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.