Recent content by jimmyoung

ahh ok, yeah, that clears up a lot of my confusion. Thank you. So with ##d\vec s,## being the displacement vector along the path could it be represented by ##dr\hat r## ? In this case since it is in the decreasing radial direction ##d\vec s = -dr\hat r## I may just be confusing these two...

Thank you for replying. However, I am still not getting it. Wouldn't dr be a negative quantity since it is pointing radially inward? So then its magnitude would still be positive? Why would I need to remove the absolute value sign? Thanks.

Homework Statement I am trying to derive an expression for the potential of a positive point charge by bringing in another positive test charge in from infinity to a point at a distance R from the point charge. Homework Equations $$V_f - V_i = - \int \vec E \cdot d \vec r \, dr$$ The Attempt...