Derivation of the formula for the electric potential of a dipole

You do not really have to know this, but I heard that some people were interested in, and wondering about, the origin of the equation for the electric potential of an electric dipole.   The starting point to derive that equation is simply to write down the individual contributions to the potential (at a given point) from the positive charge and the negative charge and add them together (superposition).  That is the exact and correct expression for the potential. In order to derive the equation which has cos(theta)/r^2,  you need make an approximation and to use 2 things that are not easy or obvious.

The first one is a binomial expansion of the denominator in the expressions for the contribution to the potential from the individual charges.  To set up the binomial expansion you first factor out an r from the denominator,  and then expand, to linear order, in d/r ,  where d is  is the separation between positive and negative charge, and r is the distance from the center point between the two charges to the place at which you are evaluating the electric field.

The second thing is to know that x/r is equal to cos(theta).   An x emerges from the binomial expansion because we have chosen to orient the dipole along the x axis.  This substitution, of cos(theta) for x/r, is the last step in the derivation. [r and theta provide a 2nd way in which to specify points in 2 dimensions, the first being x,y --the "cartesian coordinates".  In the r, theta  coordinate system r ranges from 0 to infinity (it is never negative) and theta ranges from 0 to 360 degrees (2 pi).  Theta sweeps out from the x axis  in a counterclockwise direction and thereby one covers the entire x-y plane in a new way.  Typically these are called cylindrical coordinates.   When you sketch a picture,  you may be able to see why x/r is equal to cos(theta).]

The binomial  expansion  is valid in the limit d much  less than r,  and that limits the  range of validity of the cos(theta)/r^2 of formula.  Later I could scan a page which shows the derivation;  if you would like to try it yourself this outline tells you how to go about it.