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Difference in potential between point charges

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data

    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.



    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. jcsd
  3. Apr 19, 2012 #2

    ehild

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    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
     
  4. Apr 19, 2012 #3
    yea , that is k. And why do we need to find W first? Can we just find V by using V=-∫E.dl ?
     
  5. Apr 19, 2012 #4

    ehild

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

    ehild
     
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