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.

  1. Pingback: Dipole in dielectric sphere | physics

    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.

  2. Pingback: Dipole in a spherical cavity in an infinite dielectric – Internet and Tecnnology Answers for Geeks

    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.


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