Now the scalar product(see Fig 2-3) 1·dl= dl cos6=d, QQ dr QQ 24T ∮d()=0 4丌0r Remark: the integrant is a total differential d(), and has the same value at the beginning and the end points Next, consider the field e generated by a general dis tribution of charges. Since the path integrals corre sponding to each charge are zero,so W=-fQEdl=-Q'2fEidI=0 Conclusion ∮E.dl an electrostatic field is conservative