Electric Dipole Force between Ammonia and a Proton

[Badger]

Homework Statement

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?

Homework 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

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!

 

[TSny]

The solution looks correct to me. The student should state the physical meaning of the negative sign of the answer.