91影视

Write the expression for the electric filed due to dipole.

${\mathit{E}}_{\mathbf{d}\mathbf{i}\mathbf{p}\mathbf{o}\mathbf{l}\mathbf{e}}\left(r,胃\right)\mathbf{=}\frac{\mathbf{P}}{\mathbf{4}\mathbf{蟺}{\mathbf{蔚}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\left(2cos\widehat{胃}r+sin\widehat{胃}胃\right)$ 鈥︹��.. (1)

Here, $蚁$is the dipole moment, r is the distance and r and $胃$are the spherical co-ordinates.

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

role="math" localid="1655730438252" $\mathit{F}\mathbf{=}\mathit{q}\mathit{E}$ 鈥︹�� (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),

$$\begin{gathered} r=a \\ 胃=\frac{蟺}{2} \\ 蠒=0 \end{gathered}$$

Write the expression for the force on the charge q.

$F_{1}q{E}_{dipole}$ 鈥︹�� (3)

Substitute $\frac{P}{4蟺{蔚}_0{r}^3}\left(2\cos \widehat{胃}r+\sin \widehat{胃}胃\right)$for localid="1655730630645" $E_{dipo\ln e}\left(r,胃\right)$ in equation (3).

$F_1=q\left(\frac{P}{4蟺{蔚}_0{r}^3}\left(2\cos \widehat{胃}r+\sin \widehat{胃}胃\right)\right)$ 鈥︹�� (4)

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

$$\begin{gathered} F_1=q\left(\frac{蚁}{4蟺{蔚}_0{a}^3}\left(2\cos \left(\frac{蟺}{2}\right)r+\sin \left(\frac{蟺}{2}\right)胃\right)\right) \\ =q\frac{蚁}{4蟺{蔚}_0{a}^3}\sin \frac{蟺}{2}胃 \end{gathered}$$

Simplify the above equation,

$F_1-\frac{q蚁{}^{鈭�}}{4蟺{蔚}_0{a}^3}胃$

As the dipole $蚁$is pointing in the z-direction, so the electric force,

$F_1-\frac{q蚁{}^{鈭�}}{4蟺{蔚}_0{a}^3}胃$

Therefore, the force $F_1-\frac{q蚁{}^{鈭�}}{4蟺{蔚}_0{a}^3}胃$.

b)

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

$$\begin{gathered} r=a \\ 胃=0 \\ 蠒=0 \end{gathered}$$

Then,

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

$F_2=q\left(\frac{P}{4{\mathrm{蟺蔚}}_0{\mathrm{r}}^3}\left(2\cos \widehat{胃}r+\sin \widehat{胃}胃\right)\right)$ 鈥︹��. (5)

Substitute$a$for $r$and $0^{掳}$for $胃$in equation (5).

$$\begin{gathered} F_2=q\left(\frac{P}{4{\mathrm{蟺蔚}}_0{\mathrm{r}}^3}\left(2\cos \left(0^{掳}\right)r+\sin \left(0^{掳}\right)胃\right)\right) \\ =q\left(\frac{蚁}{4蟺{蔚}_0{a}^3}\left(2\cos \left(0^{掳}\right)r\right)\right) \end{gathered}$$

As the dipole $蚁$is pointing in the z-direction, so the electric force,

$F_2=\frac{2蚁q}{4蟺{蔚}_0{a}^3}\left(z\right)$

Thus, force $F_2=\frac{2蚁q}{4蟺{蔚}_0{a}^3}\left(z\right)$.

c)

Write the expression for potential due to dipole.

$V\left(r,胃\right)=\frac{蚁\cos 胃}{4蟺{蔚}_0{a}^2}$ 鈥︹��. (6)

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

$$\begin{gathered} v_2=\frac{蚁\cos \left(0\right)}{4蟺{蔚}_0{a}^2} \\ =\frac{蚁}{4蟺{蔚}_0{a}^2} \end{gathered}$$

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

$$\begin{gathered} V_2=\frac{蚁\cos \left[0\right]}{4蟺{蔚}_0{a}^{}} \\ =\frac{蚁}{蟺{蔚}_0{a}^2} \end{gathered}$$

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.

$W=\left(V_2-V_1\right)q$ 鈥︹��. (7)

Substitute 0 for $V_1$and $\frac{蚁}{4蟺{蔚}_0{a}^2}$for $V_2$in equation (7).

$$\begin{gathered} W=\left(\frac{蚁}{4蟺{蔚}_0{a}^2}-0\right)q \\ =\frac{蚁q}{4蟺{蔚}_0{a}^2} \end{gathered}$$

Therefore, the work done is $=\frac{蚁q}{4蟺{蔚}_0{a}^2}$.