• Support PF! Buy your school textbooks, materials and every day products Here!

Force on an electric dipole

  • Thread starter physiks
  • Start date
  • #1
101
0
The PE of an electric dipole in an external E-field is
U=-p.E
where p is it's dipole moment.

I was under the impression I could find U, and then easily determine the force on the dipole using F=-∇U, to obtain
Fx=px∂Ex/∂x+py∂Ey/∂x+pz∂Ez/∂x
Fy=px∂Ex/∂y+py∂Ey/∂y+pz∂Ez/∂y
Fz=px∂Ex/∂z+py∂Ey/∂z+pz∂Ez/∂z

however my book annoyingly states that these should be

Fx=px∂Ex/∂x+py∂Ex/∂y+pz∂Ex/∂z
Fy=px∂Ey/∂x+py∂Ey/∂y+pz∂Ey/∂z
Fz=px∂Ez/∂x+py∂Ez/∂y+pz∂Ez/∂z

which they give a derivation for in a different way. However they do go on to prove later that FL=-∂U/∂L with L the direction in question. I'm now very confused. What is wrong with my approach?
 
Last edited:

Answers and Replies

  • #2
64
6
The PE of an electric dipole in an external E-field is
U=-p.E
where p is it's dipole moment.

I was under the impression I could find U, and then easily determine the force on the dipole using F=-∇U, to obtain
Fx=px∂Ex/∂x+py∂Ey/∂x+pz∂Ez/∂x
Fy=px∂Ex/∂y+py∂Ey/∂y+pz∂Ez/∂y
Fz=px∂Ex/∂z+py∂Ey/∂z+pz∂Ez/∂z

What is wrong with my approach?
If [itex]\vec{F}[/itex] = -[itex]\vec{∇}[/itex]U, and U = -[itex]\vec{p}[/itex][itex]\cdot\vec{E}[/itex], then [itex]\vec{F}[/itex] = [itex]\vec{∇}[/itex]([itex]\vec{p}[/itex][itex]\cdot\vec{E}[/itex]).

Now from vector calculus:
[itex]\vec{∇}[/itex]([itex]\vec{p}[/itex][itex]\cdot\vec{E}[/itex]) = ([itex]\vec{p}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{E}[/itex] + ([itex]\vec{E}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{p}[/itex] + [itex]\vec{p}[/itex]x([itex]\vec{∇}[/itex]x [itex]\vec{E}[/itex]) + [itex]\vec{E}[/itex]x([itex]\vec{∇}[/itex]x [itex]\vec{p}[/itex])

Now since the dipole is unbound to an [itex]\vec{E}[/itex] field in free space, we can say [itex]\vec{∇}[/itex]x [itex]\vec{p}[/itex] = 0.
Since this is an electrostatic problem, [itex]\vec{∇}[/itex]x [itex]\vec{E}[/itex] = 0 and we are left with [itex]\vec{∇}[/itex]([itex]\vec{p}[/itex][itex]\cdot\vec{E}[/itex]) = ([itex]\vec{p}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{E}[/itex] + ([itex]\vec{E}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{p}[/itex]
Now ([itex]\vec{E}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{p}[/itex] = (Ex[itex]\frac{∂}{∂x}[/itex] + Ey[itex]\frac{∂}{∂y}[/itex] + Ez[itex]\frac{∂}{∂z}[/itex])[itex]\vec{p}[/itex], which when taking the x-component of [itex]\vec{p}[/itex] is (Ex[itex]\frac{∂}{∂x}[/itex] + Ey[itex]\frac{∂}{∂y}[/itex] + Ez[itex]\frac{∂}{∂z}[/itex])px

Now because there is no bound charge, [itex]\frac{∂}{∂i}[/itex]px = [itex]\frac{∂}{∂i}[/itex]py = [itex]\frac{∂}{∂i}[/itex]pz = 0 where i = x, y, z.

All we have remaining now is [itex]\vec{∇}[/itex]([itex]\vec{p}[/itex][itex]\cdot\vec{E}[/itex]) = ([itex]\vec{p}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{E}[/itex] which is similar to the step we just did, and whose x-component is equal to: (px[itex]\frac{∂}{∂x}[/itex] + py[itex]\frac{∂}{∂y}[/itex] + pz[itex]\frac{∂}{∂z}[/itex])Ex and more generally:

j = x1, y1, z1

Fj = ([itex]\sum[/itex]i pi[itex]\frac{∂}{∂i}[/itex])Ej where i indexes across x,y,z

You're approach assumes the electric field lines in every (x, y, z) contribute to the forces in only the (x1, y1, z1) directions. The second way (correct way) takes the electric field lines corresponding to the direction of force and analyzes how they change through space along that basis.
 
Last edited:
  • #3
101
0
If [itex]\vec{F}[/itex] = -[itex]\vec{∇}[/itex]U, and U = -[itex]\vec{p}[/itex][itex]\cdot\vec{E}[/itex], then [itex]\vec{F}[/itex] = [itex]\vec{∇}[/itex]([itex]\vec{p}[/itex][itex]\cdot\vec{E}[/itex]).

Now from vector calculus:
[itex]\vec{∇}[/itex]([itex]\vec{p}[/itex][itex]\cdot\vec{E}[/itex]) = ([itex]\vec{p}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{E}[/itex] + ([itex]\vec{E}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{p}[/itex] + [itex]\vec{p}[/itex]x([itex]\vec{∇}[/itex]x [itex]\vec{E}[/itex]) + [itex]\vec{E}[/itex]x([itex]\vec{∇}[/itex]x [itex]\vec{p}[/itex])

Now since the dipole is unbound to an [itex]\vec{E}[/itex] field in free space, we can say [itex]\vec{∇}[/itex]x [itex]\vec{p}[/itex] = 0.
Since this is an electrostatic problem, [itex]\vec{∇}[/itex]x [itex]\vec{E}[/itex] = 0 and we are left with [itex]\vec{∇}[/itex]([itex]\vec{p}[/itex][itex]\cdot\vec{E}[/itex]) = ([itex]\vec{p}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{E}[/itex] + ([itex]\vec{E}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{p}[/itex]
Now ([itex]\vec{E}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{p}[/itex] = (Ex[itex]\frac{∂}{∂x}[/itex] + Ey[itex]\frac{∂}{∂y}[/itex] + Ez[itex]\frac{∂}{∂z}[/itex])[itex]\vec{p}[/itex], which when taking the x-component of [itex]\vec{p}[/itex] is (Ex[itex]\frac{∂}{∂x}[/itex] + Ey[itex]\frac{∂}{∂y}[/itex] + Ez[itex]\frac{∂}{∂z}[/itex])px

Now because there is no bound charge, [itex]\frac{∂}{∂i}[/itex]px = [itex]\frac{∂}{∂i}[/itex]py = [itex]\frac{∂}{∂i}[/itex]pz = 0 where i = x, y, z.

All we have remaining now is [itex]\vec{∇}[/itex]([itex]\vec{p}[/itex][itex]\cdot\vec{E}[/itex]) = ([itex]\vec{p}[/itex][itex]\cdot\vec{∇}[/itex])[itex]\vec{E}[/itex] which is similar to the step we just did, and whose x-component is equal to: (px[itex]\frac{∂}{∂x}[/itex] + py[itex]\frac{∂}{∂y}[/itex] + pz[itex]\frac{∂}{∂z}[/itex])Ex and more generally:

j = x1, y1, z1

Fj = ([itex]\sum[/itex]i pi[itex]\frac{∂}{∂i}[/itex])Ej where i indexes across x,y,z

You're approach assumes the electric field lines in every (x, y, z) contribute to the forces in only the (x1, y1, z1) directions. The second way (correct way) takes the electric field lines corresponding to the direction of force and analyzes how they change through space along that basis.
So the vector calculus fits in with the book, but how would I show it by explicitly writing out the components of p.E and then taking a derivative, as I would be more likely to want to use such a method in a specific problem. I can't see why there shouldn't be a way of showing it this way - maybe I'm being silly. The book itself seems to say it should work as FL=-∂U/∂L...

U=-p.E=-pxEx-pyEy-pzEz

FL=-∂U/∂L=... as before
 
Last edited:
  • #4
101
0
Anybody have any answer to my above post?
 
  • #5
64
6
Anybody have any answer to my above post?
The answer is in my previous post...the math is done component wise once the simplifications have been made.
 
  • #6
101
0
The answer is in my previous post...the math is done component wise once the simplifications have been made.
But why can't it be done component wise before making the simplifications?
 
  • #7
64
6
If the problem was not electrostatic and void of bound charges, then the equation would not simplify. You'd have a TON of terms.
 
  • #8
101
0
If the problem was not electrostatic and void of bound charges, then the equation would not simplify. You'd have a TON of terms.
What I think I was trying to say is:
If I have already calculated the potential energy function U(x,y,z) or U(r,θ,ψ), then is applying F=-∇U acceptable? The answer to this is obviously yes.

Therefore If I was given a dipole moment p and and a field E, I could calculate the force by:
- Using the relation F=(p.∇)E directly or
- Calculating U first from U=-p.E and then using F=-∇U
and the second method would be the more logical method I believe.

I think this is what bemused me about the whole situation, as I have been used to deriving forces from known potential functions. The case in my original post however was very generalised.

Thanks for your help :)
 

Related Threads on Force on an electric dipole

  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
11K
Replies
0
Views
11K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
14
Views
3K
Replies
6
Views
3K
Replies
9
Views
4K
  • Last Post
Replies
7
Views
37K
Replies
0
Views
5K
Top