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Force on an electron due to a dipole?

  1. Mar 16, 2015 #1
    1. The problem statement, all variables and given/known data
    A dipole is centered at the origin, and is composed of charged particles with charge +e and -e, separated by a distance 9 ✕ 10-10 m along the y axis. The +e charge is on the -y axis, and the -echarge is on the +y axis.
    An electron is located at <-3 ✕ 10^-8, 0, 0> m. What is the force on the electron, due to the dipole?

    2. Relevant equations
    Electric field of perpendicular charge: (9*10^9)(2qs)/(d^3)
    where q = source charge, s = distance between the dipole charges, d = distance to observed location
    F = qE, q = observed charge

    3. The attempt at a solution
    E = (9 x 10^9)(1.6 x 10^-19)(9 x 10^-10) / (3 x 10^-8)^3 = 48000
    F = (-1.6 x 10^-19)(48000) = -7.68 x 10^-15
    So in vector form, would this be <0, -7.68x10^-15, 0> ? I thought it would be <-3 x 10^-8, -7.68 x 10^-15, 0>, because the force vector would extend down from the observed location, but it turns out this is wrong... I'm semi-confident that the magnitude is correct...
     
  2. jcsd
  3. Mar 17, 2015 #2

    Simon Bridge

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    Did you try sketching how the forces from the two charges in the dipole act on the third charge?
     
  4. Mar 17, 2015 #3
    Yes I did, and the electric field is pointing in the +y direction, and the force is pointing in the -y direction... Is that correct?
     
  5. Mar 17, 2015 #4

    Simon Bridge

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    You can answer that question by considering the definition of the electric field in terms of a force.
    You know the answer to this... what sort of charge does an electron carry?
     
  6. Mar 17, 2015 #5
    Physicslove22,

    On what basis do you think that <0, -7.68x10^-15, 0> is wrong, and, even more intriguingly, on what basis do you think that <-3 x 10^-8, -7.68 x 10^-15, 0> is right?

    Chet
     
  7. Mar 17, 2015 #6
    Well an electron has a negative charge, so after being multiplied to a negative electric field, the force would be pointing in the positive direction.
     
  8. Mar 17, 2015 #7
    The force should be in the negative y direction (as you originally found). The positive charge is at negative y, and the negative charge is at positive y, so a negative charge on the x axis will be repelled downward by the negative charge and attracted downward by the positive charge.

    Chet
     
  9. Mar 17, 2015 #8
    I'm thinking about the way I would draw my answer as a vector on a graph that includes the two charges... It doesn't make sense to me that that the force vector would be <0, -7.68x10^-15, 0> so that it extends downward on the y-axis directly below the charges, instead of extending downward starting at the specified location, thus: <-3x10^-8, -7.68x10^-15, 0>...... I guess I must be missing some piece of information and it's throwing off my reasoning... :(
     
  10. Mar 17, 2015 #9
    Oh whoops! I meant negative electric field... So what about the x axis value?
     
  11. Mar 17, 2015 #10
    Have you drawn a diagram showing the lines of force resulting from the two charges of the dipole? Have you located on this diagram the test charge (electron) of interest? The lines of force diagram will tell all.

    Chet
     
  12. Mar 17, 2015 #11
    Hmm well we haven't done that in class yet, but I think I can do it... Am I correct in drawing my electric field line from the electron's location at <-3x10^-8, 48000, 0>?
     
  13. Mar 17, 2015 #12
    Actually, you can draw the lines of force schematically. You know what the lines of force look like for a single charge, and by combining the effects of the two charges, you can determine what the lines of force look like. Imagine the plus charge as a point source of flowing fluid, and the negative charge as a point sink of fluid. Figure out what the stream lines would look like. These will be the same as the lines of force in your electrostatics problem. Or google lines of force for dipole.

    Chet
     
  14. Mar 17, 2015 #13

    Simon Bridge

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    ... have you misunderstood the notation? The force does not act from the origin... that is not what the vector says. All the vector is telling you is the magnitude and direction... not location.

    <0, F, 0 > is a force magnitude F pointing in the +y direction from wherever it's point of action is.
    The force on the electron points in the -y direction, so that is <0,-eE, 0> from the location of the electron.

    It is not like a position vector which tells you where something is in relation to some origin.
     
  15. Mar 17, 2015 #14

    Ohhhhh wow I can't believe I was looking at it that way....

    And I googled the force lines and the line is headed in the -y direction... thank you both for patiently guiding me to finding my stupid mistake! o0)
     
  16. Mar 18, 2015 #15

    Simon Bridge

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    Sometimes an idea gets stuck in our heads and it can be hard to spot - this is why we need to talk to other people ... well done.
     
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