Irodov 3.47: Head to Tail Dipoles Attract – Amit & Deepali Sinha Foundation

Problem Irodov 3.47

Two water molecules are a distance $l = 10$ nm apart, and the electric dipole moment of each, of magnitude $p = 0.62 \cdot 10^{-29}$ C $\cdot$ m, points along the straight line joining them. Find the force with which the molecules interact.

Two water molecules as point dipoles a distance l apart, both moments along the line joining them, head to tail. Drawn schematically: l is about a hundred molecular sizes. — Two water molecules as point dipoles a distance $l$ apart, both moments along the line joining them, head to tail. Drawn schematically: $l$ is about a hundred molecular sizes.

Given

$l = 10$ nm
$p = 0.62\cdot 10^{-29}$ C $\cdot$ m
both moments along the joining line

Find

the interaction force

F

Solution

Step 1Field of one molecule on its axis

At $l = 10$ nm each molecule, about $0.1$ nm across, is a point dipole. The potential of the first molecule, a point dipole of moment $p$ , at distance $r$ and angle $\theta$ from its axis is

\varphi = \frac{1}{4\pi\varepsilon_0}\,\frac{p\cos\theta}{r^2}

On the axis, $\theta = 0$ , symmetry about the axis leaves only the component along it, and differentiating in $r$ gives

E_1(r) = -\frac{\partial\varphi}{\partial r} = \frac{1}{4\pi\varepsilon_0}\,\frac{2p}{r^3} \tag{1}

directed along $\vect p_1$ .

Step 2Force on the second dipole

The energy of $\vect p_2$ in the field of $\vect p_1$ is $U = -\vect p_2\cdot\vect E_1$ . With both moments along the axis, take $x$ as the distance from $\vect p_1$ along it; then $U = -pE_1(x)$ , and the force on the second molecule is minus its derivative:

\begin{aligned} F_x &= -\frac{dU}{dx} = p\,\frac{dE_1}{dx}\\ &= p\,\frac{d}{dx}\left(\frac{1}{4\pi\varepsilon_0}\,\frac{2p}{x^3}\right)\\ &= -\frac{1}{4\pi\varepsilon_0}\,\frac{6p^2}{x^4} \end{aligned}

using (1). The minus sign points the force back toward the first molecule: the head-to-tail pair attracts.

The force F on the second molecule points back toward the first, and the first feels - F: the head-to-tail pair attracts. — The force $\vect F$ on the second molecule points back toward the first, and the first feels $-\vect F$ : the head-to-tail pair attracts.

Step 3The force at the given separation

At $x = l$ , writing $6/4\pi = 3/2\pi$ , the magnitude of the force is

F = \frac{3\,(0.62\cdot 10^{-29})^2}{2\pi\,(8.85\cdot 10^{-12})\,(10\cdot 10^{-9})^4}\ \text{N}

F = \frac{3p^2}{2\pi\varepsilon_0 l^4} \approx 2.1\cdot 10^{-16}\text{ N}

(2)

The force is an attraction, by Newton’s third law equal and opposite on the two molecules.

Check

Units: $p^2/(\varepsilon_0 l^4)$ is $\text{C}^2\text{m}^2/(\text{C}^2\,\text{N}^{-1}\text{m}^{-2}\cdot\text{m}^4) = \text{N}$ . As a second check, model each dipole as charges $\pm q$ a short distance $a$ apart, $p = qa$ . The four Coulomb forces between the pairs sum to

F_x = \frac{q^2}{4\pi\varepsilon_0}\left[\frac{2}{l^2} - \frac{1}{(l+a)^2} - \frac{1}{(l-a)^2}\right] \approx -\frac{q^2}{4\pi\varepsilon_0}\,\frac{6a^2}{l^4}

to order $a^2$ : an attraction of magnitude $6p^2/(4\pi\varepsilon_0 l^4)$ , the same as (2).

Remark

Two point charges fall off as $1/l^2$ ; each dipole costs one more power, two in all, because the leading $1/l^2$ and $1/l^3$ parts of the four forces cancel. So between two fixed molecular dipoles the force falls as $1/l^4$ , and reversing one moment turns the attraction into an equal repulsion. Molecules that are free to turn spend more time in the attracting head-to-tail order, which is why the thermally averaged pull between polar molecules survives the averaging.