# Electromagnetism question (Vector integration).

1. Nov 11, 2009

### MathematicalPhysicist

1. The problem statement, all variables and given/known data
Show that the force caused by external electric field on a charge distribution is given by:
$$F=qE(0)+(p\cdot \nabla)E(0)+...$$ (it's enough to show the first two terms), where E(0) is the electric field in the origin which we choose to develop the expression around it.
p is the electric dipole.

2. Relevant equations

3. The attempt at a solution
So, $$F=\int d^3 r E(r)\rho(r)$$, from integration by parts I get:
$$F= r \rho(r) E(r) -\int r[\nabla(\rho(r)) E(r) +div(E(r))\rho(r)]d^3 r$$
Now I develop E(r) around E(0) by taylor expansion: $$E(r)=E(0)+div(E(0))r+...$$, so div(E(r))=div(E(0))+...
Now the first term in F is zero by boundary conditions, Now I know that $$p=\int d^3 r \rho(r) r$$, but if I plug div(E(r))=div(E(0))+... to the third term I get something else than the second term in the question, the first term I do get, because: $$\int r \nabla(rho) d^3 r=r\rho(r) - \int \rho(r) d^3r$$ and plugging it back and inserting E(r)=E(0)+div(E(0))r... I get that the third term is: div(E(0))p, and the second term is -2div(E(0))p+-qE(0), so in summary I get: qE(0)+p.div(E(0))+...
But it's not the same as what I need to show here, but I don't see where did I got this wrong, can someone enlight me?

2. Nov 11, 2009

### gabbagabbahey

How exactly are you coming up with this?

You also seem to be using the wrong formula for the Taylor expansion of a vector field...where are you getting this from?

3. Nov 11, 2009

### MathematicalPhysicist

Ok, I think I got my mistake, the taylor expansion should be:
$$E(r)=E(0)+r\cdot\nabla E(0)+...$$
correct?

4. Nov 11, 2009

### MathematicalPhysicist

For the first question, I guess I take $$u=E(r)\rho(r)$$ and $$v=r$$ in integration by parts, where I use (uv)'=u'v+uv'.

5. Nov 11, 2009

### gabbagabbahey

That looks better!

6. Nov 11, 2009

### gabbagabbahey

No, integration by parts doesn't work the same way in 3 dimensions as it does in one. You are trying to compute a volume integral (over $d^3\textbf{r}$), not just an integral over $r$.

In vector Calculus, there are actually six product rules and none of them help you here.

7. Nov 12, 2009

### MathematicalPhysicist

OK, thanks I see how to solve it now, it's quite immediate.