Question:A (perfect) dipole p is situated a distance z above an infinite grounded conducting plane (Fig. 4.7). The dipole makes an angle $\theta$with the perpendicular to the plane. Find the torque on p . If the dipole is free to rotate, in what orientation will it come to rest?

Consider the(perfect) dipole p is situated a distance z above an infinite grounded conducting plane (Fig. 4.7).

Write the formula of magnitude of the torque acting on the dipole.

$\overrightarrow{\mathbf{N}}=\overrightarrow{\mathbf{p}}\times {\overrightarrow{\mathbf{E}}}_{\mathbf{i}}$ …… (1)

Here, p is the position coordinate of the dipole, E is electric field at the position of the real dipole due to image dipole.

From the given data, the dipole p is situated a distance z above an infinite grounded conducting plate.

The arrangement of the dipole is on the z-axis is shown in the following figure:

Figure 1

In the figure, p is the dipole,$\theta$ is the angle made by the dipole with the z-axis.

By using method of images, the arrangement of dipole p is shown in the following figure:

Figure 2

In the figure, the distance between the two dipole is as follows:

$r=2z$

From the figure, the position coordinate of the dipole is as follows:

$\overrightarrow{p}=\left(p\cos \theta \right)\hat{r}+\left(p\sin \theta \right)\hat{\theta }$

The electric field caused by the image dipole at the location of the real dipole may be calculated using equation 3.103 as follows:

$\stackrel{\rightharpoonup }{E_i}=\frac{p}{4\pi {\epsilon }_0{r}^3}\left(\left(2\cos \theta \right)\hat{r}+\left(\sin \theta \right)\hat{\theta }\right)$

Substitute 2z for r in the equation

The expression for the torque on the dipole is as follows:

$\overrightarrow{N}=\overrightarrow{p}\times {\overrightarrow{E}}_i$

Substitute$\stackrel{\rightharpoonup }{E_i}=\frac{p}{4\pi {\epsilon }_0{\left(2z\right)}^3}\left(\left(2\cos \theta \right)\hat{r}+\left(\sin \theta \right)\hat{\theta }\right)$ for ${\overrightarrow{E}}_i$ and $\left(p\cos \theta \right)\hat{r}+\left(p\sin \theta \right)\hat{\theta }$for p in the equation $\overrightarrow{N}=\overrightarrow{p}\times {\overrightarrow{E}}_i$and solve for .

Substitute $\frac{p}{4\pi {\epsilon }_0{\left(2z\right)}^3}\left(\left(2\cos \theta \right)\hat{r}+\left(\sin \theta \right)\hat{\theta }\right)$ for ${\overrightarrow{E}}_i$ in the equation $\left(p\cos \theta \right)\hat{r}+\left(p\sin \theta \right)\hat{\theta }$and solve for the magnitude of the torque on the dipole.

$$\begin{gathered} \overrightarrow{N}=\left(\left(p\cos \theta \right)\hat{r}+\left(p\sin \theta \right)\hat{\theta }\right)\times \left(\frac{p}{4\pi {\epsilon }_0{\left(2\pi \right)}^3}\left(\left(2\cos \theta \right)\hat{r}+\left(\sin \theta \right)\hat{\theta }\right)\right) \\ =\frac{p^2}{4\pi {\epsilon }_0{\left(2z\right)}^3}\left(\left(\left(\cos \theta \right)\hat{r}+\left(\sin \theta \right)\hat{\theta }\right)\times \left(\left(2\cos \theta \right)\hat{r}+\left(\sin \theta \right)\hat{\theta }\right)\right) \\ =\frac{p^2}{4\pi {\epsilon }_0{\left(2z\right)}^3}\left(\left(\cos \theta \sin \theta \right)\varphi +\left(2\sin \theta \cos \theta \right)\left(-\hat{\varphi }\right)\right) \\ =-\left(\frac{P^2\sin \theta \cos \theta }{4\pi {\epsilon }_0{\left(2z\right)}^3}\right)\hat{\varphi } \end{gathered}$$

Therefore, $\frac{p^2\sin 2\theta }{4\pi {\epsilon }_0\left(8z^3\right)}$the magnitude of the torque acting on the dipole is and the direction of the torque is out of the page.

For $0<\theta <\frac{\pi }{2}$, the torque on the dipole is positive. As a result, the dipole will turn in the opposite direction of its stable orientation of .

For $\frac{\pi }{2}<\theta <\pi$, the dipole's torque is negative. As a result, the dipole will turn in the direction of the stable orientation of $\theta =\pi$.