In Figure 14.84 there is a permanent dipole on the left with dipole moment ${\mathbf{\mu }}_{\mathbf{1}}\mathbf{=}{\mathbf{Qs}}_{\mathbf{1}}$ and a neutral atom on the right with polarizability$\mathit{\alpha }$ , so that it becomes an induced dipole with dipole moment ${\mathit{\mu }}_{\mathbf{2}}\mathbf{=}{\mathbf{Qs}}_{\mathbf{2}}\mathbf{=}\mathit{\alpha }{\mathit{E}}_{\mathbf{1}}$, where${\mathit{E}}_{\mathbf{1}}$ is the magnitude of the electric field produced by the permanent dipole. Show that the force the permanent dipole exerts on the neutral atom is$\mathit{F}\mathbf{\approx }{\left(\frac{1}{4\pi {\epsilon }_0}\right)}^{\mathbf{2}}\frac{\mathbf{12}\mathbf{\alpha }{\mathbf{\mu }}_{\mathbf{1}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{7}}}$

The given data can be listed below as:

• The dipole moment of the left permanent dipole is,${\mathrm{\mu }}_1={\mathrm{Qs}}_1$ .
• The dipole moment of the induced dipole is, ${\mathrm{\mu }}_2={\mathrm{qs}}_2={\mathrm{\alpha E}}_1$.

The dipole moment is described as the product of the charge and the distance of the separation of the dipole. Moreover, dipole moment is also described as the pair of the opposite and the equal charges.

The equation of the electric field because of the permanent dipole at the atom’s center is expressed as:

${\mathrm{E}}_1=\mathrm{k}\frac{2{\mathrm{\mu }}_1}{{\mathrm{r}}^3}$ …(i)

Here, ${\mathrm{E}}_1$is the magnitude of the electric field because of the permanent dipole at the atom’s center, $\mathrm{k}$is the electric field’s constant, ${\mathrm{\mu }}_1$is the left dipole moment and $\mathrm{r}$is the distance between the dipoles.

The magnitude of the electric field due to the atom’s induced dipole is expressed as:

${\mathrm{E}}_2=\mathrm{k}\frac{{\mathrm{\mu }}_2}{{\mathrm{r}}^3}$

Here, ${\mathrm{E}}_2$is the magnitude of the electric field due to the atom’s induced dipole, $\mathrm{k}$is the electric field’s constant, ${\mathrm{\mu }}_2$is the dipole moment of the induced dipole and $r$is the distance between the dipoles.

Substitute the values in the above equation.

${\mathrm{E}}_2=\mathrm{k}\frac{{\mathrm{\alpha E}}_1}{{\mathrm{r}}^3}$

Here,$\mathrm{\alpha }$ is the polarizability of the right neutral atom.

Substitute the values of equation (i) in the above equation.

$\begin{array}{c}{\mathrm{E}}_2=\mathrm{k}\frac{\mathrm{\alpha }}{{\mathrm{r}}^3}\cdot \mathrm{k}\frac{{\mathrm{\mu }}_1}{{\mathrm{r}}^3}\\ ={\mathrm{k}}^2\frac{2{\mathrm{\alpha \mu }}_1}{{\mathrm{r}}^6}\end{array}$

The equation of the force on the permanent dipole is expressed as:

$\mathrm{F}=-{\mathrm{Qk}}^2\frac{2{\mathrm{\alpha \mu }}_1}{{(\mathrm{r}+{\mathrm{s}}_1/2)}^6}+{\mathrm{Qk}}^2\frac{2{\mathrm{\alpha \mu }}_1}{{(\mathrm{r}-{\mathrm{s}}_1/2)}^6}$

Here, $\mathrm{F}$is the force on the permanent dipole, $\mathrm{Q}$is the charge of the left dipole, ${\mathrm{s}}_1$is the distance between the charges of the left dipole, $\mathrm{k}$is the electric field’s constant, $\mathrm{\alpha }$, is the polarizability of the right neutral atom, ${\mathrm{\mu }}_2$is the dipole moment of the induced dipole and $\mathrm{r}$is the distance between the dipoles.

According to the question $\mathrm{r}>>{\mathrm{s}}_1$, then with the help of the binomial theorem, the above equation can be written as:

$F=\frac{Q{k}^{2}2\alpha {\mu }_1}{r^6}\left[\frac{1}{{(1-s_1/2r)}^6}-\frac{1}{{(1+s_1/2r)}^6}\right]$…(ii)

According to the binomial theorem, the first denominator can be expressed as:

$\begin{array}{c}{\left(1-\frac{{\mathrm{s}}_1}{2\mathrm{r}}\right)}^{-6}\approx 1-\left(\frac{-6{\mathrm{s}}_1}{2\mathrm{r}}\right)\\ \approx 1+\frac{3{\mathrm{s}}_1}{\mathrm{r}}\end{array}$ …(iii)

According to the binomial theorem, the second denominator can be expressed as:

$\begin{array}{c}{\left(1+\frac{{\mathrm{s}}_1}{2\mathrm{r}}\right)}^{-6}\approx 1-\left(\frac{6{\mathrm{s}}_1}{2\mathrm{r}}\right)\\ \approx 1-\frac{3{\mathrm{s}}_1}{\mathrm{r}}\end{array}$…(iv)

Substitute the values of equation (iii) and (iv) in the equation (ii).

$\begin{array}{c}\mathrm{F}=\frac{{\mathrm{Qk}}^{2}2{\mathrm{\alpha \mu }}_1}{{\mathrm{r}}^6}\left[1+\frac{3{\mathrm{s}}_1}{\mathrm{r}}-1+\frac{3{\mathrm{s}}_1}{\mathrm{r}}\right]\\ =\frac{{\mathrm{Qk}}^{2}2{\mathrm{\alpha \mu }}_1}{{\mathrm{r}}^6}\cdot \frac{6{\mathrm{s}}_1}{\mathrm{r}}\\ =\frac{{\mathrm{Qs}}_1{\mathrm{k}}^{2}12{\mathrm{\alpha \mu }}_1}{{\mathrm{r}}^7}\end{array}$

Substitute $\frac{1}{4{\mathrm{\pi \epsilon }}_0}$for $\mathrm{k}$and${\mathrm{Qs}}_1$for${\mu }_1$in the above equation.

role="math" localid="1661325916339" $\mathrm{F}\approx {\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)}^2\frac{12{\mathrm{\alpha \mu }}_1^2}{{\mathrm{r}}^7}$

Thus, the force the permanent dipole exerts on the neutral atom is$\mathrm{F}\approx {\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)}^2\frac{12{\mathrm{\alpha \mu }}_1^2}{{\mathrm{r}}^7}$ .