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Homework Help: Electric Potential due to a dipole

  1. Sep 17, 2012 #1


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    1. The problem statement, all variables and given/known data

    NOTE: Coordinates are in centimeters.

    A dipole consists of two point charges: +q = 0.911 μC at (-2.53,0) and -q at (2.53,0).

    a) Calculate V, the potential created by the dipole at (31.9,82).

    2. Relevant equations
    V = k \frac{q}{r} \\
    \\ r_{1} = \sqrt{(x-.0253)^2 + (y)^2}
    \\ r_{2} = \sqrt{(x+.0253)^2 + (y)^2}
    \\ V = \sum_{q} V_{due \ to \ q} = V_{+} + V_{-}
    \\ \ \ \ = k \frac{q_{+}}{r_{2}} + k \frac{q_{-}}{r_{1}}

    3. The attempt at a solution
    Using the equations above, where x = .319 meters (distance from (0,0) to point for potential) and y = .82 meters. I got [tex]r_{1} = .871m[/tex] and [tex]r_{2} = .889m[/tex].

    I converted the microColumbs to Columbs so I know it can't be that.

    Plugging in to the electric potential formula, I got:
    V=(8.99e9)*(9.11e-7)/(.871) - (8.99e9)*(9.11e-7)/(.889)
    V=190 Volts

    I put this answer into the assignment and it's wrong. I'm on my last trial, I can't seem to figure out what I'm doing wrong. The school is closed tommorow so I can't go to tutoring.

    Please help in whatever way you can, and any hints would be appreciated.
    Last edited: Sep 18, 2012
  2. jcsd
  3. Sep 18, 2012 #2


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    Homework Helper

    Which charge is nearer to the point P?
  4. Sep 18, 2012 #3


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    I see. I redone the calculations and just got the negative of my previous answer:

    I'm unsure of whether to put in the negative, isn't electric potential a scalar?

    EDIT: I put in the negative and I still got it wrong. :_(

    I still need to figure this out though. Any more help you can give will be appreciated.
    Last edited: Sep 18, 2012
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