Electrostatics: work moving point dipole

[Bapelsin]

Homework Statement

Two electrical dipoles with dipole moments $$\vec{p}_0=p_0\hat{y}$$ and $$\vec{p}_1=p_1\hat{y}$$ are located in the xy-plane. $$\vec{p}_0$$ i located at the origin and $$\vec{p}_1$$ is initially in (x,y)=(a,0). What work is required to move $$\vec{p}_1$$ (with unhanged directon) to the point (x,y)=(0,a)?

Homework Equations

Dipole potential: $$\phi_D(\vec{r})=\frac{\vec{p}\cdot\hat{r}}{4\pi\epsilon_0r^2}$$
Work: $$W=Q\phi$$

The Attempt at a Solution

Taking the difference of the potential in the two cases:

$$\left(\phi_{D, p_o}+\frac{p_1\hat{y}\cdot a\hat{x}}{4\pi\epsilon_0a^2}\right) - \left(\phi_{D,p_0}+\frac{p_1\hat{y}\cdot a\hat{y}}{4\pi\epsilon_0a^2}\right)=-\frac{p_1}{4\pi\epsilon_0a}$$

Here comes the step that I'm not sure about. The work is the potential times the point charge - how is it when we have a point dipole? Attempt:

$$W=Q\phi = -\frac{p_0}{a}\cdot\frac{p_1}{4\pi\epsilon_0a}=-\frac{p_0p_1}{4\pi\epsilon_0a^2}$$

Does this make sense? Any help appreciated!

Thanks!

 

[kuruman]

Start with the expression for the potential energy of one dipole in the external electric field of the other dipole

$$U=-\vec{p}_{1}\cdot\vec{E}_{2}$$

where

$$E_{2}=\frac{1}{4 \pi\epsilon_{0}}\large(\frac{(3\vec{p}_{2} \cdot \vec{r})\vec{r}}{r^{5}}-\frac{\vec{p}_2}{r^3} \large)$$

** Addition on edit **
Here, vector r is the position of dipole 1 relative to dipole 2, i.e. dipole 2 is at the origin.