Electromagnetism question (Vector integration).

[MathematicalPhysicist]

Homework Statement

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.

Homework Equations

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?

 

[gabbagabbahey]

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

How exactly are you coming up with this?

Now I develop E(r) around E(0) by taylor expansion: E(r)=E(0)+div(E(0))r+...

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

 

[MathematicalPhysicist]

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

 

[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'.

 

[gabbagabbahey]

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

That looks better!

 

[gabbagabbahey]

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'.

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.

 

[MathematicalPhysicist]

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