# Griffiths 4.12

1. Apr 5, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
This question refers to Griffiths E and M book.

Of course I am stuck with the integral.

I chose z to be in the same direction as the polarization.

The integral is then
$$P \int d\Omega r'^2 dr'\frac{\left(\mathbf{r} -\mathbf{r'}\right)\cdot \hat{\mathbf{z}}}{\left(\mathbf{r} -\mathbf{r'}\right)^2}$$

BTW, how do you get the script r that Griffiths uses in latex?

2. Relevant equations

3. The attempt at a solution

2. Apr 6, 2008

### pam

Your denominator should be $$({\bf r-r')}^3$$.

3. Apr 6, 2008

### Mindscrape

Yes, the integral is a pain in the neck, but you should notice that it is the same integral as a uniformly charged sphere. I would probably just use mathematica, but if you don't have that luxury then I believe you can use trig substitutions and partial fractions to get it done.