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## Homework Statement

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.

## Homework 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)

## 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