91Ó°ÊÓ

The provided surface charge density is$\mathrm{σ}\left(\mathrm{θ}\right)=k\cos \mathrm{θ}$

The radius of the sphere is R.

The potential of the provided surface charge density is given by

$\mathbf{v}\mathbf{=}\frac{\mathbf{k}}{\mathbf{3}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{°ù³¦´Ç²õθ}$ ....(1)

Here, ${\mathrm{Î\mu }}_0$is the permittivity of free space, r and $\mathrm{θ}$are spherical polar coordinates.

Since $\mathrm{r}\mathrm{³¦´Ç²õθ}=\mathrm{z},$the potential of the spherical shell can be written as

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

Here, z is the Cartesian co-ordinate.

The dipole moment of the spherical shell is

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

The expression for the electric field is given as

$\stackrel{→}{E}=-\stackrel{→}{∇}V$

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

localid="1657538019274" $$\begin{gathered} \stackrel{→}{\mathrm{E}}=-\stackrel{→}{∇}\left(\frac{\mathrm{kz}}{3{\mathrm{Î\mu }}_0}\right) \\ =-\frac{\mathrm{k}}{3{\mathrm{Î\mu }}_0}\hat{\mathrm{z}} \end{gathered}$$

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

$$\begin{gathered} \stackrel{→}{p}=\frac{4\mathrm{π}{R}^{3}k}{3}\hat{z} \\ k\hat{z}=\frac{3\hat{p}}{4\mathrm{π}{R}^3} \end{gathered}$$

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

$$\begin{gathered} \stackrel{→}{E}=-\frac{1}{3{\mathrm{Î\mu }}_0}\frac{3\stackrel{→}{p}}{4\mathrm{Ï€}{R}^3} \\ =-\frac{\stackrel{→}{p}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{R}^3} \end{gathered}$$

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

$∫\stackrel{→}{E}d\mathrm{τ}=\stackrel{→}{E}×\frac{4}{3}\mathrm{π}{R}^3$

Here, $d\mathrm{Ï„}$is the infinitesimal volume element.

Substitute the expression for the electric field

localid="1657537878656" $$\begin{gathered} ∫\stackrel{→}{E}d\mathrm{Ï„}=\left(-\frac{\stackrel{→}{p}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{R}^3}\right)×\frac{4}{3}\mathrm{Ï€}{R}^3 \\ =-\frac{\stackrel{→}{p}}{3{\mathrm{Î\mu }}_0} \end{gathered}$$

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