- #1
jelliDollFace
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Homework Statement
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
Homework 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
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