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Need help explaining two equations for similar events

  1. Sep 20, 2008 #1
    Ok, while rummaging through the internet, trying to find out the magnitude of water's pull on charged objects, I came across two different equations. One seemed more helpful, because it seems specific to water polarity, and the other seemed a bit generic.

    Helpful one: F= q [(2kp)/m^3] A reply from yahoo answers
    F is force in newtons
    q is the charge of affected ion
    k is the coulomb force constant
    p is water's dipole moment (C.m.)
    m is the distance between the centers of the two objects

    Generic one: E = k (q/r^2) (unit vector) Wikipedia: electric fields

    When looking closer, it seems that the two are the same, but what is the deal with one's distance being cubed and the other squared? Then, only one with unit vector multiplied in, and other has water's dipole moment.

    Are they the same? Is the unit vector water's dipole moment? Is the cubed distance an error, or a fundamental difference?
  2. jcsd
  3. Sep 21, 2008 #2


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    Welcome to PF!

    Hi phox2! Welcome to PF! :smile:

    yahoo is using dipole moment, which is coulomb.metres, and dividing by metres^3;

    wikipedia is using charge, which is coulombs, and dividing by metres^2 …

    same thing! :wink:
  4. Sep 21, 2008 #3
    Oh...wow...hehe. Thanks, I'm kind of new to this subject in physics, so it's kinda hard dealing with what a part if an equation does, along with figuring out which unit is what.

    Wait, so based on that, then the charge of the (+) side of water, because the ion is chlorine, is the dipole moment. Does that mean that the dipole moment is the charge of both the (+) and (-) side of water, no matter which which is exerting force? Maybe the charge of the (+) side of the dipole is the dipole moment, and the (-) side is 1, because the dipole moment is the multiple of both charge strength?
    Last edited: Sep 21, 2008
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