(a) For the configuration in Prob. 4.5, calculate the forceon ${\overrightarrow{\mathrm{p}}}_2$due to ${\overrightarrow{\mathrm{p}}}_1$and the force on ${\overrightarrow{\mathrm{p}}}_1$due to ${\overrightarrow{\mathrm{p}}}_2$. Are the answers consistent with Newton's third law?

(b) Find the total torque on ${\overrightarrow{\mathrm{p}}}_2$ with respect to the center of${\overrightarrow{\mathrm{p}}}_1$and compare it with

the torque on${\overrightarrow{\mathrm{p}}}_1$ about that same point. [Hint:combine your answer to (a) with

the result of Prob. 4.5.]

There are two dipoles having moments ${\overrightarrow{p}}_1$and${\overrightarrow{p}}_2$perpendicular to each other.

The force on a dipole having moment $\overrightarrow{p}$in the presence of an electric field $\overrightarrow{E}$is

$\overrightarrow{\mathbf{F}}\mathbf{=}\left(\overrightarrow{p}.\overrightarrow{\nabla }\right)\overrightarrow{\mathbf{E}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The cross product of dipole moment of the second dipole and the electric field due to the first dipole is

${\overrightarrow{\mathbf{P}}}_{\mathbf{2}}\mathbf{\times }{\overrightarrow{\mathbf{E}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{-}{\mathbf{p}}_{\mathbf{1}}{\mathbf{p}}_{\mathbf{2}}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here,${\epsilon }_0$ is the permittivity of free space.

Let the second dipole moment be

$\overrightarrow{p}=p_2\hat{y}$

The electric field due to the first dipole at the position of the second dipole is

$\overrightarrow{E}=-\frac{p_1}{4\pi {\epsilon }_0{y}^3}\hat{z}$

Substitute this in equation (1) to get the force on the second dipole due to the first dipole

localid="1657775080935" $$\begin{gathered} \overrightarrow{F}=p_2\frac{\partial }{\partial y}\left(-\frac{p_1}{4\pi {\epsilon }_0{y}^3}\hat{z}\right) \\ =\frac{3p_1{p}_2}{4\pi {\epsilon }_0{y}^4}\hat{z} \\ =\frac{3p_1{p}_2}{4\pi {\epsilon }_0{r}^4}\hat{z} \end{gathered}$$

To find the electric field due to the second dipole, it is kept at the origin and pointed towards z axis. The expression for the electric field due to the first dipole at the position of the second dipole is

$\overrightarrow{E}=\frac{p_2}{4\pi \epsilon }\left[\frac{3xz\hat{x}+3yz\hat{y}-\left(x^2+y^2-2z^2\right)\hat{z}}{{\left(x^2+y^2+z^2\right)}^{5/2}}\right]$

Thus from equation (1), he force on the first dipole due to the second dipole is

${\overrightarrow{F}}_2={\overline{)-p_1\frac{\partial {\overrightarrow{E}}_2}{\partial y}}}_{x=0,y=0.z=r}$

Substitute the values in the above equation006E

localid="1657775867177" $$\begin{gathered} {\overrightarrow{F}}_2=-p_1\frac{\partial }{\partial y}\frac{p_2}{4\pi {\epsilon }_0}{\overline{)\left[\frac{3xz\hat{x}+3yz\hat{y}-\left(x^2+y^2-2z^2\right)\hat{z}}{{\left(x^2+y^2+z^2\right)}^{5/2}}\right]}}_{x=0,y=0.z=-r} \\ =\frac{-p_1{p}_2}{4\pi {\epsilon }_0}{\left[\frac{-5}{2}2y\frac{3xz\hat{x}+3yz\hat{y}-\left(x^2+y^2-2z^2\right)\hat{z}}{{\left(x^2+y^2+z^2\right)}^{7/2}}+\frac{3yz\hat{y}-2y\hat{z}}{{\left(x^2+y^2+z^2\right)}^{5/2}}\right]}_{x=0,y=0.z=r} \\ =\frac{3p_1{p}_2}{4\pi {\epsilon }_0{r}^4}\hat{y} \end{gathered}$$

$\hat{y}$in the turned coordinate is equal to $\hat{z}$in the original coordinate.

Thus, the forces are$\frac{3p_1{p}_2}{4\pi {\epsilon }_0{r}^4}\hat{z}$ and $-\frac{3p_1{p}_2}{4\pi {\epsilon }_0{r}^4}\hat{z}$. They are equal and opposite and hence are consistent with Newton's third law of motion.

The expression for the torque on the second dipole is

$\hat{\tau }={\overrightarrow{p}}_2\times {\overrightarrow{E}}_1+\overrightarrow{r}\times {\overrightarrow{F}}_2$

Here, $\overrightarrow{r}=-r\hat{y}$

Use equation (2) and substitute values in the above equation

$$\begin{gathered} \overrightarrow{\tau }=\frac{-p_1{p}_2}{4\pi {\epsilon }_0{r}^3}\hat{x}+r\hat{y}\times \frac{3p_1{p}_2}{4\pi {\epsilon }_0{r}^4}\hat{z} \\ =\left(\frac{-p_1{p}_2}{4\pi {\epsilon }_0{r}^3}+\frac{3p_1{p}_2}{4\pi {\epsilon }_0{r}^3}\right)\hat{x} \\ =\frac{2p_1{p}_2}{4\pi {\epsilon }_0{r}^3}\hat{x} \end{gathered}$$

This is equal and opposite to the torque on the first dipole.