In Fig. 4.6,${\overrightarrow{\mathbf{P}}}_{\mathbf{1}}$and${\overrightarrow{\mathbf{P}}}_2$are (perfect) dipoles a distance rapart. What is

the torque on${\overrightarrow{\mathbf{P}}}_{\mathbf{1}}$due to${\overrightarrow{\mathbf{P}}}_{\mathbf{2}}$? What is the torque on${\overrightarrow{\mathbf{P}}}_{\mathbf{2}}$due to${\overrightarrow{\mathbf{P}}}_{\mathbf{1}}$? [In each case, I want the torque on the dipole about its own center.If it bothers you that the answers are not equal and opposite, see Prob. 4.29.]

There are two dipoles having dipole moments ${\overrightarrow{P}}_1$and${\overrightarrow{P}}_2$ .

The electric field due to a dipole having dipole moment $p$is

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

Here, ${\epsilon }_0$is the permittivity of free space and$r$ and $\theta$ are spherical polar coordinates.

The field due to ${\overrightarrow{P}}_1$at ${\overrightarrow{P}}_2$which is at a distance $r$from ${\overrightarrow{P}}_1$and $\theta =\frac{\mathrm{\pi }}{2}$is

$$\begin{gathered} {\overrightarrow{E}}_1=\frac{P_1}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}\left(2\cos \frac{\mathrm{\pi }}{2}\stackrel{⏜}{r}+\sin \frac{\mathrm{\pi }}{2}\stackrel{⏜}{\theta }\right) \\ =\frac{p_1}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}{\theta }^3 \end{gathered}$$

The field thus points downwards and makes an angle $90°$with $\overrightarrow{P_2}$.

Thus the expression for the torque on ${\overrightarrow{p}}_2$is

$$\begin{gathered} {\tau }_2=p_2{E}_1\sin 90° \\ =p_2{E}_1 \end{gathered}$$

Substitute the expression for electric field in the above equation and get

${\tau }_2=\frac{p_1{p}_2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}$

The torque points into the screen.

The field due to ${\overrightarrow{p}}_2$at ${\overrightarrow{p}}_1$which is at a distance $r$from ${\overrightarrow{p}}_2$and $\theta =\mathrm{\pi }$is

$$\begin{gathered} {\overrightarrow{E}}_2=\frac{P_2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}\left(2\mathrm{cos\pi }\stackrel{⏜}{r}+\mathrm{sin\pi }\stackrel{⏜}{\theta }\right) \\ =\frac{2p}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}\stackrel{⏜}{r} \end{gathered}$$

The field thus points towards the right and makes an angle $90°$with ${\overrightarrow{p}}_1$.

Thus the expression for the torque on ${\overrightarrow{p}}_1$is

$$\begin{gathered} {\tau }_1=p_1{E}_2\sin 90° \\ =p_1{E}_2 \end{gathered}$$

Substitute the expression for electric field in the above equation and get

${\tau }_1=\frac{2p_1{p}_2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}$

The torque points into the screen.

Thus, the torque on ${\overrightarrow{p}}_1$due to ${\overrightarrow{p}}_2$is $\frac{2p_1{p}_2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}$and the torque on ${\overrightarrow{p}}_2$ due to${\overrightarrow{p}}_1$is $\frac{p_1{p}_2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^3}$.