Line integrals and Potentials

The potential of a point charge q at a distance r from it is given by , where K is . There can be no doubt that the potential is positive, since it is defined as the work done to bring a unit charge from infinity to that point. Since the electrostatic force is repulsive, the work done by us is positive. However, let us consider the definition of potential . is , where is the unit vector in the r direction, which points radially outwards. What is ? If we take to be the elemental vector when we move from infinity to the point r, we get , so that the calculated value of the potential is -Kq/r instead of Kq/r. Where is the fallacy in this argument?

The problem is that dl is a symbol, so that in spherical co-ordinates it is always given by whichever direction we go. Clearly, for a line integral we cannot say if we are going from a smaller to a higher value. For instance, for a regular integral we have , where we take b > a. Since b and a are real numbers, we have an order relation among them, so that we can say if b > a or a > b. This is not possible for vectors, and it is not possible to do for real numbers if a and b stand for as yet unknown numbers which we cannot know a priori.

Thus, we have which is the correct value.

Note that you will have an analogous problem with gravitation. While the gravitational force is attractive, and the potential is thus negative, naive application of calculus will give a positive potential. Methods such as those outlined above will have to be used to get the correct sign.

Tags: potential of point charge, proof