Dipole embedded in dielectric sphere is this the best way to do this?

1. Mar 25, 2012

VortexLattice

1. The problem statement, all variables and given/known data

Hey all, I want to find the potential due to a point-like (mathematical, not physical) dipole inside and outside of a sphere made of dielectric material with dielectric ε. It's of radius R. Outside the sphere is vacuum.

2. Relevant equations

3. The attempt at a solution

Inside there sphere is easy (I think). For the potential of a dipole, we just use the standard

$V(r,\theta) = \frac{1}{4\pi \epsilon} \frac{\vec{p}\bullet\vec{x}}{\left|x\right|^3}$

with the dielectric's $\epsilon$, of course. Outside the sphere, I try to build a solution using Legendre polynomials. I want them to match the boundary condition of the potential on the surface of the sphere. To find this, I note that for a point on the surface of the sphere, x is $\vec{x} = sin(\theta)cos(\phi)\hat{x} + sin(\theta)sin(\phi)\hat{y} + cos(\theta)\hat{z}$ and I say the dipole p is pointing in the z direction, with magnitude p. So on the surface of the sphere:

$V(R,\theta) = \frac{1}{4\pi \epsilon}\frac{p cos(\theta)}{R^3}$

So this is what I need my solution outside to match at r = R.

So outside I want something of the form $V(r,\theta) = \sum_{m = 0} ^\infty [(A_m r^m + B_m R^{-(m+1)})P_m(cos(\theta))]$

But I'm looking at the outside, as r goes to infinity, so I know A = 0 for all m. So I get rid of these, and do the standard little trick to figure out the coefficients B:

$B_m = \frac{2m + 1}{2}R^{m + 1} \int_0^{\pi} P_m(cos(\theta)) \frac{1}{4\pi\epsilon}\frac{p cos(\theta)}{R^3} sin(\theta) d\theta$

Is this all I can do? Do I just find the coefficients and then that's as far as I can go? I don't think I can integrate the Legendre polynomials generally.

Also, is there a nicer way of doing it? I was thinking of something involving that trick of viewing the dipole as two point charges very close together (at a distance d), finding the field of each by using method of images, and then taking the limit as d->0. But doing method of images with dielectrics is tricky. Any advice?

Thanks!!

2. Mar 26, 2012

francesco85

Hi!

I don't know whether there exists a nicer solution; anyway I try to propose mine; let me know if you agree with it. As a convention, I will use spherical coordinates, with the z-axis as the axis of the dipole.
As you said, you can write the potential outside the sphere in the way you wrote it (with all A's equal to zero, A_0=0 is a convention).
Then you impose the border condition at the surface of the sphere: continuity of the component of the electric field parallel to the tangent of the surface and continuity of the component of the electric induction perpendicular to the surface:

$\vec{E}_{||}^{\text{out}}=\vec{E}_{||}^{\text{in}}$

$\vec{D}_{\bot}^{\text{out}}=\vec{D}_{\bot}^{\text{in}}$

So you have expressed the border condition inside the dielectric as a field which depends on the parameters B's. We now want to find the solution inside the dielectric.
Now, call $\vec{E}^{\text{part}}$ the particular solution of the electric field generated by an electric dipole (the usual one that is found on the textbooks rescaled by ε) and define (for points inside the sphere)

$\vec{E}^{\text{new}}=\vec{E}^{\text{in}}-\vec{E}^{\text{part}}$

Now, $\vec{E}^{\text{new}}$ is the electric field inside the sphere in absence of free charges inside the sphere with border conditions (written in a very rough way)

$\vec{E}^{\text{new}}|_R=\vec{E}^{\text{in}}|_R-\vec{E}^{\text{part}}|_R$

that is, the border conditions depend on the B's.

Now, since, for what concerns the interior of the sphere, $\vec{E}^{\text{new}}$ should not have any singularity (it is the electric field in absence of charges), you can write the potential inside the sphere in the same form with the Legendre polynomials; the point is that now you save only the terms for which the electric field is nonsingular in the limit r->0. You have then an expression for the electric field in terms of unknown parameters C's (the equivalent of the A's); by matching the two expressions of the electric field at the border I think you would get the answer for $\vec{E}^{\text{new}}$; you then easily conclude.
Best,
Francesco

Last edited: Mar 26, 2012
3. Mar 26, 2012

VortexLattice

I'm sorry, it's not really clear to me what you're doing. What would be your actual expression for the potential?

4. Mar 26, 2012

sunjin09

You are trying to work out the Green's function for the poisson equation in and outside a dielectric sphere. Solution is in terms of spherical functions r^l*Y_{l,m} and r^{-l-1}Y_{l,m}. You'll need spherical expansion of the dipole and match the boundary conditions at the boundary, as well as at infinity. No closed form solution exist for general case. The more general Green's function for helmholtz equation is standard in textbooks on EM Green's functions, derived from Mie scattering theory. If you are looking at a conducting sphere, method of images gives closed form solution

Last edited: Mar 26, 2012
5. Mar 26, 2012

francesco85

If I haven't made any mistake in my computations, I obtain (in spherical coordinates and with the axis of the dipole as the z axis)

1) outside the sphere the electric field is

$E_r=\frac{2p}{2+\epsilon}\frac{1}{r^3}$cos$(\theta)$

$E_\theta=\frac{p}{2+\epsilon}\frac{1}{r^3}$sin$(\theta)$

2) inside the sphere the electric field is

$E_r=\frac{2p}{\epsilon}\frac{1}{r^3}$cos$(\theta)-\frac{p}{R^3}(\frac{1}{\epsilon}+\frac{1}{2+ε})$cos$(\theta)$

$E_\theta=-\frac{p}{\epsilon}\frac{1}{r^3}$sin$(\theta)+\frac{p}{R^3}(\frac{1}{\epsilon}+\frac{1}{2+\epsilon})$sin$(\theta)$

Hope this is right and helpful.

f.

EDIT: as a first check, it seems that
a) inside the sphere the electric field is the one of the electric dipole + a zero-divergence field (the first Maxwell equation is solved);

b) idem outside the sphere;

c) the border conditions are satisfied;

the only thing that is left to check is that the second Maxwell equation is satisfied (the one wit rot$\vec{E}=\vec{0}$), which I haven't done yet; please can you check? Thanks

Last edited: Mar 26, 2012
6. Mar 26, 2012

VortexLattice

Hmmmm, interesting. All the stuff on yours checks out, and I'd really love it if worked... However, that means one of us is wrong because solutions to Laplace's equation are unique. So I don't mean to be challenging, but can you see what's wrong with mine?

7. Mar 27, 2012

francesco85

I'm not so sure about what I am going to say, but anyway I try to pose two questions:

1) how do you know that inside the sphere the electric potential is the potential of the dipole? Until you don't impose border conditions on the sphere, I think that you should write the solution as V=potential of the dipole+ function which satisfies the Laplace equation

2) how do you impose the border condition between the interior of the sphere and the exterior of the sphere? At a first sight, it seems that you should impose some conditions on the derivatives of the potential.

Let me know! Best,
Francesco

8. Mar 27, 2012

fzero

If you note that

$$\cos\theta = P_1(\cos\theta),$$

you can use orthogonality to perform this integral (there's only one nonzero term). It seems to agree with Francesco.

9. Mar 27, 2012

VortexLattice

Whoa! I was kind of wondering about that but I didn't actually write them out and try. How can you be sure all the others integrate to zero?

That would be dandy...let me write it out.