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$$W=-\int_\infty^\textbf{r}\textbf{F}\cdot\textbf{dl} =-Q\int_\infty^\textbf{r}\textbf{E}\cdot\textbf{dl} = Q(V(\textbf{r})-V(\infty)) =QV(\textbf{r})$$

I'm trying to understand how this definition of work turns into a positive

$$W=QV(\textbf{r}).$$

The integration of $-Q\int_\infty^r \textbf {E} \cdot \textbf{dl}$ would just flip the bounds of integration and then the equation would become $W=Q(V(\infty)-V(\textbf{r}))$ which would simplify to $W=-QV(\textbf{r})$? Isn't this correct. Why does Griffiths have it as a positive? What did I miss?

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Keep in mind that electric field is negative gradient of potential.

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