# Potential of a uniformly polarized sphere

1. Mar 25, 2017

### Dustgil

1. The problem statement, all variables and given/known data
Calculate the potential of a uniformly polarized sphere directly from eq. 9

2. Relevant equations

$$V(r)=k \int \frac {P(r') \cdot \hat{r}} {r^2} d\tau$$
3. The attempt at a solution
P is a constant and can be factored out. Since r is taken, call the radius of the sphere R and and an arbitrary radial length l. Then by the law of cosines we can express the denominator for every volume element as.

$$r^2=l^2+z^2-2lzcos\theta$$

This is where I'm stuck. What can I do with r hat? Nothing? I understand that its a unit vector in the direction of of the volume element to the point I'm trying to evaluate at, but I'm not sure if I need to change anything about it or not. z is constant so in theory once I have that changed I can evaluate the integral. Is there an easier way?

2. Mar 25, 2017

### TJGilb

It says uniformly polarized, but what is the direction of polarization? Keep in mind it's actually a vector field, and you are dotting it with $\hat r$.

3. Mar 25, 2017

### kuruman

Your expression for the potential should be
$$V(r)=k \int \frac {\vec{P}(r') \cdot (\vec{r}-\vec{r}')} {|\vec{r}-\vec{r}'|^{3/2}} d^3 r'$$
where $\vec{r}$ is the position of the point of interest and $\vec{r}'$ is the position of volume element $d^3 r'$. Note the integration over primed coordinates. I suggest that you write each of the vectors in the integrand in terms of unit vectors, then assemble the integrand.

4. Mar 26, 2017

### Dustgil

Okay so P is uniform in the z direction. z hat dotted with r hat is equal to the cosine of the angle between the two, which I call phi. This angle is the top angle in the triangle that i utilized with the law of cosines to find r^2. Using the law of cosines again, I can relate phi to the rest of the triangle and plug in for r.

$$l^{2}=z^2+r^2-2zrcos\phi$$

$$cos\phi = \frac {l^2-z^2-r^2} {-2zr}$$

With some simplication,

$$kP \int \frac {z - lcos\theta} {(z^2+l^2-2zlcos\theta)^{3/2}}l^2sin\theta dld\theta d\psi$$

How's that integral look?

5. Mar 26, 2017

### kuruman

You are cutting corners that will lead you into trouble. The integration is over primed coordinates which your integrand does not show.
1. Use spherical coordinates $(r, \theta, \phi)$ for $\vec{r}$ and $(r', \theta ', \phi ')$ for $\vec{r}'$.
2. Write expressions for $\vec{r}$ and $\vec{r}'$ in terms of these coordinates in the Cartesian representation. Without loss of generality, you can set $\phi=0$ and $\phi ' = 0$.
3. Find an expression for $(\vec{r}-\vec{r}')$.
4. Find an expression for $|\vec{r}-\vec{r}'|^{3/2}$.
5. Take the dot product $\vec{P}(r') \cdot (\vec{r}-\vec{r}')$.
6. Assemble the integrand. It should be in terms of $(r, \theta, \phi)$ and $(r', \theta ', \phi ')$.
7. Integrate over primed coordinates. I would do $\theta '$ first.

6. Mar 26, 2017

### TJGilb

kuruman is right. I didn't notice at first but your equation is slightly off. But, if that's what the book is giving you/asking you to use then so be it.