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This thing has been confusing me for some time now. When I try to find the potential energy of a system of two charges by evaluating the work done to bring them into a certain configuration, I get a negative value. Please correct me. Here's my work. $$ U(r)=\int_{\infty}^{r}\vec{F}_{ext}.\vec{dr}$$ $$=\int_{\infty}^{r}\frac{kQq}{r^2}dr$$ (since $F_{ext}$ and $dr$ are in the same direction) $$=kQq\left(-\frac{1}{r}+\frac{1}{\infty}\right)$$ $$=-\frac{kQq}{r}$$

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If you are bringing a positive charge in toward another positive charge, the electric force on the moving charge is in the direction of a positive (dr), but the force you must exert is in the direction of a negative (dr). You are doing positive work in the direction of motion and the potential energy goes up. In the integral, you have used the electric force rather than your force.

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  • $\begingroup$ Thank you for the clarification! I got it now. I was thinking that a positive function would never integrate to a negative value since integration is summation. I was going in the decreasing direction of r. Thank you once again! $\endgroup$ Oct 18, 2021 at 9:46
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The external force is the negative of the field force. f.dr is work done by the field, -f.dr is the work done against

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  • $\begingroup$ Thank you for the answer! I think R.W. Bird gave an unambiguous reply, so I'm accepting that one. $\endgroup$ Oct 18, 2021 at 9:51

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