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Electric Dipole's Force of Ammonia on a Proton

  1. Sep 23, 2007 #1
    1. The problem statement, all variables and given/known data
    An ammonia molecule (NH_3) has a permanent electric dipole moment 5.0 * 10^-30 Cm . A proton is 2.50 nm from the molecule in the plane that bisects the dipole.

    What is the electric force of the molecule on the proton?

    2. Relevant equations
    E_dipole = K 2p / r^3 (on axis of electric dipole)
    E_dipole = -K p / r^3 (in the plane perpendicular to an electric dipole)
    E_dipole = (F on q) / q

    3. The attempt at a solution
    p = 5.0E-30 Cm
    r = 2.50E-19 m
    Use second equation because the question implies its perpendicular to the electric dipole of ammonia?
    q of proton = q_p = 1.60E-19 C

    F_(molecule on proton) = F = E_dipole * | q of proton |
    F = [-K * (p / r^3)] * |q_p|
    F = [(-9.00E9) * (5.0E-30) / (2.5E-9)^3)] * 1.60E-19
    F = -4.60E-13 N

    Help would be awesome!
     
  2. jcsd
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