91Ó°ÊÓ

The charge density of the electron cloud is

$\mathrm{ÒÏ}=Ar.....\left(1\right)$

Here, Ais the proportionality constant and r is the distance from the centre.

The electric field on the a spherical Gaussian surface of radius ris

$\mathbf{E}\mathbf{=}\frac{{\mathbf{Q}}_{\mathbf{enc}}}{\mathbf{4}{\mathbf{Ï„Ï„Î\mu }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here, ${\mathrm{Q}}_{\mathrm{enc}}$is the charge enclosed by the Gaussian surface.

The infinitesimal volume element in spherical polar coordinates with just radial dependence is

$\mathbf{»åÏ„}\mathbf{=}\mathbf{4}{\mathbf{Ï„Ï„°ù}}^{\mathbf{2}}\mathbf{dr}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

The expression for the charge enclosed is

$Q_{enc}={∫}_0^r\mathrm{ÒÏ}\left(r\text{'}\right)d\mathrm{Ï„}\text{'}$

Substitute the expression for the charge density from equation (1) and volume element from equation (3)

$$\begin{gathered} Q_{enc}=4\mathrm{π}{∫}_0^{r}Ar\text{'}r{\text{'}}^{2}dr\text{'} \\ =\mathrm{π}A{r}^4 \end{gathered}$$

Substitute expression for ${\mathrm{Q}}_{\mathrm{enc}}$in equation (2)

$$\begin{gathered} E=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}\mathrm{Ï€}A{r}^4 \\ =\frac{A{r}^2}{4{\mathrm{Î\mu }}_0} \end{gathered}$$

Rearranging this

$r=\sqrt{\frac{4{\mathrm{Î\mu }}_{0}E}{A}}$

The expression for the dipole moment is

$p=qr$

Substitute the expression for rand get

$p=q\sqrt{\frac{4\mathrm{Î\mu }E}{A}}$

Thus, the dipole moment is proportional to $\sqrt{E}$.

For the electric field to be proportional to the dipole moment, the electric field has to be proportional to r. Thus, the charge density has to be constant.