91Ó°ÊÓ

Write the expression for the due to dipole.

${\mathbf{V}}_{\mathbf{dip}\mathbf{}\mathbf{ole}}\mathbf{=}\frac{\mathbf{P}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{θ}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}$ …… (1)

Here, P is the dipole moment, ${\mathrm{Î\mu }}_0$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.

${\mathrm{V}}_{\mathrm{monop}\mathrm{ole}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\left(\frac{\mathrm{Q}}{\mathrm{r}}\right)$

Substitute -q in above equation then,

${\mathrm{V}}_{\mathrm{monop}\mathrm{ole}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\left(\frac{-q}{\mathrm{r}}\right)$

Now, differentiate the above equation with respect to .

$$\begin{gathered} {\stackrel{→}{\mathrm{E}}}_{\mathrm{mom}\mathrm{opole}}=-\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\left(\frac{-\mathrm{q}}{\mathrm{r}}\right)\right)\left(\hat{\mathrm{r}}\right) \\ =\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0}\left(-\frac{1}{{\mathrm{r}}^2}\right)\left(\hat{\mathrm{r}}\right) \\ =-\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\left(\hat{\mathrm{r}}\right) \end{gathered}$$

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.

${\mathrm{V}}_{\mathrm{dipole}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{p}.\hat{\mathrm{r}}}{{\mathrm{r}}^2}$

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

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

Thus the electric potential due to dipole is,

${\mathrm{V}}_{\mathrm{dipole}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{qa³¦´Ç²õθ}}{{\mathrm{r}}^2}$

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

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

Substitute $\frac{4}{{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{qa³¦´Ç²õθ}}{{\mathrm{r}}^2}\mathrm{for}{\mathrm{V}}_{\mathrm{dipole}}\mathrm{in}\mathrm{equation}{\mathrm{E}}_{\mathrm{r}}=-\frac{∂{\mathrm{V}}_{\mathrm{dipole}}}{\mathrm{dr}}.$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{r}}=-\frac{∂}{∂\mathrm{r}}\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{qa}\mathrm{³¦´Ç²õθ}}{{\mathrm{r}}^2}\right) \\ =-\frac{\mathrm{qa}\mathrm{³¦´Ç²õθ}}{4{\mathrm{Ï€Î\mu }}_0}\left(-\frac{2}{{\mathrm{r}}^3}\right) \\ =\frac{2\mathrm{qa}\mathrm{³¦´Ç²õθ}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3} \end{gathered}$$

Write the expression for the $\mathrm{θ}$component of the electric field.

$E_0=-\frac{1}{r}\frac{∂V_{dipole}}{∂\mathrm{θ}}$

Substitute localid="1657514776765" $\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{qa³¦´Ç²õθ}}{{\mathrm{r}}^2}\mathrm{for}{\mathrm{V}}_{\mathrm{dipole}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}$

$$\begin{gathered} {\mathrm{E}}_0=-\frac{1}{\mathrm{r}}\frac{∂}{∂\mathrm{θ}}\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{qa}\mathrm{³¦´Ç²õθ}}{{\mathrm{r}}^2}\right) \\ =-\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{qa}}{{\mathrm{r}}^3}\right)\left(-\mathrm{²õ¾\pm ²Ôθ}\right) \\ =\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\right)\left(\frac{\mathrm{qa}\mathrm{²õ¾\pm ²Ôθ}}{{\mathrm{r}}^3}\right) \end{gathered}$$

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

${\stackrel{→}{\mathrm{E}}}_{\mathrm{dip}\mathrm{ole}}={\mathrm{E}}_{\mathrm{r}}\hat{\mathrm{r}}+{\mathrm{E}}_{\mathrm{θ}}\widehat{\mathrm{θ}}$

$$\begin{gathered} \mathrm{Substitute}\frac{2\mathrm{qa}\mathrm{³¦´Ç²õθ}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}\mathrm{for}{\mathrm{E}}_{\mathrm{r}}\mathrm{and}\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\right)\left(\frac{\mathrm{qa}\mathrm{²õ¾\pm ²Ôθ}}{{\mathrm{r}}^3}\right)\mathrm{for}{\mathrm{E}}_{\mathrm{θ}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ {\stackrel{→}{\mathrm{E}}}_{\mathrm{dip}\mathrm{ole}}={\mathrm{E}}_{\mathrm{r}}\hat{\mathrm{r}}+{\mathrm{E}}_{\mathrm{θ}}\widehat{\mathrm{θ}} \\ =\left(\frac{2\mathrm{qacos}\mathrm{θ}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}\right)\hat{\mathrm{r}}+\left(\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\right)\left(\frac{\mathrm{qa}\mathrm{²õ¾\pm ²Ôθ}}{{\mathrm{r}}^3}\right)\right)\widehat{\mathrm{θ}} \\ =\left(\frac{\mathrm{qa}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}\right)\left(2\mathrm{³¦´Ç²õθ}\hat{\mathrm{r}}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right) \end{gathered}$$

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

$\stackrel{→}{\mathrm{E}}\left(\mathrm{r},\mathrm{θ}\right)={\stackrel{→}{\mathrm{E}}}_{\mathrm{monopole}}+{\stackrel{→}{\mathrm{E}}}_{\mathrm{dipole}}$

$$\begin{gathered} \mathrm{Substitute}\left(\frac{\mathrm{qa}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}\right)\left(2\cos \mathrm{θ}\hat{\mathrm{r}}+\mathrm{²õ¾\pm ²Ôθ}\widehat{\mathrm{θ}}\right)\mathrm{for}{\stackrel{→}{\mathrm{E}}}_{\mathrm{dipole}}\mathrm{and}\frac{-\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\left(\hat{\mathrm{r}}\right)\mathrm{for}{\stackrel{→}{\mathrm{E}}}_{\mathrm{monopole}}\mathrm{in}\mathrm{above} \\ \mathrm{equation}. \end{gathered}$$

$$\begin{gathered} \stackrel{→}{\mathrm{E}}\left(\mathrm{r},\mathrm{θ}\right)={\stackrel{→}{\mathrm{E}}}_{\mathrm{monopole}}+{\stackrel{→}{\mathrm{E}}}_{\mathrm{dipole}} \\ =\left(-\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\left(\hat{\mathrm{r}}\right)\right)+\left(\left(\frac{\mathrm{qa}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}\right)\left(2\mathrm{³¦´Ç²õθ}\hat{\mathrm{r}}+\mathrm{²õ¾\pm ²Ôθ}\widehat{\mathrm{θ}}\right)\right) \\ =\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0}\left(-\frac{1}{{\mathrm{r}}^2}\left(\hat{\mathrm{r}}\right)+\left(\frac{\mathrm{a}}{{\mathrm{r}}^3}\right)\left(2\cos \mathrm{θ}\hat{\mathrm{r}}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right)\right) \end{gathered}$$

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

$=\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0}\left(-\frac{1}{{\mathrm{r}}^2}\left(\hat{\mathrm{r}}\right)+\left(\frac{\mathrm{a}}{{\mathrm{r}}^3}\right)\left(2\cos \mathrm{θ}\hat{\mathrm{r}}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right)\right)$