91Ó°ÊÓ

Write the expression for the electric filed due to dipole.

${\mathit{E}}_{\mathbf{d}\mathbf{i}\mathbf{p}\mathbf{o}\mathbf{l}\mathbf{e}}\left(r,\mathrm{θ}\right)\mathbf{=}\frac{\mathbf{P}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\left(2cos\widehat{\mathrm{θ}}r+sin\widehat{\mathrm{θ}}\mathrm{θ}\right)$ …….. (1)

Here, $\mathrm{ÒÏ}$is the dipole moment, r is the distance and r and $\mathrm{θ}$are the spherical co-ordinates.

Write the expression of the electric force in terms of the charge and electric filed.

role="math" localid="1655730438252" $\mathit{F}\mathbf{=}\mathit{q}\mathit{E}$ …… (2)

Here, F is the force, q is the charge and E is the electric field.

a)

The dipole is facing along z-direction.

From equation (1),

$$\begin{gathered} r=a \\ \mathrm{θ}=\frac{\mathrm{π}}{2} \\ \mathrm{ϕ}=0 \end{gathered}$$

Write the expression for the force on the charge q.

$F_{1}q{E}_{dipole}$ …… (3)

Substitute $\frac{P}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^3}\left(2\cos \widehat{\mathrm{θ}}r+\sin \widehat{\mathrm{θ}}\mathrm{θ}\right)$for localid="1655730630645" $E_{dipo\ln e}\left(r,\mathrm{θ}\right)$ in equation (3).

$F_1=q\left(\frac{P}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^3}\left(2\cos \widehat{\mathrm{θ}}r+\sin \widehat{\mathrm{θ}}\mathrm{θ}\right)\right)$ …… (4)

Substitute $a$for $r$and $\frac{\mathrm{π}}{2}$for $\mathrm{θ}$in equation (4)

$$\begin{gathered} F_1=q\left(\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3}\left(2\cos \left(\frac{\mathrm{Ï€}}{2}\right)r+\sin \left(\frac{\mathrm{Ï€}}{2}\right)\mathrm{θ}\right)\right) \\ =q\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3}\sin \frac{\mathrm{Ï€}}{2}\mathrm{θ} \end{gathered}$$

Simplify the above equation,

$F_1-\frac{q\mathrm{ÒÏ}{}^{âˆ\S }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3}\mathrm{θ}$

As the dipole $\mathrm{ÒÏ}$is pointing in the z-direction, so the electric force,

$F_1-\frac{q\mathrm{ÒÏ}{}^{âˆ\S }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3}\mathrm{θ}$

Therefore, the force $F_1-\frac{q\mathrm{ÒÏ}{}^{âˆ\S }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3}\mathrm{θ}$.

b)

From the equation (1), the electric field due to dipole at $\left(0,0,a\right)$.

$$\begin{gathered} r=a \\ \mathrm{θ}=0 \\ \mathrm{ϕ}=0 \end{gathered}$$

Then,

Write the expression for the force $F_2$on the charge $q$.

$F_2=q\left(\frac{P}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}\left(2\cos \widehat{\mathrm{θ}}r+\sin \widehat{\mathrm{θ}}\mathrm{θ}\right)\right)$ ……. (5)

Substitute$a$for $r$and $0^{°}$for $\mathrm{θ}$in equation (5).

$$\begin{gathered} F_2=q\left(\frac{P}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}\left(2\cos \left(0^{°}\right)r+\sin \left(0^{°}\right)\mathrm{θ}\right)\right) \\ =q\left(\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3}\left(2\cos \left(0^{°}\right)r\right)\right) \end{gathered}$$

As the dipole $\mathrm{ÒÏ}$is pointing in the z-direction, so the electric force,

$F_2=\frac{2\mathrm{ÒÏ}q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3}\left(z\right)$

Thus, force $F_2=\frac{2\mathrm{ÒÏ}q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3}\left(z\right)$.

c)

Write the expression for potential due to dipole.

$V\left(r,\mathrm{θ}\right)=\frac{\mathrm{ÒÏ}\cos \mathrm{θ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2}$ ……. (6)

Write the expression for the potential V₁ at (0, 0, a) due to dipole.

$$\begin{gathered} v_2=\frac{\mathrm{ÒÏ}\cos \left(0\right)}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2} \\ =\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2} \end{gathered}$$

Write the expression for the potential $v_2$at $\left(a,0,0\right)$due to dipole.

$$\begin{gathered} V_2=\frac{\mathrm{ÒÏ}\cos \left[0\right]}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^{}} \\ =\frac{\mathrm{ÒÏ}}{\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2} \end{gathered}$$

Now, calculate work done in moving charge from $\left(0,0,a\right)$to $\left(a,0,0\right)$.

Therefore,

Write the expression for the work done.

$W=\left(V_2-V_1\right)q$ ……. (7)

Substitute 0 for $V_1$and $\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2}$for $V_2$in equation (7).

$$\begin{gathered} W=\left(\frac{\mathrm{ÒÏ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2}-0\right)q \\ =\frac{\mathrm{ÒÏ}q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2} \end{gathered}$$

Therefore, the work done is $=\frac{\mathrm{ÒÏ}q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2}$.