Chapter 23, Problem 46P

(III) The dipole moment, considered as a vector, points from the negative to the positive charge. The water molecule, Fig. 23–32, has a dipole moment $\stackrel{\to }{\text{p}}$ which can be considered as the vector sum of the two dipole moments ${\stackrel{\to }{\text{p}}}_{\text{1}}$ and ${\stackrel{\to }{\text{p}}}_{\text{2}}$ as shown. The distance between each H and the O is about 0.96 × 10⁻¹⁰ m; the lines joining the center of the O atom with each H atom make an angle or 104° as shown, and the net dipole moment has been measured to be p = 6.1 × 10⁻³⁰C · m. (a) Determine the effective charge q on each H atom. (b) Determine the electric potential, far from the molecule, due to each dipole, ${\stackrel{\to }{\text{p}}}_{\text{1}}$ and ${\stackrel{\to }{\text{p}}}_{\text{2}}$, and show that

$V\text{=}\frac{1}{4\pi {ϵ}_0}\frac{p\cos \theta }{r^2}$

where p is the magnitude or the net dipole moment, $\stackrel{\to }{\text{p}}$ = ${\stackrel{\to }{\text{p}}}_{\text{1}}$ + ${\stackrel{\to }{\text{p}}}_{\text{2}}$, and V is the total potential due to both ${\stackrel{\to }{\text{p}}}_{\text{1}}$ and ${\stackrel{\to }{\text{p}}}_{\text{2}}$. Take V = 0 at r = ∞.

FIGURE 23–32 Problem 46.