Question

The ammonia molecule NH₃ has a permanent electric dipole moment equal to 1.52 D, where 1 D = 1 debye unit = 3.34x10^-30 C·m. Calculate the electric potential due to an ammonia molecule at a point 50.0 nm away along the axis of the dipole. Use V = 0 at infinity.  

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Answer 1

We have to find V in R. We'll suppose d<<r.

p=moment of the dipole

$\vec{p}=q\vec{d}$

q=charge

d=distance between the charges forming the dipole

The potential created by the dipole at distance r from the charges (considering the general case, in P; CP is another r):

$V=\frac{kpcos\theta }{r^{2}}\: \left ( 1 \right )$

$k=\frac{1}{4\pi \varepsilon _{0}}$

k=9*10⁹ Nm²/C²

r=50*10^-9 m

p~5.08*10^-30 C*m

In the case of point R, alpha=0 --> cos=1.

Replace the values in eq(1) and get V~18.3*10^-6V.