# Uniformly Polarized sphere

1. May 9, 2007

### stunner5000pt

1. The problem statement, all variables and given/known data
Find the electric field of a uniformly polarized sphere of radius R

2. Relevant equations
$$V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \oint_{S} \frac{\sigma_{b}}{r} da' + \int_{V} \frac{\rho_{b}}{r} d\tau'$$

3. The attempt at a solution
well obviously there is no volume charge density rho
but there is a surface charge density
$\sigma_{b} = P \cos\theta$

now to calculate the potentail we gotta use that above formula
Suppose r > R

then
$$V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \int \frac{P \cos\theta}{r'} da'$$

now the squigly r is found using the cosine law right...?

$r' = \sqrt{R^2 + r^2 - 2Rr\cos\theta'}$
and
$da' = R^2 \sin\theta' d\theta d\phi$
So then
$$V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \int_{0}^{2\pi}\int_{0}^{\pi} \frac{P \cos\theta}{\sqrt{R^2 + r^2 - 2Rr\cos\theta'}} \cos\theta' R^2 \sin\theta' d\theta' d\phi$$

cos theta prime because we care about the Z components only
is that right???
and the limits of integrate for the theta would be from 0 to pi
and for the phi is 0 to 2pi??

(o by the way how do i put the squigly r??)

Last edited: May 9, 2007
2. May 10, 2007

### siddharth

Instead of evaluating the integral, you could also find the potential by the separation of variables technique. Since there's no volume charge inside the sphere as you pointed out, laplace's equation will be satisfied.

So, you can find find $$V(r,\theta)$$ by the separation of variables technique. Since the boundary condition is $$V(R,\theta)=P \cos \theta$$, the coefficients will be easy to find by the orthogonality property of the legendre polynomials.

3. May 10, 2007

### Meir Achuz

A uniformly polarized sphere has a dipole moment p=P*volume.
The outside potential and field are just that of dipole p.
phi inside the sphere can be found by matching the Legendre poynomial expansion for phi at the surface r=R.

4. May 10, 2007

### maverick280857

Inside a uniformly polarized sphere, the field is uniform.
Outside, the field is like the field of a dipole placed symmetrically about the center with dipole moment equal to the polarization vector times the volume of the sphere.

This can be shown by integration or by separation of variables. The latter method (as stated by siddharth) is neater. You can look at Griffiths for the solution to Laplace Equation in spherical coordinates.