Question

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?

Short answer

Answer

The value of the magnitude of the torque acting on the dipole is $\frac{p^2\sin 2\theta }{4\pi {\epsilon }_0\left(8z^3\right)}$.

Step-by-step solution

Write the given data from the question.

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

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

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.

Determine the magnitude of the torque acting on the 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

localid="1658382834921" $$\begin{gathered} \stackrel{\rightharpoonup }{E_i}=\frac{p}{4\pi {\epsilon }_0{r}^3}\left(\left(2\cos \theta \right)\hat{r}+\left(\sin \theta \right)\hat{\theta }\right) \\ \stackrel{\rightharpoonup }{E_i}=\frac{p}{4\pi {\epsilon }_0{\left(2z\right)}^3}\left(\left(2\cos \theta \right)\hat{r}+\left(\sin \theta \right)\hat{\theta }\right) \end{gathered}$$.

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 $\theta =\pi$.