# Dipole in dielectric sphere

Required math: calculus

Required physics: electrostatics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.34.

Another example of solving a boundary value problem in a system with linear dielectrics. Suppose we have an ideal dipole ${\mathbf{p}}$ at the centre of a sphere of dielectric (dielectric constant ${\epsilon_{r}}$) and radius ${R}$. What is the potential at any point (inside or outside the sphere)?

We can use a similar approach to that of the problem of a dielectric cylinder in an electric field. We first specify the boundary conditions. At the surface of the sphere, the potential must be continuous, so we have

$\displaystyle V_{in}(R)=V_{out}(R) \ \ \ \ \ (1)$

As we saw in the cylinder problem, the condition on the normal derivative of the potential is, since on the outside of the sphere, ${\epsilon=\epsilon_{0}}$:

$\displaystyle \epsilon\left.\frac{\partial V_{in}}{\partial r}\right|_{r=R}=\epsilon_{0}\left.\frac{\partial V_{out}}{\partial r}\right|_{r=R} \ \ \ \ \ (2)$

Since there is no external field, we must have ${V\rightarrow0}$ as ${r\rightarrow\infty}$. Finally, as ${r\rightarrow0}$, the potential must behave like that of an ideal dipole, so, assuming that ${\mathbf{p}}$ points in the ${+z}$ direction:

$\displaystyle \lim_{r\rightarrow0}V_{in}(r)=\frac{p\cos\theta}{4\pi\epsilon r^{2}} \ \ \ \ \ (3)$

Note that we’re using ${\epsilon}$ rather than ${\epsilon_{0}}$ in this formula, since the dipole is inside the dielectric and the potential is reduced by a factor of ${\epsilon_{r}}$, where ${\epsilon=\epsilon_{0}\epsilon_{r}}$.

With these conditions, we must solve Laplace’s equation in spherical coordinates, so we can quote the general form of the solution:

$\displaystyle V(r,\theta)=\sum_{l=0}^{\infty}\left(A_{l}r^{l}+\frac{B_{l}}{r^{l+1}}\right)P_{l}(\cos\theta) \ \ \ \ \ (4)$

Inside the sphere, we have

$\displaystyle V_{in}=\frac{p\cos\theta}{4\pi\epsilon r^{2}}+\sum_{l=0}^{\infty}A_{l}r^{l}P_{l}\left(\cos\theta\right) \ \ \ \ \ (5)$

Outside, we have

$\displaystyle V_{out}=\sum_{l=0}^{\infty}\frac{B_{l}}{r^{l+1}}P_{l}\left(\cos\theta\right) \ \ \ \ \ (6)$

Applying the continuity condition 1 and equating coefficients of the ${P_{l}}$, we get

 $\displaystyle \frac{p}{4\pi\epsilon R^{2}}+A_{1}R$ $\displaystyle =$ $\displaystyle \frac{B_{1}}{R^{2}}\ \ \ \ \ (7)$ $\displaystyle A_{l}R^{l}$ $\displaystyle =$ $\displaystyle \frac{B_{l}}{R^{l+1}}\quad(l\ne1) \ \ \ \ \ (8)$

We can therefore express the ${B_{l}}$ in terms of ${A_{l}}$:

 $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle \frac{p}{4\pi\epsilon}+A_{1}R^{3}\ \ \ \ \ (9)$ $\displaystyle B_{l}$ $\displaystyle =$ $\displaystyle R^{2l+1}A_{l}\quad\left(l\ne1\right) \ \ \ \ \ (10)$

Using the condition on the derivatives 2 and equating coefficients as before, we get

 $\displaystyle -\frac{p}{2\pi R^{3}}+\epsilon A_{1}$ $\displaystyle =$ $\displaystyle -\epsilon_{0}\frac{2B_{1}}{R^{3}}\ \ \ \ \ (11)$ $\displaystyle \epsilon A_{l}lR^{l-1}$ $\displaystyle =$ $\displaystyle -\epsilon_{0}\frac{\left(l+1\right)B_{l}}{R^{l+2}}\quad\left(l\ne1\right) \ \ \ \ \ (12)$

Substituting for ${B_{l}}$ into the second equation gives

$\displaystyle \epsilon A_{l}lR^{2l+1}=-\epsilon_{0}\left(l+1\right)A_{l}R^{2l+1} \ \ \ \ \ (13)$

The only way this can be satisfied is if ${A_{l}=0}$ for ${l\ne1}$, so we get ${A_{l}=B_{l}=0}$ for ${l\ne1}$.

For the ${l=1}$ case, we can solve the two equations in ${A_{1}}$ and ${B_{1}}$ to get (using ${\epsilon=\epsilon_{0}\epsilon_{r}}$):

 $\displaystyle A_{1}$ $\displaystyle =$ $\displaystyle \frac{p}{2\pi\epsilon R^{3}}\left[\frac{\epsilon-\epsilon_{0}}{\epsilon+2\epsilon_{0}}\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2p}{4\pi R^{3}\epsilon}\left(\frac{\epsilon_{r}-1}{\epsilon_{r}+2}\right)\ \ \ \ \ (15)$ $\displaystyle B_{1}$ $\displaystyle =$ $\displaystyle \frac{p}{4\pi\epsilon}\left[1+2\frac{\epsilon_{r}-1}{\epsilon_{r}+2}\right]\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{p}{4\pi\epsilon_{0}}\left(\frac{3}{\epsilon_{r}+2}\right) \ \ \ \ \ (17)$

The potential is

 $\displaystyle V_{in}(r)$ $\displaystyle =$ $\displaystyle \frac{p\cos\theta}{4\pi\epsilon r^{2}}+\frac{2pr\cos\theta}{4\pi R^{3}\epsilon}\left(\frac{\epsilon_{r}-1}{\epsilon_{r}+2}\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{p\cos\theta}{4\pi\epsilon r^{2}}\left[1+2\frac{r^{3}}{R^{3}}\left(\frac{\epsilon_{r}-1}{\epsilon_{r}+2}\right)\right]\ \ \ \ \ (19)$ $\displaystyle V_{out}(r)$ $\displaystyle =$ $\displaystyle \frac{p\cos\theta}{4\pi\epsilon_{0}r^{2}}\left(\frac{3}{\epsilon_{r}+2}\right) \ \ \ \ \ (20)$

## 10 thoughts on “Dipole in dielectric sphere”

1. growescience

The ${costheta}$ shouldn’t have appeared in the equation for ${A_{1}}$ and ${B_{1}}$ since we are equating coefficients of ${costheta}$. I’ve fixed the equation now.

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2. Tim

Why is it that $V_{in} \propto 1/r^{2}$ is ok? This will blow up as $r$ approaches 0.

1. growescience

I think it’s because the dipole is ideal, in the sense that it’s a point object. In any real system, we’d need to take account of the non-zero size of the charged object. You get the same problem with the potential and electric field of point charges.

1. growescience

In a dielectric, we have to use the permittivity ${epsilon}$ of the dielectric rather than ${epsilon_{0}}$ which is the permittivity of free space (that is, vacuum). See here for more details.

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1. gwrowe Post author

I assume you’re referring to eqn 17 (please use equation numbers in comments!). It is equal to what you say, but if you then use ${\epsilon=\epsilon_{0}\epsilon_{r}}$, it simplifies to eqn 17.