An ideal electric dipole is situated at the origin, and points in the direction, as in Fig. 3.36. An electric charge is released from rest at a point in the x-y plane. Show that it swings back and forth in a semi-circular arc, as though it were apendulum supported at the origin.

An electric dipole with dipole moment $\overrightarrow{p}$pointing in the $\hat{z}$direction.

A charge q released in the x-y plane.

The field of a dipole having dipole moment $\overrightarrow{p}$is

$\overrightarrow{\mathbf{E}}\mathbf{=}\frac{\mathbf{p}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\left(2\mathrm{cos\theta }\hat{r}+\mathrm{sin\theta }\hat{\theta }\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The net force on a pendulum of mass mis

$\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{-}\mathbf{mg}\hat{\mathbf{z}}\mathbf{-}\mathbf{T}\hat{\mathbf{r}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here, gis the acceleration due to gravity and Tis the tension in the string supporting the pendulum.

The unit vector$\hat{z}$expressed in terms of $\hat{r}$and $\hat{\theta }$

$\hat{\mathbf{z}}\mathbf{=}\mathbf{cos\theta }\hat{\mathbf{r}}\mathbf{-}\mathbf{sin\theta }\hat{\mathbf{\theta }}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Following equation (1), the expression for force on a charge qdue to a dipole having dipole moment $\overrightarrow{p}$is

$$\begin{gathered} \overrightarrow{F}=q\overrightarrow{E} \\ =\frac{qp}{4\pi {\epsilon }_0{r}^3}\left(2\cos \theta \hat{r}+\sin \theta \hat{\theta }\right)....\left(4\right) \end{gathered}$$

The equilibrium condition of a pendulum is,

$$\begin{gathered} T-mg\cos \varphi =\frac{m{v}^2}{1} \\ T=mg\cos \varphi =\frac{m{v}^2}{1}......\left(5\right) \end{gathered}$$

Here, v is the velocity of the pendulum.

From conservation of energy,

$$\begin{gathered} mg/\cos \varphi =\frac{1}{2}m{v}^2 \\ {v}^2=2g/\cos \varphi \end{gathered}$$

From Fig. 3.36,

$\cos \varphi =-\cos \theta$

Substitute the expression for $v^2$and $\cos \varphi$in equation (5),

$$\begin{gathered} T=-mg\cos \theta -\frac{m2gl\cos \theta }{l} \\ =-mg\cos \theta -2mg\cos \theta \\ =3mg\cos \theta \end{gathered}$$

Substitute this expression in equation (2) and use equation (3)

$$\begin{gathered} \overrightarrow{F}=-mg\left(\cos \theta \hat{r}-\sin \theta \hat{\theta }\right)+3mg\cos \theta \hat{r} \\ =mg\left(2\cos \theta \hat{r}+\sin \theta \hat{\theta }\right) \end{gathered}$$

This is similar to the force expressed in equation (4).

Hence, the charge also oscillates like a pendulum.