Difference in potential between point charges

1. Apr 18, 2012

nobrainer612

1. The problem statement, all variables and given/known data

Below is a graphic representation of the problem.
With a point charge at the origin, calculate the difference in potential of point B relative to point A using the path integral.

2. Relevant equations

As I recalled, to find the electric potential difference, Vab = - ∫ E ° dl (° = dot product)
Potential difference for point charge: V = Q/ 4πεr
Vab = Vb - Va
Electric field for point charge should be, E = Q/4πε(R^2)

3. The attempt at a solution

I am not sure what is meant by path integral. But I will try to do it but not sure the correctness:
Since there is one point charge, and it is located at the origin. SO
V = -∫E°dl = -∫(Q/4πε(R^2) ) dr

V = Q/4πεR + C

I assume zero potential at infinity. If V(∞) = 0 , C = 0.

At point A, potential difference will be:
|r - r1| = |(0,0,0) - (1,0,0)| = |(-1,0,0)| = 1
V(1,0,0) = [1/ 4π(10^-9 /36π) ] *[q/1]

At point B, potential difference will be:
|r - r2| = |(0,0,0) - (-2,0,0)| = |(2,0,0)| = 2
V(-2,0,0) = [1/ 4π(10^-9 /36π) ] * [q/2]

So for the difference in potential of point B relative to point A using the path integral, I just do V(-2,0,0) - V(1,0,0)?

Please tell me if I am doing the correct step. This is very important. Thank you

2. Apr 19, 2012

ehild

You get the work as the negative potential difference, but calculate it with the line integral this case. Integrate the electric field strength between the points along any line. (A straight line is the simplest).

What do you mean on [1/ 4π(10^-9 /36π) ]? Is it k=9.10^9?

ehild

3. Apr 19, 2012

nobrainer612

yea , that is k. And why do we need to find W first? Can we just find V by using V=-∫E.dl ?

4. Apr 19, 2012

ehild

I meant work on unit positive charge: ∫E.dl=-ΔV

ehild