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Where between two charges does voltage = 0?

  1. Feb 5, 2015 #1
    1. The problem statement, all variables and given/known data
    screen_shot_2015-02-05_at_1.45.32_am.png
    If Q1 in the above figure is twice Q2 and both are positive, where can a point of zero potential be found?


    2. Relevant equations
    V = kq/r

    3. The attempt at a solution
    I know that eventually I'll have to set it up so that kq/r = kq/r, but my problem is, how do you know where this point will be? How do you know whether it will be between the two points or outside the two points? Thanks!
     
  2. jcsd
  3. Feb 5, 2015 #2

    rcgldr

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    If it was outside the points horizontally, then you have the sum of two forces pointing in the same direction, so it can't be outside the points. You might want to use the terms k Q1/R1 and k Q2/(8-R1).

    If both charges are positive, it couldn't be between the two points either. Although there would be a point somewhere between the two charges where the force on a positively charged object would be zero, the potential would be less at some vertical distance above that point, so the potential wouldn't be zero anywhere except at an infinite distance away from both point charges.

    The question was mis-worded and corrected below, so that one of the charges is negative and the other positive. In this case there would be a point where the potential is zero.
     
    Last edited: Feb 5, 2015
  4. Feb 5, 2015 #3
    But I thought voltage was a scalar?
     
  5. Feb 5, 2015 #4

    rcgldr

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    Voltage is defined as the difference in potential per unit charge between two points, say A and B. The voltage from A to B is the negative of the voltage from B to A, so in that sense it has a direction. Link to article, look at the lower half of the page where infinity is used as a reference point:

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potpoi.html

    As corrected in my previous post, there's isn't any point within finite distance of two positive point charges where the potential is zero (it can be considered to be zero at infinity).
     
    Last edited: Feb 5, 2015
  6. Feb 5, 2015 #5
    SORRY for the confusion, I accidentally copied and pasted the wrong question. The actual problem is,

    If Q1 is negative and twice Q2, which is positive, where will the potential be zero?
     
  7. Feb 5, 2015 #6

    rcgldr

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    Now it makes sense to use: k q1/(r) + k q2 / (8-r) = 0.
     
    Last edited: Feb 5, 2015
  8. Feb 5, 2015 #7

    BvU

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    Your original equation still holds. As does your attempt (albeit slightly updated).
    Work it out and simplify to - 2/r1 + 1/r2 = 0
    Where r1 is the distance to Q1 and r2 is the distance to Q2.
    Now you should be able to see that rcq's last hint helps you find a point on the line connecting Q1 and Q2 (his/her r is the distance to Q1 and distance to Q2 is 8-r only on this connecting line).

    Depending on how you interpret the wording of the exercise, one point with potential zero is already an adequate answer.

    But if you want to be complete, your answer provides all the points (also off-axis) where the potential is zero. The so called 'locus' for which ## -{2\over |\vec r_1|} + {1\over |\vec r_2|} = 0 ## . A very relevant exercise in analytical geometry !
     
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