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Electric Potential of 3 point charges

  1. Feb 5, 2010 #1
    1. The arrangement of charges shown in the figure is called a linear electric quadrupole. The positive charges are located at +-s. Notice that the net charge is zero.
    Find an expression for the electric potential on the x-axis at distances y>>s.

    2. Relevant equations
    V= [tex]\sum[/tex]1/4pi [tex]\epsilon[/tex]0 * qi/ri

    3. The attempt at a solution
    Im just not sure how to go about modeling this equation.. otherwise I don't know were to start.
    I have tried: 1/4pi[tex]\epsilon[/tex]0 *Q/y thinking that it was a simple question, and I now understand that I should pick a s-value and use that to come up with the radius. However, I am stuck on how.
    Last edited: Feb 5, 2010
  2. jcsd
  3. Feb 5, 2010 #2


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    Let's start by finding expressions for the ri.

    Can you express them in terms of y and/or s?
  4. Feb 5, 2010 #3

    You have the definition of the potential. Say I sit at x=0, y=10. Ignore the constants but not the charge and distance. You should be able to write down the potential at this point, (remember the distance from x=0, y=10 to each charge are 10 + s, 10, 10-s. Your general formula must take into account the fact that the charges are not at the same place.

    V(y=10) = 1/(10 + s) -2/10 + 1/(10 - s)

    In general V(y) = 1/(y + s) - 2/y + 1/(y - s).
  5. Feb 6, 2010 #4
    so I change the constants to -s or s depending on which direction on the direction on the y axis? that makes sense.
    What I tried was V = K [tex]\sum[/tex]Q/(y+s) -2Q/(y) + Q/(y-s) , and it said that the answer does not depend on the variable s. Would this have something to do with Q= 2qs^2, being the electric quadrupole moment.
    Last edited: Feb 6, 2010
  6. Feb 6, 2010 #5


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    You solved the problem for a point on the y-axis, like Spinnor said in his example.

    However, you want to take a point on the x-axis.
    Did you read my post (#2) ?
  7. Feb 7, 2010 #6
    ahhhh, okay so now I see so I get:
    K* Q/(y^3)
    which is the answer masteringphysics took!
    Thanks So much!!!
  8. Feb 7, 2010 #7


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    I assume that you worked this out using the Pythagorean theorem, and found that s cancels out
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