Dipole moment electric potential

[jelliDollFace]

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

 

[jelliDollFace]

how do i factor in the angle into the dipole moment for part a?

 

[jelliDollFace]

anyone, its been 4 days, please help

 

[scrplyr]

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 that's how i solved it, and my answers came out right.