Question

A point charge qis situated a large distance rfrom a neutral atom of

polarizability $\mathit{\alpha }$.Find the force of attraction between them.

Short answer

The force on charge q situated at a distance rfrom a neutral atom of

polarizability $\alpha$is $\frac{2\alpha q^2}{16{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0^2{\mathrm{r}}^5}$.

Step-by-step solution

Given data

A point charge qis situated a large distance rfrom a neutral atom of

polarizability $\alpha$.

Electric field due to a dipole

The magnitude of electric field due to a dipole having dipole moment $\rho$at spherical polar coordinate$\theta =\mathrm{\pi }$

${\mathbf{E}}_{\mathbf{d}}\mathbf{=}\frac{\mathbf{2}\mathbf{p}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here, ${\epsilon }_0$is the permittivity of free space and r is the distance from the centre of the dipole.

Derivation of force on a charge due to a dipole

The electric field due to the charge q at the position of the dipole is

$E_q=\frac{q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$

The expression for the dipole moment is

$p=\alpha E_q$

Substitute the expression for $E_q$in the above equation we get

$p=\frac{\alpha q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$

Substitute this expression for dipole moment in equation (1)

$$\begin{gathered} E_d=\frac{2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}\frac{\alpha q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} \\ =\frac{2\alpha q}{16{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0^2{\mathrm{r}}^5} \end{gathered}$$

The expression for the force on the charge is

$F=q{E}_d$

Substitute the expression for $E_d$in the above equation and get

role="math" localid="1657529076590" $F=\frac{2\alpha q^2}{16{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0^2{\mathrm{r}}^5}$

Thus, the force on the charge due to the dipole is $F=\frac{2\alpha q^2}{16{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0^2{\mathrm{r}}^5}$.