Because ∇ x E = 0, it is possible to write the electric field as a gradient of some scalar. This is true for any vector whose line integral around a close loop is 0 (path indepdendence). Because the line integral is path indepdendent we define V(r) = - ∫ E . dl . It is then easy to derive E = -∇V. I think these two equations provide the best insight into the relation between the electric field and the electric potential. Remember that in the definition of the electric potential there is a choice of reference point that is arbitrary. Thus any two V's differing only in reference point correspond to the same E.