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## Summary:

- Given the potential ##V(\textbf{r})=-\int_\infty^r \textbf{E}\cdot d\textbf{l}##, and given that ##\textbf{E}## is antiparallel to ##d\textbf{l}##, should I evaluate the positive integral ##V(\textbf{r})=\int_\infty^r E\ dr##?

The final result will only differ in its sign, but this is crucial. Having a positively, radially oriented electric field ##\textbf{E}##, I understand that the sign of the integral should be positive (## - (- A) = A##), but it is not! How and why is this the case? A line integral where the vector field is antiparallel to the displacement vector should be negative; is this negativity already taken into account by the conventional minus sign in ##V(\textbf{r})=-\int_\Lambda \textbf{E}\cdot d\textbf{l}##?

Thank you very much.

Thank you very much.