Question

Show that the interaction energy of two dipoles separated by a displacement $\mathit{r}$ is

$\mathbf{U}=\frac{1}{\mathbf{4}\pi {\epsilon }_0}\frac{1}{{\mathbf{r}}^3}[{\mathbf{p}}_1\cdot {\mathbf{p}}_2-\mathbf{3}({\mathbf{p}}_1\cdot \widehat{\mathbf{r}})({\mathbf{p}}_2\cdot \widehat{\mathbf{r}})]$

[Hint: Use Prob. 4.7 and Eq. 3.104.]

Short answer

The value of the interaction energy between the two dipoles is $\frac{1}{4\pi {\epsilon }_0{r}^3}\{p_1\cdot p_2-3(p_1\cdot \stackrel{\text{V}}{r})(p_2\cdot \stackrel{\text{V}}{r})\}$.

Step-by-step solution

Write the given data from the question.

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

Determine the formulaofinteraction energy between the two dipoles.

Write the formula of interaction energy between the two dipoles.

$\mathbf{U}=-{\mathbf{p}}_2\cdot \mathbf{E}\left(\mathbf{r}\right)$…… (1)

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

Step 3:Determine theinteraction energy between the two dipoles.

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:

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

The electric dipole moment is expressed as follows:

$\begin{array}{c}\text{P}=(\text{P}\cdot \stackrel{\text{V}}{r})\stackrel{\text{V}}{r}+(P\cdot \stackrel{\text{V}}{\theta })\stackrel{\text{V}}{\theta }\\ =\text{P}\cos \theta \stackrel{\text{V}}{r}-\text{P}\sin \text{θ}\stackrel{\text{V}}{\theta }\end{array}$

Then, solve further as:

$\begin{array}{c}3(\text{P}\cdot \stackrel{\text{V}}{r})\stackrel{\text{V}}{r}-\text{P}=3\left[\text{Pcosθ}\stackrel{\text{V}}{r}\right]-\text{Pcosθ}\stackrel{\text{V}}{r}+\text{P}\sin \text{θ}\stackrel{\text{V}}{\theta }\\ =2\text{Pcosθ}\stackrel{\text{V}}{r}+P\sin \text{θ}\stackrel{\text{V}}{\theta }\\ =\text{P}[2\cos \text{θ}\stackrel{\text{V}}{r}+\sin \text{θ}\stackrel{\text{V}}{\theta }]\end{array}$ …… (3)

From the equation (2) and (3).

$\text{E}\left(r\right)=\frac{{\mu }_0}{4\pi r^3}[3\text{P}\cdot \stackrel{\text{V}}{r}]\stackrel{\text{V}}{r}-\text{P}$

Now the electric field due to dipole is,

$\begin{array}{c}\text{E}=\frac{1}{4\pi {\epsilon }_0}\frac{1}{r^3}\{3\left[(\text{p}\cdot (-\stackrel{\text{V}}{r}))\right](-\stackrel{\text{V}}{r})-\text{P}\}\\ =\frac{1}{4\pi {\epsilon }_0}\frac{1}{r^3}[3(\text{p}\cdot \stackrel{\text{V}}{r})-\text{p}]\end{array}$

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:

$E\left(r\right)=\frac{1}{4\pi {\epsilon }_0{r}^3}\{3(p_1\cdot \stackrel{\text{V}}{r})\stackrel{\text{V}}{r}-{\text{p}}_{\text{1}}\}$

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 $\frac{1}{4\pi {\epsilon }_0{r}^3}\{3(p_1\cdot \stackrel{\text{V}}{r})\stackrel{\text{V}}{r}-{\text{p}}_{\text{1}}\}$for $E\left(r\right)$ into equation (1).

$\begin{array}{c}U=-p_2\cdot \left\{\frac{1}{4\pi {\epsilon }_0{r}^3}(3(\stackrel{\text{V}}{P_1}\cdot \widehat{r})\widehat{r}-p_1)\right\}\\ =\frac{1}{4\pi {\epsilon }_0{r}^3}\{p_1\cdot p_2-3(\stackrel{\text{V}}{p_1}\cdot \widehat{r})(p_2\cdot \widehat{r})\}\end{array}$

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

$\frac{1}{4\pi {\epsilon }_0{r}^3}\{p_1\cdot p_2-3(p_1\cdot \stackrel{\text{V}}{r})(p_2\cdot \stackrel{\text{V}}{r})\}$