91影视

The dipole moment of the dipole situated at the origin is $\stackrel{鈬赌}{p}=p\hat{z}$.

A charged particle of mass m, charge q performs circular motion of radius r .

The electric field of a dipole having dipole moment $\stackrel{鈬赌}{p}$

$$\begin{gathered} \stackrel{\mathbf{鈬赌}}{\mathbf{E}}\mathbf{=}\mathbf{3}\mathbf{(}\stackrel{\mathbf{鈬赌}}{\mathbf{p}}\mathbf{脳}\hat{\mathbf{r}}\mathbf{)}\hat{\mathbf{r}}\mathbf{-}\stackrel{\mathbf{鈬赌}}{\mathbf{p}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{p}}{\mathbf{4}{\mathbf{蟿蟿蔚}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\mathbf{(}\mathbf{2}\mathbf{肠辞蝉胃}\hat{\mathbf{r}}\mathbf{+}\mathbf{蝉颈苍胃}\widehat{\mathbf{胃}}\mathbf{)} \end{gathered}$$ .... (1)

The expression for unit vector $\hat{r}$in terms of $\hat{x}$, $\hat{y}$and $\hat{z}$

$\hat{\mathbf{r}}\mathbf{=}\mathbf{蝉颈苍胃cos}\hat{\mathbf{x}}\mathbf{+}\mathbf{蝉颈苍胃sin}\hat{\mathbf{y}}\mathbf{+}\mathbf{肠辞蝉胃}\hat{\mathbf{z}}$ ... ( 2)

The expression for unit vector $\widehat{胃}$in terms of $\hat{x}$, $\hat{y}$and $\hat{z}$

$\widehat{\mathbf{胃}}\mathbf{=}\mathbf{肠辞蝉胃cosf}\hat{\mathbf{x}}\mathbf{+}\mathbf{肠辞蝉胃sinf}\hat{\mathbf{y}}\mathbf{-}\mathbf{蝉颈苍胃}\hat{\mathbf{z}}$ ...(3)

From symmetry, the electric field has to be perpendicular to the z axis. Thus from equation (1),

$$\begin{gathered} \stackrel{鈫�}{E}.\hat{z}=0 \\ 3\left(\stackrel{鈬赌}{p}.\hat{r}\right)\hat{r}.\hat{z}-\stackrel{鈬赌}{p}.\hat{z}=0 \\ 3{\cos }^2胃-1=0 \\ \cos 胃=-\frac{1}{\sqrt{3}} \end{gathered}$$

Thus,

$$\begin{gathered} \sin 胃=\sqrt{1-{\cos }^2胃} \\ =\frac{\sqrt{2}}{3} \\ \tan 胃=\frac{\sin 胃}{\cos 胃} \\ =-\sqrt{2} \end{gathered}$$

Substitute the expressions of $\sin 胃$, $\cos 胃$and unit vectors from equation (3) and (4) into equation (2),

$$\begin{gathered} \stackrel{鈬赌}{E}=\frac{p}{4{\mathrm{蟺蔚}}_0{\mathrm{r}}^3}\sqrt{\frac{2}{3}}\left[\begin{array}{c}-\sqrt{2}\left(\sin 胃\cos 蠒\hat{x}+\sin 胃\sin 蠒\hat{y}+\cos 胃\hat{z}\right)\\ +\left(\cos 胃\cos 蠒\hat{x}+\cos 胃\sin 蠒\hat{y}+\sin 胃\hat{z}\right)\end{array}\right] \\ =\frac{p}{4{\mathrm{蟺蔚}}_0{\mathrm{r}}^3}\sqrt{\frac{2}{3}}\left[\left(-\sqrt{2}\sqrt{\frac{2}{3}}-\frac{1}{\sqrt{3}}\right)\left(\cos 蠒\hat{x}+\sin 蠒\hat{y}\right)\right] \\ =-\frac{\sqrt{2}p\hat{s}}{4{\mathrm{蟺蔚}}_0{\mathrm{r}}^3} \end{gathered}$$

But,

$r=\frac{s}{\sin 胃}$

Thus,

$\stackrel{鈬赌}{E}=-\frac{p\hat{s}}{3\sqrt{3}{\mathrm{蟺蔚}}_0{\mathrm{s}}^3}$

Equate the electric force to the centripetal force for circular motion,

$$\begin{gathered} \frac{qp}{3\sqrt{3}{\mathrm{蟺蔚}}_0{\mathrm{s}}^3}=\frac{m{v}^2}{s} \\ v=\frac{1}{s}\sqrt{\frac{qp}{3\sqrt{3}{\mathrm{蟺蔚}}_0\mathrm{m}}} \end{gathered}$$

Thus, the velocity of the charged particle is $\frac{1}{s}\sqrt{\frac{qp}{3\sqrt{3}{\mathrm{蟺蔚}}_0\mathrm{m}}}$.

The expression for the angular momentum is

L = smv

Substitute the expression for and get

localid="1657691428418" $$\begin{gathered} L=sm脳\frac{1}{s}\sqrt{\frac{qp}{3\sqrt{3}{\mathrm{蟺蔚}}_0\mathrm{m}}} \\ =\sqrt{\frac{qpm}{3\sqrt{3}{\mathrm{蟺蔚}}_0}} \end{gathered}$$

Thus, the angular momentum of the charge is $=\sqrt{\frac{qpm}{3\sqrt{3}{\mathrm{蟺蔚}}_0}}$.

The expression for the energy is

$E=\frac{1}{2}m{v}^2+qV$

Here, V is the electric potential given by

$$\begin{gathered} V=\frac{p\cos 胃}{4{\mathrm{蟺蔚}}_0{\mathrm{r}}^2} \\ =\frac{p}{4{\mathrm{蟺蔚}}_0\sqrt{3}\left(3/2\right){\mathrm{s}}^2} \end{gathered}$$

Substitute the expression for and get

= 0

Thus, the energy of the charged particle is zero.