Question

A stationary electric dipole $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}\mathbf{=}\mathbf{p}\hat{\mathbf{z}}$is situated at the origin. A positive

point charge q(mass m) executes circular motion (radius s) at constant speed

in the field of the dipole. Characterize the plane of the orbit. Find the speed, angular momentum and total energy of the charge.

Short answer

The velocity of the charged particle executing circular motion under the influence of a dipole at origin having dipole moment $\stackrel{\rightharpoonup }{p}=p\hat{z}$is $\frac{1}{s}\sqrt{\frac{qp}{3\sqrt{3}{\mathrm{\pi \epsilon }}_0\mathrm{m}}}$. Its $\sqrt{\frac{qp}{3\sqrt{3}{\mathrm{\pi \epsilon }}_0}}$angular momentum is and its energy is zero.

Step-by-step solution

Given data

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

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

Electric field of a dipole and unit vectors

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

$$\begin{gathered} \stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}\mathbf{=}\mathbf{3}\mathbf{(}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}\mathbf{\times }\hat{\mathbf{r}}\mathbf{)}\hat{\mathbf{r}}\mathbf{-}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{p}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\mathbf{(}\mathbf{2}\mathbf{cos\theta }\hat{\mathbf{r}}\mathbf{+}\mathbf{sin\theta }\hat{\mathbf{\theta }}\mathbf{)} \end{gathered}$$ .... (1)

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

$\hat{\mathbf{r}}\mathbf{=}\mathbf{sin\theta cos}\hat{\mathbf{x}}\mathbf{+}\mathbf{sin\theta sin}\hat{\mathbf{y}}\mathbf{+}\mathbf{cos\theta }\hat{\mathbf{z}}$ ... ( 2)

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

$\hat{\mathbf{\theta }}\mathbf{=}\mathbf{cos\theta cosf}\hat{\mathbf{x}}\mathbf{+}\mathbf{cos\theta sinf}\hat{\mathbf{y}}\mathbf{-}\mathbf{sin\theta }\hat{\mathbf{z}}$ ...(3)

Velocity, angular momentum and energy of a charged particle under the influence of a dipole

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

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

Thus,

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

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

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

But,

$r=\frac{s}{\sin \theta }$

Thus,

$\stackrel{\rightharpoonup }{E}=-\frac{p\hat{s}}{3\sqrt{3}{\mathrm{\pi \epsilon }}_0{\mathrm{s}}^3}$

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

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

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

The expression for the angular momentum is

L = smv

Substitute the expression for and get

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

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

The expression for the energy is

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

Here, V is the electric potential given by

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

Substitute the expression for and get

$E=\frac{1}{2}m\times \frac{1}{s^2}\frac{qp}{3\sqrt{3}{\mathrm{\pi \epsilon }}_0\mathrm{m}}+q\frac{p}{4{\mathrm{\pi \epsilon }}_0\sqrt{3}\left(3/2\right){\mathrm{s}}^2}$

= 0

Thus, the energy of the charged particle is zero.