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My concrete problem is, the final sign on the integral.
Purcell and Morin, Electricity and Magnetism third (3rd) edition on page 12. They show a simple calculation of the work required to bring two equal charges closer together, having one steady (e.g. q1) and bringing the other (e.g. q2) from infinity to a closer finite distance $r_{12}$.

The force that has to be applied to move one charge toward the other is equal and opposite to the Coulomb force. Therefore,

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Note that because r is changing from $\infty$ to $r_{12}$ , the differential $dr$ is negative. We know that the overall sign of the result is correct, because the work done on the system must be positive for charges of like sign; they have to be pushed together (consistent with the minus sign in the applied force). Both the displacement and the applied force are negative in this case, resulting in positive work being done on the system.

Conceptually (physically) I think it is clear: external force is opposite to the coulomb force, therefore external force is negative. Displacement is from $\infty$ to $r_{12}$ so $dr$ is also negative, giving a dot product of two negative entities, therefore the work is positive.

What is the mathematical justification then for the minus sign on the integral? I am missing something with the indices?, I believe the fact that dr is negative justifies writing the indices from infinity to $r_{12}$ without having to use another minus sign.

I have reviewed several posts on this topic without reaching peace of mind. Examples of posts:
Derivation of a work done by forces on charge
Why the work done is positive when bringing 2 opposite charges together?

Qmechanic
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NullMalUnendlich
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1 Answers1

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So charge $q_1$ is fixed and is the origin of the coordinate system radially outwards being chosen as positive.
Charge $q_2$ is in the field of charge $q_1$.
If both charges have the same sign then charge $q_2$ feels a force which is radially outwards from charge $q_1$, ie in the positive direction.
Thus the external applied force on $q_2$ must be in the opposite direction (negative) and hence the negative sign in the bracket.
Opposite sign charges and the bracket becomes positive, ie external applied force in the positive direction.
The sign of $dr$ is entirely dictated by the limits of integration.
Equal sign charges lead to a negative sign within the bracket and moving the charge $q_2$ from infinity towards charge $q_1$ means that the increments $dr$ are negative.
(minus bracket) $\times$ (minus increment) gives a positive amount of work done by the external applied force.

Farcher
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