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Suppose we have the following situation:

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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?

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    $\begingroup$ Because if two line integrals are each path independent, then their sum is path independent. $\endgroup$ – G. Smith Aug 4 at 23:03
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Mathematically, you can use Stokes theorem to show that because the curl of the electric field is zero (just one of Maxwell's equations), that the integral you indicate is also independent of the path between your two end points (: apologies for the lack of equations

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