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Say I had a charge +Q and I moved closer to a charge +100Q why is the energy needed independent (if I went some strange shape or a straight line) of the path +Q takes as long as it starts and ends at the same place. Many thanks!

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    $\begingroup$ This is true if and only if the forces acting on the particles are conservative forces. Can you be a bit more specific about what you want to know about that? $\endgroup$
    – ACuriousMind
    Dec 28, 2016 at 13:41

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In the following answer, I assumed that your question was about a charge in electrostatic field, without any other interaction.

The independence of the work on the charge from the traveled path comes from the fact that the electrostatic field is conservative.

This is stated in the Maxwell Equation (in electrostatic) about the curl of the electric field: $$ \nabla \times E = 0 $$ with $E$ electrostatic field and $\nabla$ vector differential operator. This equivalent to say that the electrostatic field can admit a potential $\phi$ and there is the following relation: $$ E = - \nabla \phi. $$ Stated this, you can prove that every path integral of a conservative field depends only on the initial and final position of the path, as stated in the Gradient theorem.

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