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.
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)
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