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Homework Help: Dipole moment electric potential

  1. Oct 11, 2008 #1
    1. The problem statement, all variables and given/known data

    what is the potential 18 cm from a dipole moment 2.6nCm at
    a) 42 deg to axis
    b) on the perpendicular bisector

    note: dipole separation << 18cm

    2. Relevant equations

    electric dipole moment, p = qd where q is charge, d is distance
    electric potential for point charge, V = kq/r where k is 9*10^9 and r is distance

    3. The attempt at a solution

    a) 42 degrees

    i think my eq may be wrong, but...

    p =qd sin (theta) where theta = 42deg
    so q = p/dsin(theta)

    so V = [k(p/dsin(theta)]/r

    so using sin(42) i got V = 1.08 kV and using cos(42) i got V = 0.971 kV --> both incorrect

    i'm guessing my eq for electric dipole moment is wrong, i was thinking along the lines of the torque eq, rFsin(theta)

    b) not attempted yet, but what is the perpendicular bisector? any tips much appreciated
  2. jcsd
  3. Oct 12, 2008 #2
    how do i factor in the angle into the dipole moment for part a?
  4. Oct 14, 2008 #3
    anyone, its been 4 days, plz help
  5. Oct 15, 2008 #4
    for part a, i just mulitplied the V you get in the beginning (on the dipole axis) by cos(angle given).
    for b, the perpendicular bisector is 90 degrees from the, and if u multiply V by cos(90 deg) you get zero.
    at least thats how i solved it, and my answers came out right.
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