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

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


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?


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.