A solid sphere, radius R, is centered at the origin. The "northern" hemisphere carries a uniform charge density ${\mathit{\rho }}_{\mathbf{0}}$, and the "southern" hemisphere a uniform charge density $\mathbf{-}{\mathit{\rho }}_{\mathbf{0}}$• Find the approximate field $\mathit{E}\left(r,\theta \right)$for points far from the sphere $\left(r>>R\right)$.

The location of charge distribution and sphere is shown in below figure.

Here, P is the point at which electric field to be determined and z,r are the distances.

Write the expression for the dipole.

$\mathit{p}\mathbf{=}\mathit{r}\mathbf{\text{'}}\mathbf{\int }\left(a"\right)\mathit{\zeta }\mathbf{\text{'}}$ …… (1)

Here, r' is the distance, $\rho \left(r\text{'}\right)$is the charge density.

Diploes always point the positive charge and thus substitute z for r', ${\rho }_0$for $\rho \left(r\text{'}\right)$and $r^2\sin \theta drd\theta d\varphi$for $d\zeta$in equation (1),

$p=\int z{\rho }_0{r}^2\sin \theta drd\theta d\varphi$ …… (2)

Determine value of z in terms of r.

$$\begin{gathered} \cos \theta =\frac{z}{2r} \\ z=2r\cos \theta \end{gathered}$$

Now, substitute $2r\cos \theta$for z.

To solve the integration, take the limits $\theta$from 0 to $\frac{\pi }{2}$, r from 0 to R and from 0 to $2\pi$.

$$\begin{gathered} p=2{\rho }_0\underset{0}{\overset{R}{\int }}{r}^{3}dr\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \sin \theta d\theta \underset{0}{\overset{2\pi }{\int }}d\varphi \\ =2{\rho }_0{\left[\frac{r^4}{4}\right]}_0^R{\left[\frac{-\cos 2\theta }{4}\right]}_0^{\frac{\pi }{2}}{\left[\varphi \right]}_0^{2\pi } \\ =-\frac{R^4}{4}\left(\frac{-1-1}{4}\right)\left(2\pi \right) \\ =\frac{{\rho }_0{R}^4\pi }{2} \end{gathered}$$

Now, write the formula for electric field in terms of dipole.

$E_{dipole}\left(r,\theta \right)=\frac{p}{4\pi {\epsilon }_0{r}^3}\left(2\cos \hat{\theta }r+\sin \hat{\theta }\theta \right)$

Substitute $\frac{{\rho }_0{R}^4\pi }{2}$for $\rho$in above equation.

$$\begin{gathered} E_{dipole}\left(r,\theta \right)=\frac{\left({\displaystyle \frac{{\rho }_0{R}^4\pi }{2}}\right)}{4\pi {\epsilon }_0{r}^3}\left(2\cos \hat{\theta }+\sin \hat{\theta }\theta \right) \\ =\frac{{\rho }_0{R}^4}{8\pi {\epsilon }_0{r}^3}\left(2\cos \hat{\theta }r+\sin \hat{\theta }\theta \right) \end{gathered}$$

Hence, the magnitude of electric field is $\frac{{\rho }_0{R}^4}{8{\epsilon }_0{r}^3}\left(2\cos \hat{\theta }r+\sin \hat{\theta }\theta \right)$.