Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

When an electric dipole of moment $\mathbf{P}$ is located in a non-uniform electric field $\mathbf{E}$, there is an net force exerted on it.

However, the formula of the force in some books is read $\mathbf{F}=\nabla(\mathbf{P}·\mathbf{E})$, while in other books, it is $\mathbf{F}=(\mathbf{P}·\nabla)\mathbf{E}$. Obviously, the two formula are not the same. So, which one is true?

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Both formulas are equivalent, if you are in the electrostatic approximation and your dipole vector does not depend on the position $\mathbf{r}$.

Let's consider the expression $\mathbf{F}=\nabla_{\mathbf{r}}(\mathbf{p} \cdot \mathbf{E})$ which can be easily obtained from the potential energy function

$U=-\mathbf{p} \cdot \mathbf{E}$

and its relation with the force $\mathbf{F}=\nabla_\mathbf{r} U$. Now, recall the vector identity

$\nabla_\mathbf{r}(\mathbf{a}\cdot \mathbf{b})= (\mathbf{a} \cdot \nabla_\mathbf{r}) \mathbf{b}+(\mathbf{b} \cdot \nabla_\mathbf{r}) \mathbf{a} + \mathbf{a} \times (\nabla_\mathbf{r} \times \mathbf{b})+ \mathbf{b} \times (\nabla_\mathbf{r} \times \mathbf{a})$

for $\mathbf{a}=\mathbf{a}(\mathbf{r})$ and $\mathbf{b}=\mathbf{b}(\mathbf{r})$ two arbitrary vectors. For $\mathbf{p}=\mathbf{a} \neq \mathbf{p}(\mathbf{r})$ [independent of the position] and $\mathbf{b}=\mathbf{E}(\mathbf{r}$) we have

$\nabla_\mathbf{r}(\mathbf{p}\cdot \mathbf{E})= (\mathbf{p} \cdot \nabla_\mathbf{r}) \mathbf{E}+(\mathbf{E} \cdot \nabla_\mathbf{r}) \mathbf{p} + \mathbf{p} \times (\nabla_\mathbf{r} \times \mathbf{E})+ \mathbf{E} \times (\nabla_\mathbf{r} \times \mathbf{p})$

As the dipole vector does not depend on the position we can drop the second and the fourth terms. In the electrostatic approximation, Faraday's law reads $\partial_t \mathbf{B}=\mathbf{0}\Leftrightarrow \nabla_\mathbf{r} \times \mathbf{E}(\mathbf{r})=\mathbf{0} $ [this is known as ''Carn's law''] so that the electric field is irrotational and the curl vanishes. Then we can drop the third term and

$\nabla_\mathbf{r}(\mathbf{p}\cdot \mathbf{E})= (\mathbf{p} \cdot \nabla_\mathbf{r}) \mathbf{E}$

so that your definitions agree.

share|improve this answer
    
Thanks for your answer! But I think the potential energy function $U=-\mathbf{p} \cdot \mathbf{E}$ is only valid in the uniform electric field. How can you prove it for a dipole in the non-uniform field? –  hlew Nov 18 '12 at 9:10
    
Yes, you can use it for non-uniform electric fields as long as the characteristic lengthscale of variation of the latter is bigger than the dipole length, so that the dipole sees smooth changes in the electric field strength and direction. In other words, for the dipole the electric field is locally constant. –  DaniH Nov 18 '12 at 11:23
    
For the dipole the electric field is locally constant, means that the positive charge and negative charge of the dipole feel the same electric field. So that there is no net force! How to explain this? –  hlew Nov 18 '12 at 11:57
1  
Perhaps I was not clear... I am not saying that it has to be constant but rather that the variations of the electric field are smooth. What I have in mind is that you can make a valid Taylor expansion of $\mathbf{E}(\mathbf{r}_\pm)$ around the center of the dipole $\mathbf{r}_0$. Indeed you can derive the expression for the force using this method, too. –  DaniH Nov 18 '12 at 15:31
1  
The expression for the potential energy, nevertheless, is valid for all electric fields although in the books you find the derivation just for the case of constant external electric field. –  DaniH Nov 18 '12 at 15:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.