Chapter 23 Problem 46

Problem

H20

The dipole moment, considered as a vector, points from the negative to the positive charge. The water molecule has a dipole moment ${\vec {p}}$ which is due to two dipole moments ${\vec {p}}_{1}$ and ${\vec {p}}_{2}$ as shown. The distance between each H and the O is about $0.96\times 10^{-10}{\textrm {m}}$ m the lines joining the center of the O atom with each H atom make an angle of 104° as shown, and the net dipole moment has been measured to be $p=6.1\times 10^{-30}{\textrm {C}}\cdot {\textrm {m}}$.

(a) Determine the effective charge q on each H atom.

(b) Determine the electric potential, far from the molecule, due to each dipole, ${\vec {p}}_{1}$ and ${\vec {p}}_{2}$, and show that

$V={\frac {1}{4\pi \epsilon _{0}}}{\frac {p\cos \theta }{r^{2}}}$

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

Solution

around the H20 in polar coordinates

(a)

Since the interatomic distance and the charge distrubution is the same, $\left|{\vec {p_{1}}}\right|=\left|{\vec {p_{2}}}\right|$

$\left|{\vec {p}}\right|=p=2\left|{\vec {p_{1}}}\right|\cos 52^{\circ }$

$\left|{\vec {p_{1}}}\right|=\left|{\vec {p_{2}}}\right|={\frac {p}{2\cos 52^{\circ }}}=qd$

$q={\frac {p}{2d\cos 52^{\circ }}}={\frac {6.1\times 10^{-30}{\textrm {C}}\cdot {\textrm {m}}}{2(0.96\times 10^{-10}{\textrm {m}})\cos 52^{\circ }}}=5.2\times 10^{-20}{\textrm {C}}\approx 0.32\mathrm {e^{-}}$

(b)

using the polar coordinates, any point of interest can be represented as $(r,\theta )$. Since we know the angle between ${\vec {p}}_{1}$ and ${\vec {p}}_{2}$, it is easy to relate the point of interest to the dipole moments. Since the potential is a scalar quantity, the potential at point of interest is simply the scalar summation of the potential due to ${\vec {p}}_{1}$ and ${\vec {p}}_{2}$

$V=V_{p_{1}}+V_{p_{2}}$

$={\frac {1}{4\pi \epsilon _{0}}}{\frac {p_{1}\cos(52^{\circ }-\theta )}{r}}+{\frac {1}{4\pi \epsilon _{0}}}{\frac {p_{2}\cos(52^{\circ }+\theta )}{r}}$ $={\frac {1}{4\pi \epsilon _{0}r}}{\frac {p}{2\cos(52^{\circ })}}\left[\cos(52^{\circ }-\theta )+\cos(52^{\circ }+\theta )\right]$ $={\frac {1}{4\pi \epsilon _{0}r}}{\frac {p}{2\cos(52^{\circ })}}\left[\cos(52^{\circ })\cos(\theta )+\sin(52^{\circ })\sin(\theta )+\cos(52^{\circ })\cos(\theta )-\sin(52^{\circ })\sin(\theta )\right]$ $={\frac {1}{4\pi \epsilon _{0}}}{\frac {p\cos(\theta )}{r}}$