- #26

- 582

- 111

Just to emphasize. This has nothing to do with electrostatics. The maths you need to understand is that for any vector:

##\int_a^b \vec v \cdot d \vec r = - \int_b^a \vec v \cdot d \vec r = \int_b^a -\vec v \cdot d \vec r ##

Where, as previously explained we are keeping the same direction of ##d \vec r## throughout.

If you change the direction of ##d \vec r##, I.e. ##\vec r## points from ##b## to ##a## then that changes the sign again.

So I understand this bit. Perhaps I should rephrase my misunderstanding in a clearer way.

So we will end up with [itex] Work = \int_{R}^{\infty} \vec F_{elec} \cdot d \vec r [/itex], right? My question is then:

**how do we reconcile/ deal with the fact that**[itex] \vec F_{elec} [/itex] is in the opposite direction to [itex] d \vec r [/itex] (because we have started out by defining [itex] d \vec r [/itex] inwards)? If we were to treat it as such, then we would end up with [itex] Work = \frac{-Qq}{4 \pi \epsilon_{0} R} [/itex]