91Ó°ÊÓ

Consider the electric field of dipole moment $P$ is at the origin, which point out towards $z$direction.

Write the formula of interaction energy between the two dipoles.

$\mathbf{U}=−{\mathbf{p}}_2â‹\dots \mathbf{E}\left(\mathbf{r}\right)$…… (1)

Here, ${\mathrm{p}}_2$ is dipole moment and $\mathrm{E}\left(\mathrm{r}\right)$ is electric field.

The electric dipole of dipole moment $p$is at the origin, which point out towards z direction as shown in following figure.

Figure 1

Determine the electric field of a dipole as follows:

Determine the electric field due to a dipole is expressed as follows:

$\text{E}\left(r\right)=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{P}{r^3}(2\cos \mathrm{θ}\stackrel{\text{V}}{r}+\sin \text{θ}\stackrel{\text{V}}{\mathrm{θ}})$ …… (2)

The electric dipole moment is expressed as follows:

$\begin{array}{c}\text{P}=(\text{P}â‹\dots \stackrel{\text{V}}{r})\stackrel{\text{V}}{r}+(Pâ‹\dots \stackrel{\text{V}}{\mathrm{θ}})\stackrel{\text{V}}{\mathrm{θ}}\\ =\text{P}\cos \mathrm{θ}\stackrel{\text{V}}{r}−\text{P}\sin \text{θ}\stackrel{\text{V}}{\mathrm{θ}}\end{array}$

Then, solve further as:

$\begin{array}{c}3(\text{P}â‹\dots \stackrel{\text{V}}{r})\stackrel{\text{V}}{r}−\text{P}=3\left[\text{±Ê³¦´Ç²õθ}\stackrel{\text{V}}{r}\right]−\text{±Ê³¦´Ç²õθ}\stackrel{\text{V}}{r}+\text{P}\sin \text{θ}\stackrel{\text{V}}{\mathrm{θ}}\\ =2\text{±Ê³¦´Ç²õθ}\stackrel{\text{V}}{r}+P\sin \text{θ}\stackrel{\text{V}}{\mathrm{θ}}\\ =\text{P}[2\cos \text{θ}\stackrel{\text{V}}{r}+\sin \text{θ}\stackrel{\text{V}}{\mathrm{θ}}]\end{array}$ …… (3)

From the equation (2) and (3).

$\text{E}\left(r\right)=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}{r}^3}[3\text{P}â‹\dots \stackrel{\text{V}}{r}]\stackrel{\text{V}}{r}−\text{P}$

Now the electric field due to dipole is,

$\begin{array}{c}\text{E}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^3}\{3\left[(\text{p}â‹\dots (−\stackrel{\text{V}}{r}))\right](−\stackrel{\text{V}}{r})−\text{P}\}\\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^3}[3(\text{p}â‹\dots \stackrel{\text{V}}{r})−\text{p}]\end{array}$

The minus sign indicates that $r$points towards$P$.

Draw the circuit diagram shows the interaction between two dipoles, which are separated by a distancelocalid="1658226423292" $r$.

Determine the electric field due to $P_1$ is expressed as follows:

$E\left(r\right)=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^3}\{3(p_1â‹\dots \stackrel{\text{V}}{r})\stackrel{\text{V}}{r}−{\text{p}}_{\text{1}}\}$

Here,${\text{P}}_1$,${\text{P}}_{\text{2}}$ are the dipole moments of the two dipoles.

So, the interaction energy of two dipoles is expressed.

Determine the interaction energy between the two dipoles.

Substitute $\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^3}\{3(p_1â‹\dots \stackrel{\text{V}}{r})\stackrel{\text{V}}{r}−{\text{p}}_{\text{1}}\}$for $E\left(r\right)$ into equation (1).

$\begin{array}{c}U=−p_2â‹\dots \left\{\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^3}(3(\stackrel{\text{V}}{P_1}â‹\dots \widehat{r})\widehat{r}−p_1)\right\}\\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^3}\{p_1â‹\dots p_2−3(\stackrel{\text{V}}{p_1}â‹\dots \widehat{r})(p_2â‹\dots \widehat{r})\}\end{array}$

Therefore, the value of the interaction energy between the two dipoles is .

$\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^3}\{p_1â‹\dots p_2−3(p_1â‹\dots \stackrel{\text{V}}{r})(p_2â‹\dots \stackrel{\text{V}}{r})\}$