In Ex. 3.9, we obtained the potential of a spherical shell with surface

charge$\mathbf{\sigma }\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\mathbf{kcos\theta }$. In Prob. 3.30, you found that the field is pure dipole outside; it's uniforminside (Eq. 3.86). Show that the limit $\mathbf{R}\mathbf{\to }\mathbf{0}$reproduces the deltafunction term in Eq. 3.106.

The provided surface charge density is$\sigma \left(\theta \right)=k\cos \theta$

The radius of the sphere is R.

The potential of the provided surface charge density is given by

$\mathbf{v}\mathbf{=}\frac{\mathbf{k}}{\mathbf{3}{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{rcos\theta }$ ....(1)

Here, ${\epsilon }_0$is the permittivity of free space, r and $\theta$are spherical polar coordinates.

Since $\mathrm{r}\mathrm{cos\theta }=\mathrm{z},$the potential of the spherical shell can be written as

$\mathbf{v}\mathbf{=}\frac{\mathbf{kz}}{\mathbf{3}{\mathbf{\epsilon }}_{\mathbf{0}}}$

Here, z is the Cartesian co-ordinate.

The dipole moment of the spherical shell is

$\overrightarrow{\mathbf{p}}\mathbf{=}\frac{\mathbf{4}\mathbf{\pi }{\mathbf{R}}^{\mathbf{3}}\mathbf{k}}{\mathbf{3}}\widehat{\mathbf{z}\mathbf{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}....\left(2\right)$

The expression for the electric field is given as

$\overrightarrow{E}=-\overrightarrow{\nabla }V$

Substitute the value of the potential from equation (1) in the above equation.

localid="1657538019274" $$\begin{gathered} \overrightarrow{\mathrm{E}}=-\overrightarrow{\nabla }\left(\frac{\mathrm{kz}}{3{\mathrm{\epsilon }}_0}\right) \\ =-\frac{\mathrm{k}}{3{\mathrm{\epsilon }}_0}\hat{\mathrm{z}} \end{gathered}$$

From the expression of the dipole moment in equation (2)

$$\begin{gathered} \overrightarrow{p}=\frac{4\pi R^{3}k}{3}\hat{z} \\ k\hat{z}=\frac{3\hat{p}}{4\pi R^3} \end{gathered}$$

Substitute the expression in the electric field equation and get the following

$$\begin{gathered} \overrightarrow{E}=-\frac{1}{3{\epsilon }_0}\frac{3\overrightarrow{p}}{4\pi R^3} \\ =-\frac{\overrightarrow{p}}{4\pi {\epsilon }_0{R}^3} \end{gathered}$$

The electric field derived above blows up for $R\to 0$. But the volume integral of the electric field gives

$\int \overrightarrow{E}d\tau =\overrightarrow{E}\times \frac{4}{3}\pi R^3$

Here, $d\tau$is the infinitesimal volume element.

Substitute the expression for the electric field

localid="1657537878656" $$\begin{gathered} \int \overrightarrow{E}d\tau =\left(-\frac{\overrightarrow{p}}{4\pi {\epsilon }_0{R}^3}\right)\times \frac{4}{3}\pi R^3 \\ =-\frac{\overrightarrow{p}}{3{\epsilon }_0} \end{gathered}$$

Thus, the delta function term from Eq. 3.106 is reproduced.