Electromagnetism question (Vector integration).

In summary, 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.
  • #1
MathematicalPhysicist
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Homework Statement


Show that the force caused by external electric field on a charge distribution is given by:
[tex]F=qE(0)+(p\cdot \nabla)E(0)+...[/tex] (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, [tex]F=\int d^3 r E(r)\rho(r)[/tex], from integration by parts I get:
[tex]F= r \rho(r) E(r) -\int r[\nabla(\rho(r)) E(r) +div(E(r))\rho(r)]d^3 r[/tex]
Now I develop E(r) around E(0) by taylor expansion: [tex]E(r)=E(0)+div(E(0))r+...[/tex], so div(E(r))=div(E(0))+...
Now the first term in F is zero by boundary conditions, Now I know that [tex]p=\int d^3 r \rho(r) r[/tex], 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: [tex]\int r \nabla(rho) d^3 r=r\rho(r) - \int \rho(r) d^3r[/tex] 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?
 
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  • #2
MathematicalPhysicist said:
So, [tex]F=\int d^3 r E(r)\rho(r)[/tex], from integration by parts I get:
[tex]F= r \rho(r) E(r) -\int r[\nabla(\rho(r)) E(r) +div(E(r))\rho(r)]d^3 r[/tex]

How exactly are you coming up with this?:confused:

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

You also seem to be using the wrong formula for the Taylor expansion of a vector field...where are you getting this from?
 
  • #3
Ok, I think I got my mistake, the taylor expansion should be:
[tex]E(r)=E(0)+r\cdot\nabla E(0)+...[/tex]
correct?
 
  • #4
For the first question, I guess I take [tex]u=E(r)\rho(r)[/tex] and [tex]v=r[/tex] in integration by parts, where I use (uv)'=u'v+uv'.
 
  • #5
MathematicalPhysicist said:
Ok, I think I got my mistake, the taylor expansion should be:
[tex]E(r)=E(0)+r\cdot\nabla E(0)+...[/tex]
correct?

That looks better!:smile:
 
  • #6
MathematicalPhysicist said:
For the first question, I guess I take [tex]u=E(r)\rho(r)[/tex] and [tex]v=r[/tex] 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 [itex]d^3\textbf{r}[/itex]), not just an integral over [itex]r[/itex].

In vector Calculus, there are actually six product rules and none of them help you here.
 
  • #7
OK, thanks I see how to solve it now, it's quite immediate.
 

What is electromagnetism?

Electromagnetism is a branch of physics that deals with the study of the interactions between electrically charged particles. It includes the study of electric fields, magnetic fields, and how they are related to each other.

What is a vector?

A vector is a mathematical quantity that has both magnitude (size) and direction. In electromagnetism, vectors are commonly used to represent electric and magnetic fields.

What is integration?

Integration is a mathematical process that involves finding the area under a curve, or the sum of infinitesimal parts of a function. In the context of electromagnetism, integration is used to calculate the total electric or magnetic field at a given point.

How is vector integration used in electromagnetism?

Vector integration is used in electromagnetism to calculate the total electric or magnetic field at a given point by integrating the contributions from all the individual elements of the field. This is particularly useful in cases where the field is not constant or has complex geometries.

What are some applications of electromagnetism?

Electromagnetism has many practical applications, including electric motors, generators, transformers, and telecommunications. It also plays a crucial role in technologies such as MRI machines, power plants, and electric vehicles.

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