# Force on an electric dipole

1. Dec 11, 2012

### aftershock

1. The problem statement, all variables and given/known data

Two electric dipoles, oriented as shown in the figure, are separated by a distance r.

What is the force on p1 due to p2

2. Relevant equations

F = (p dot ∇)E

3. The attempt at a solution

I'm confused on the p dot ∇ part.

How is the divergence of p not just zero?

It seems p1 would just be some constant in the x hat direction.

2. Dec 11, 2012

### diazona

Think about that carefully. $\mathbf{p}\cdot\nabla$ is not the divergence of $\mathbf{p}$!

3. Dec 12, 2012

### aftershock

I'm not sure I understand that. ∇⋅p would be divergence? And dot products are commutative.

4. Dec 12, 2012

### diazona

This isn't really a dot product, though, it's a dot "application." That is, it still means that $\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$, but you can't assume that e.g. $a_x b_x$ means "$a_x$ multiplied by $b_x$," because you are dealing with things that don't get multiplied. Think about it: does $x \frac{\partial}{\partial x}$ mean $x$ multiplied by $\frac{\partial}{\partial x}$? Does $\frac{\partial}{\partial x} x$ mean $\frac{\partial}{\partial x}$ multiplied by $x$?

5. Dec 12, 2012

### aftershock

Oh ok so if the vector is in front of del i'm basically taking the components and multiplying them by differential operators, where if the vector was to the right of del i would be taking derivatives of the components?

6. Dec 12, 2012

### diazona

Yep, that's the idea.

A good way to think about it is that the components of $\mathbf{p}$ are multiplicative operators. Just like the differential operator $\frac{\partial}{\partial x}$ takes a function $f(x)$ and turns it into $f'(x)$, a multiplicative operator $x$ takes a function $f(x)$ and turns it into $xf(x)$. The composition of two operators is also an operator, so $p_x \frac{\partial}{\partial x}$ is an operator that means "take the derivative and then multiply by a number."

7. Dec 12, 2012

### aftershock

Thanks, really helped me.