{\mathrm{r}}+2}\right)$and the electric potential inside the sphere is $\frac{\mathrm{pcos\theta }}{4{\mathrm{\pi \epsilon }}_0{r}^2{\epsilon }_r}\left[1+2\frac{\left({\mathrm{\epsilon }}_{\mathrm{r}}-1\right)}{\left({\mathrm{\epsilon }}_{\mathrm{r}}+1\right)}\frac{{\mathrm{r}}^3}{{\mathrm{R}}^3}\right]$.