• Support PF! Buy your school textbooks, materials and every day products Here!

Difference in potential between point charges

  • #1

Homework Statement



Below is a graphic representation of the problem.
19l16b.jpg

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
 

Answers and Replies

  • #2
ehild
Homework Helper
15,396
1,802
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
yea , that is k. And why do we need to find W first? Can we just find V by using V=-∫E.dl ?
 
  • #4
ehild
Homework Helper
15,396
1,802
I meant work on unit positive charge: ∫E.dl=-ΔV

ehild
 
Top