In Ex. 3.9, we derived the exact potential for a spherical shell of radius R , which carries a surface charge $\mathit{\sigma }\mathbf{=}\mathit{k}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$.

(a) Calculate the dipole moment of this charge distribution.

(b) Find the approximate potential, at points far from the sphere, and compare the exact answer (Eq. 3.87). What can you conclude about the higher multipoles?

The radius of the spherical shell is R .

The surface charge $\sigma =k\cos \theta$,

(a)

The expression for the dipole moment is given by,

$\mathbf{p}\mathbf{=}\mathbf{qd}$

Here,q is the charge and d is the distance.

The dipole moment in Z direction is given by,

$$\begin{gathered} \mathbf{P}\mathbf{=}\mathbf{p}\hat{\mathbf{z}} \\ \mathbf{P}\mathbf{=}\int \mathbf{z\sigma }·\mathbf{da} \end{gathered}$$

Substitute$R\cos \theta$ for z , $k\cos \theta$for $\sigma$and $R^2\sin \theta d\theta d\varphi$for $da$into above equation.

$$\begin{gathered} P={\int }_0^{2\pi }{\int }_0^{\pi }\left(R\cos \theta \right)\left(k\cos \theta \right)R^2\sin \theta d\theta d\varphi \\ P=R^{2}k{\left[\varphi \right]}_0^{2\pi }{\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta \\ P=R^{2}k\times \left[2\pi -0\right]{\left[-\frac{{\cos }^3\theta }{3}\right]}_0^{\pi } \end{gathered}$$

Apply the limits,

$$\begin{gathered} P=-\frac{2\pi R^{2}k}{3}\left({\cos }^3\pi -{\cos }^{3}0\right) \\ P=-\frac{2\pi R^{2}k}{3}\left(-1-1\right) \\ P=\frac{4\pi R^{2}k}{3} \end{gathered}$$

Hence the dipole moment of the charge distribution is$\frac{4\pi R^{2}k}{3}$ .

(b)

The potential due to dipole is given by,

$V=\frac{1}{4\pi {\epsilon }_0}\frac{P\hat{r}}{r^2}$ …… (1)

Here the dipole in the direction of z is $P=\frac{4\pi R^{2}k}{3}\hat{z}$.

Substitute $\frac{4\pi R^{2}k}{3}\hat{z}$for P into equation (1).

$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_0}\frac{4\pi R^{2}k}{3r^2}\left(\hat{z}·\hat{r}\right) \\ V=\frac{k{R}^2}{3{\epsilon }_0{r}^2}\cos \theta \end{gathered}$$

By comparing the potential at point far from the sphere is exactly same as equation 3.87. and it is concluded that the potential for high multipole is zero.

Hence the potential far from the sphere is $\frac{k{R}^2}{3{\epsilon }_0{r}^2}\cos \theta$and potential for higher multipoles is zero.