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..

[tex]W_D=-p \cdot E[/tex]

What is the force between them? [itex]F_{1,2}[/itex]

Homework Equations

Force:
[tex]F=-\nabla W_D[/tex]

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

a_r is the unit vector from p1 to p2.

The Attempt at a Solution

Using the following vector identity..
[tex]\nabla (a \cdot b)=(a \cdot \nabla)b+(b \cdot \nabla)a+a×(\nabla × b)+b×(\nabla × a)[/tex]

[tex]-\nabla (-p \cdot E)=\nabla (p \cdot E)=(p \cdot \nabla)E+(E \cdot \nabla)p+p×(\nabla × E)+E×(\nabla ×p)[/tex]

So I know [itex] \nabla × E =0 [/itex] and [itex]E×(\nabla ×p)=(E \cdot p)\nabla-(E \cdot \nabla)p[/itex]

So plugging these in, I get.

[tex]F=(p \cdot \nabla)E+(E \cdot p)\nabla[/tex]

The prof gives the answer as, [tex]F=(p \cdot \nabla)E[/tex]

So that means [itex](E \cdot p)\nabla=0[/itex] But I cannot figure out why.

 

[gabbagabbahey]

Its late and I'm tired a lazy!

Sounds French.

[itex]E×(\nabla ×p)=(E \cdot p)\nabla-(E \cdot \nabla)p[/itex]

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. :\