91Ó°ÊÓ

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

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

$\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{=}\frac{\mathbf{P}}{\mathbf{4}\mathbf{Ï„}\mathbf{Ï„}{\mathbf{Î\mu }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\left(2cos\mathrm{θ}\stackrel{Áåœ}{r}+sin\mathrm{θ}\stackrel{Áåœ}{\mathrm{θ}}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here, ${\mathrm{Î\mu }}_0$is the permittivity of free space and$r$ and $\mathrm{θ}$ are spherical polar coordinates.

The field due to ${\stackrel{→}{P}}_1$at ${\stackrel{→}{P}}_2$which is at a distance $r$from ${\stackrel{→}{P}}_1$and $\mathrm{θ}=\frac{\mathrm{π}}{2}$is

$$\begin{gathered} {\stackrel{→}{E}}_1=\frac{P_1}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}\left(2\cos \frac{\mathrm{Ï€}}{2}\stackrel{Áåœ}{r}+\sin \frac{\mathrm{Ï€}}{2}\stackrel{Áåœ}{\mathrm{θ}}\right) \\ =\frac{p_1}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}{\mathrm{θ}}^3 \end{gathered}$$

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

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

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

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

${\mathrm{Ï„}}_2=\frac{p_1{p}_2}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}$

The torque points into the screen.

The field due to ${\stackrel{→}{p}}_2$at ${\stackrel{→}{p}}_1$which is at a distance $r$from ${\stackrel{→}{p}}_2$and $\mathrm{θ}=\mathrm{π}$is

$$\begin{gathered} {\stackrel{→}{E}}_2=\frac{P_2}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}\left(2\mathrm{³¦´Ç²õÏ€}\stackrel{Áåœ}{r}+\mathrm{²õ¾\pm ²ÔÏ€}\stackrel{Áåœ}{\mathrm{θ}}\right) \\ =\frac{2p}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}\stackrel{Áåœ}{r} \end{gathered}$$

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

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

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

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

${\mathrm{Ï„}}_1=\frac{2p_1{p}_2}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}$

The torque points into the screen.

Thus, the torque on ${\stackrel{→}{p}}_1$due to ${\stackrel{→}{p}}_2$is $\frac{2p_1{p}_2}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}$and the torque on ${\stackrel{→}{p}}_2$ due to${\stackrel{→}{p}}_1$is $\frac{p_1{p}_2}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^3}$.