# Path independence of the line integral of the electrostatic field created by an arbitrary charge distribution

Suppose we have the following situation:

A positive arbitrarily shaped charge distribution. The line integral of the total electric field, generated by the whole distribution, from $$A$$ to $$B$$ is independent of the path joining them . Why is it so?

The line integral of $$E_{1}\,$$, the $$E$$-field generated by $$dq_{1}$$, depends solely on the radial distance $$\Delta r_{1}$$, here measured from $$dq_{1}$$, that the path covers. The same goes for the line integral of $$E_{2}$$.

$$E_{1}+E_{2}$$ is not radial, so why is its line integral path independent?

• Because if two line integrals are each path independent, then their sum is path independent. – G. Smith Aug 4 at 23:03