Vector Equation for Force Between Electric Dipoles

[Xyius]

(All the variables are vectors, I just didn't feel like fumbling around with the LateX code to make them vectors. Its late and I'm tired a lazy!)

Homework Statement

(This is paraphrased)
There are two dipoles with arbitrary direction to each other. You know the energy between the dipoles is..

$$W_D=-p \cdot E$$

What is the force between them? $F_{1,2}$

Homework Equations

Force:
$$F=-\nabla W_D$$

Electric Field of a Dipole:
$$E=\frac{1}{4\pi \epsilon_0 r^3}[3p \cdot a_r-p]$$

a_r is the unit vector from p1 to p2.

The Attempt at a Solution

Using the following vector identity..
$$\nabla (a \cdot b)=(a \cdot \nabla)b+(b \cdot \nabla)a+a×(\nabla × b)+b×(\nabla × a)$$

$$-\nabla (-p \cdot E)=\nabla (p \cdot E)=(p \cdot \nabla)E+(E \cdot \nabla)p+p×(\nabla × E)+E×(\nabla ×p)$$

So I know $\nabla × E =0$ and $E×(\nabla ×p)=(E \cdot p)\nabla-(E \cdot \nabla)p$

So plugging these in, I get.

$$F=(p \cdot \nabla)E+(E \cdot p)\nabla$$

The prof gives the answer as, $$F=(p \cdot \nabla)E$$

So that means $(E \cdot p)\nabla=0$ But I cannot figure out why.

 

[gabbagabbahey]

Its late and I'm tired a lazy!

Sounds French.

$E×(\nabla ×p)=(E \cdot p)\nabla-(E \cdot \nabla)p$

A vector differential operator does not a vector make. You cannot use the familiar BAC-CAB identity when one of the operands is not really a vector. Instead, what would you expect the curl of a point dipole to be?

 

[Xyius]

A vector differential operator does not a vector make. You cannot use the familiar BAC-CAB identity when one of the operands is not really a vector. Instead, what would you expect the curl of a point dipole to be?

Hmm... I think I would expect it to be zero. But I am not sure. :\