Irodov 3.72: An Induced Dipole Rides the Gradient – Amit & Deepali Sinha Foundation

Problem Irodov 3.72

A polar molecule with electric dipole moment $\vect p$ and a non-polar molecule of polarizability $\beta$ are a large distance $l$ apart, and $\vect p$ points along the line joining them. Find the magnitude of the force with which the two molecules interact.

The polar molecule, a point dipole p pointing along the line of centers, and the non-polar molecule of polarizability β a distance l away. — The polar molecule, a point dipole $\vect p$ pointing along the line of centers, and the non-polar molecule of polarizability $\beta$ a distance $l$ away.

Given

dipole moment $\vect p$ along the line of centers
polarizability $\beta$
separation $l$ , large

Find

the magnitude

F

of the interaction force

Solution

Step 1The field of the polar molecule on its axis

At a great distance the polar molecule acts as a point dipole, with potential $\varphi = p\cos\vartheta/(4\pi\varepsilon_0 r^2)$ at distance $r$ and angle $\vartheta$ from $\vect p$ . On the axis, $\vartheta = 0$ , the field points along $\vect p$ and is minus the radial derivative of the potential:

E = -\frac{\partial}{\partial r}\,\frac{p}{4\pi\varepsilon_0 r^2} = \frac{2p}{4\pi\varepsilon_0 r^3} \tag{1}

Step 2The moment the field induces

The non-polar molecule acquires a moment along the field it sits in, proportional to it, with the polarizability as the factor:

\vect p' = \beta\varepsilon_0\vect E \tag{2}

At the second molecule $\vect E$ points along the axis, so $\vect p'$ does too.

Step 3The force on the induced dipole

A dipole in a non-uniform field feels the force $\vect F = (\vect p'\cdot\nabla)\vect E$ . Here $\vect p'$ lies along the axis, and on the axis $\vect E$ has only its axial component, so only the derivative of $E$ along the axis survives, taken at $r = l$ :

F_r = p'\,\frac{dE}{dr} = \beta\varepsilon_0 E\,\frac{dE}{dr}

Differentiating (1) and substituting both factors at $r = l$ :

F_r = \beta\varepsilon_0\cdot\frac{2p}{4\pi\varepsilon_0 l^3}\cdot\left(-\frac{6p}{4\pi\varepsilon_0 l^4}\right) = -\frac{12\,\beta p^2}{16\pi^2\varepsilon_0 l^7}

F = \frac{3\beta p^2}{4\pi^2\varepsilon_0 l^7}

(3)

The minus sign in $F_r$ says the force points toward decreasing $r$ : the non-polar molecule is pulled toward the polar one, and by Newton’s third law the polar molecule is pulled toward it with the same magnitude.

The induced moment p' points along p, and the force F on the non-polar molecule points back toward the polar one. — The induced moment $\vect p'$ points along $\vect p$ , and the force $\vect F$ on the non-polar molecule points back toward the polar one.

Check

The energy route gives the same force. An induced dipole has the interaction energy $W = -\beta\varepsilon_0 E^2/2$ , half of $-\vect p'\cdot\vect E$ because the moment grows with the field, and the force along $r$ is minus its derivative:

F_r = -\frac{dW}{dr} = \beta\varepsilon_0 E\,\frac{dE}{dr}

which is the expression differentiated above. Dimensions: with $\beta$ a volume, $\beta p^2/(\varepsilon_0 l^7)$ is $\text{m}^3\,\text{C}^2\text{m}^2$ over $\text{C}^2\text{N}^{-1}\text{m}^{-2}\,\text{m}^7$ , which is newtons.

Remark

The force contains $p^2$ , so reversing $\vect p$ changes nothing: the induced moment always lines up with the field that makes it, and the pair always attracts. The steep $1/l^7$ law, an interaction energy falling as $1/l^6$ , is the signature of an induced dipole: the moment itself falls as $1/l^3$ and the field gradient acting on it as $1/l^4$ . The same $1/l^6$ appears in the van der Waals attraction between molecules.