Three point charges are located as shown in Fig. 3.38, each a distance

afrom the origin. Find the approximate electric field at points far from the origin.

Express your answer in spherical coordinates, and include the two lowest orders in the multi-pole expansion.

Write the expression for the due to dipole.

${\mathbf{V}}_{\mathbf{dip}\mathbf{}\mathbf{ole}}\mathbf{=}\frac{\mathbf{P}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{\theta }}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}$ …… (1)

Here, P is the dipole moment, ${\epsilon }_0$is the permittivity for the free space and r is the distance.

Write the expression for the relation between the electric filed and electric potential.

$\mathbf{E}\mathbf{=}\frac{\mathbf{dV}}{\mathbf{dX}}$ …… (2)

Here, V is the potential.

Write the expression for net charge.

$$\begin{gathered} \mathrm{Q}=-\mathrm{q}-\mathrm{q}+\mathrm{q} \\ =-\mathrm{q} \end{gathered}$$

Write the expression for the potential due to monopole.

${\mathrm{V}}_{\mathrm{monop}\mathrm{ole}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{Q}}{\mathrm{r}}\right)$

Substitute -q in above equation then,

${\mathrm{V}}_{\mathrm{monop}\mathrm{ole}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{-q}{\mathrm{r}}\right)$

Now, differentiate the above equation with respect to .

$$\begin{gathered} {\overrightarrow{\mathrm{E}}}_{\mathrm{mom}\mathrm{opole}}=-\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{-\mathrm{q}}{\mathrm{r}}\right)\right)\left(\hat{\mathrm{r}}\right) \\ =\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left(-\frac{1}{{\mathrm{r}}^2}\right)\left(\hat{\mathrm{r}}\right) \\ =-\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left(\hat{\mathrm{r}}\right) \end{gathered}$$

Write the expression for net dipole moment of this configuration.

$$\begin{gathered} \mathrm{p}=\mathrm{qa}\hat{\mathrm{z}}+-\mathrm{qa}\hat{\mathrm{y}}+-\mathrm{qa}\left(-\hat{\mathrm{y}}\right) \\ \mathrm{p}=\mathrm{qa}\hat{\mathrm{z}} \end{gathered}$$

Write the expression for potential due to dipole.

${\mathrm{V}}_{\mathrm{dipole}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{p}.\hat{\mathrm{r}}}{{\mathrm{r}}^2}$

The dot product of the dipole moment with unit vector is given by,

$$\begin{gathered} p.\hat{r}=qa\hat{z}.\hat{r} \\ =qa\cos \theta \end{gathered}$$

Thus the electric potential due to dipole is,

${\mathrm{V}}_{\mathrm{dipole}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qacos\theta }}{{\mathrm{r}}^2}$

Write the expression for the r component of the electric field.

${\mathrm{E}}_{\mathrm{r}}=-\frac{\partial {\mathrm{V}}_{\mathrm{dip}\mathrm{ole}}}{\mathrm{dr}}$

Substitute $\frac{4}{{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qacos\theta }}{{\mathrm{r}}^2}\mathrm{for}{\mathrm{V}}_{\mathrm{dipole}}\mathrm{in}\mathrm{equation}{\mathrm{E}}_{\mathrm{r}}=-\frac{\partial {\mathrm{V}}_{\mathrm{dipole}}}{\mathrm{dr}}.$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{r}}=-\frac{\partial }{\partial \mathrm{r}}\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qa}\mathrm{cos\theta }}{{\mathrm{r}}^2}\right) \\ =-\frac{\mathrm{qa}\mathrm{cos\theta }}{4{\mathrm{\pi \epsilon }}_0}\left(-\frac{2}{{\mathrm{r}}^3}\right) \\ =\frac{2\mathrm{qa}\mathrm{cos\theta }}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3} \end{gathered}$$

Write the expression for the $\theta$component of the electric field.

$E_0=-\frac{1}{r}\frac{\partial V_{dipole}}{\partial \theta }$

Substitute localid="1657514776765" $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qacos\theta }}{{\mathrm{r}}^2}\mathrm{for}{\mathrm{V}}_{\mathrm{dipole}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}$

$$\begin{gathered} {\mathrm{E}}_0=-\frac{1}{\mathrm{r}}\frac{\partial }{\partial \mathrm{\theta }}\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qa}\mathrm{cos\theta }}{{\mathrm{r}}^2}\right) \\ =-\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qa}}{{\mathrm{r}}^3}\right)\left(-\mathrm{sin\theta }\right) \\ =\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)\left(\frac{\mathrm{qa}\mathrm{sin\theta }}{{\mathrm{r}}^3}\right) \end{gathered}$$

Write the expression for the total electric filed due to dipole at distance from the origin.

${\overrightarrow{\mathrm{E}}}_{\mathrm{dip}\mathrm{ole}}={\mathrm{E}}_{\mathrm{r}}\hat{\mathrm{r}}+{\mathrm{E}}_{\mathrm{\theta }}\hat{\mathrm{\theta }}$

$$\begin{gathered} \mathrm{Substitute}\frac{2\mathrm{qa}\mathrm{cos\theta }}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}\mathrm{for}{\mathrm{E}}_{\mathrm{r}}\mathrm{and}\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)\left(\frac{\mathrm{qa}\mathrm{sin\theta }}{{\mathrm{r}}^3}\right)\mathrm{for}{\mathrm{E}}_{\mathrm{\theta }}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ {\overrightarrow{\mathrm{E}}}_{\mathrm{dip}\mathrm{ole}}={\mathrm{E}}_{\mathrm{r}}\hat{\mathrm{r}}+{\mathrm{E}}_{\mathrm{\theta }}\hat{\mathrm{\theta }} \\ =\left(\frac{2\mathrm{qacos}\mathrm{\theta }}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}\right)\hat{\mathrm{r}}+\left(\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)\left(\frac{\mathrm{qa}\mathrm{sin\theta }}{{\mathrm{r}}^3}\right)\right)\hat{\mathrm{\theta }} \\ =\left(\frac{\mathrm{qa}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}\right)\left(2\mathrm{cos\theta }\hat{\mathrm{r}}+\sin \mathrm{\theta }\hat{\mathrm{\theta }}\right) \end{gathered}$$

Write the expression for total electric field at a distance r from origin.

$\overrightarrow{\mathrm{E}}\left(\mathrm{r},\mathrm{\theta }\right)={\overrightarrow{\mathrm{E}}}_{\mathrm{monopole}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{dipole}}$

$$\begin{gathered} \mathrm{Substitute}\left(\frac{\mathrm{qa}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}\right)\left(2\cos \mathrm{\theta }\hat{\mathrm{r}}+\mathrm{sin\theta }\hat{\mathrm{\theta }}\right)\mathrm{for}{\overrightarrow{\mathrm{E}}}_{\mathrm{dipole}}\mathrm{and}\frac{-\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left(\hat{\mathrm{r}}\right)\mathrm{for}{\overrightarrow{\mathrm{E}}}_{\mathrm{monopole}}\mathrm{in}\mathrm{above} \\ \mathrm{equation}. \end{gathered}$$

$$\begin{gathered} \overrightarrow{\mathrm{E}}\left(\mathrm{r},\mathrm{\theta }\right)={\overrightarrow{\mathrm{E}}}_{\mathrm{monopole}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{dipole}} \\ =\left(-\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left(\hat{\mathrm{r}}\right)\right)+\left(\left(\frac{\mathrm{qa}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}\right)\left(2\mathrm{cos\theta }\hat{\mathrm{r}}+\mathrm{sin\theta }\hat{\mathrm{\theta }}\right)\right) \\ =\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left(-\frac{1}{{\mathrm{r}}^2}\left(\hat{\mathrm{r}}\right)+\left(\frac{\mathrm{a}}{{\mathrm{r}}^3}\right)\left(2\cos \mathrm{\theta }\hat{\mathrm{r}}+\sin \mathrm{\theta }\hat{\mathrm{\theta }}\right)\right) \end{gathered}$$

Thus, the electric filed at a distance far from the origin is,

$=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left(-\frac{1}{{\mathrm{r}}^2}\left(\hat{\mathrm{r}}\right)+\left(\frac{\mathrm{a}}{{\mathrm{r}}^3}\right)\left(2\cos \mathrm{\theta }\hat{\mathrm{r}}+\sin \mathrm{\theta }\hat{\mathrm{\theta }}\right)\right)$