Now the scalar product (see Fig 2-3) COS thus QQ dr QQ 24π∈0 ∮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=-4QE·d=-QzE1;:d=0 Conclusion ∮E:dl=0, an electrostatic field is conservative