# Difference in potential between point charges

## 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

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ehild
Homework Helper
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

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

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

ehild