Q & A for Dipole problem (thanks to Katia for the questions):

> - I started with the usual expression for a potential in term of the 
> charge distribution (without the part with the integral on a surface) 
> and expanded the term 1/distance in term of the legendre polynomials. 

This is a good thing, and follows the general procedure for problems with 
azimuthal symmetry.

> - I identified the term due to p, using the definition I know from the 
> past year (p calculated for a charge distribution) , for x greater than 
> x'.

Well, this is not what we need to do. Just continue with the approach you 
had, and if p = qd, where q are the two charges placed at z=+d/2 and -d/2, 
and take the limit where d -> 0 and q -> infinity in such a way that p 
remains constant. Then the first order term will survive and all higher 
order terms will vanish. You will get the desired result. Now you can 
consider a dipole which is oriented in an arbitrary direction at some 
other point.

> - do we have to calculate the potential for x'greater than x?

No, not needed when the dipole is at the origin and d -> 0.

> - to calculate the field due to p, do we have to calculate the field 
> only due to the dipole term?

Yes, because as you see from the previous parts only the dipole part 
survives. 

> if yes, it would be easier to choose p 
> along the z axis: is that ok for you?

Yes, it would be easier to choose p along the z axis, but no, that would 
not be OK for me!