Electric dipole in an electric field problem

In summary, the homework statement is that a small object with an electric dipole moment is placed in a nonuniform electric field. The net force on the dipole is F=-pEcos\theta acting in the direction of the increasing field.
  • #1
jhess12
11
0

Homework Statement


A small object with electrc dipole moment [tex]\overrightharpoonup{p}[/tex] is placed in a nonuniform electric field [tex]\overrightarrow{E}[/tex] =E(x)[tex]\hat{i}[/tex]. That is, the field is in the x direction and its magnitude depends on the coordinate x. Let [tex]\theta[/tex] represent the angle between the dipole moment and the x direction. (a) Prove that the net force on the dipole is F=p([tex]\frac{dE}{dx}[/tex])cos[tex]\theta[/tex] acting in the direction of the increasing field.



Homework Equations


U=-pEcos[tex]\theta[/tex]
p[tex]\equiv[/tex]2aq


The Attempt at a Solution



im not asking for anyone to do the problem, but I don't even know where to start. if someone could just please maybe help me understand the problem better and help me get started i would much appreciate it.
 
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  • #2
You have the potential U. The force is the negative of the gradient of the potential. Is that a good starting point?
 
  • #3
so F=-U, so F=pE[tex]cos\theta[/tex], and (dE/dx) is the direction of the electric field as it changes with the coordinate x because the electric field is nonuniform right?
 
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  • #4
The potential is a scalar function. It's -p.E ('.'=dot product). In your case since the directions of the vectors are fixed, you can as you have, write this as -|p||E(x)|cos(theta). So U is a function of x. F is not equal to -U. It's equal to minus the GRADIENT of U. How do you compute a gradient?
 
  • #5
i just don't know. i think that it might be E=[tex]\delta[/tex]V/[tex]\delta[/tex]x
 
  • #6
The gradient of a function U is (dU/dx,dU/dy,dU/dz) (where the derivatives are partial derivatives).
 
  • #7
im sorry, i just don't understand, i see all these equation for W(work)=U and i can see that there should be some way for me to solve this problem because as you have explained it to me, it actually seems very simple, except that i don't know how to put the gradient into the equation. i know that W=-[tex]\int[/tex]F.ds=-[tex]\int[/tex]qE.ds where F and E and ds are vectors but...im sorry
 
  • #8
Ok, so if W=-integral(F*ds) then dW/ds=-F. Or F=-dU/ds. Apply that to this problem. In three dimensions you want to think of a gradient rather than a simple derivative, but if that is driving you crazy, forget about it for now. Treat it as a one dimensional problem, but afterwards think about why dU/dy=0 and dU/dz=0 mean F_y=0 and F_z=0.
 
  • #9
are du/dy=0 and du/dx=0 because they are perpendicular to the electric field? but i think i understand some. and since i am trying to find F_x right? then i would evaluate W=[tex]\int[/tex]pEsin[tex]\theta[/tex]-find the derivative and i would have my answer right? i worked it out except for the p part of the equation. i know that p is a constant so to find the derivative?
 
  • #10
I give up. Can somebody else take this post please? I really give up.
 
  • #11
sorry don't bother
 
  • #12
Given the potential U... the force is:

[tex]\vec{F} = -\bigtriangledown{U}[/tex]

In other words: Fx = -dU/dx. Fy = -dU/dy. Fz = -dU/dz.

Once you know U, you can immediately get the force...

What is U?

As Dick mentions:

[tex]U = -\vec{p}\cdot\vec{E}[/tex]

so

[tex]U = -(pcos(\theta), psin(\theta),0)\cdot (E(x),0,0)[/tex]

so that gives [tex]U = -E(x) pcos(\theta)[/tex]

now you can directly get the components of the force using Fx = -dU/dx. Fy = -dU/dy. Fz = -dU/dz.

the important thing here is that E(x) is only a function of x... it is independent of y and z.

So you should be able to get your result directly using Fx = -dU/dx
 
  • #13
thank you so much, I am sorry i didnt get it before, now that you explain it, it seems so simple like i thought, i just don't know why i didnt get it before. sorry dick.
 
  • #14
jhess12 said:
thank you so much, I am sorry i didnt get it before, now that you explain it, it seems so simple like i thought, i just don't know why i didnt get it before. sorry dick.

Sometimes having somebody else say the same thing makes all the difference. Sorry, I lost patience as well.
 

1. What is an electric dipole?

An electric dipole is a pair of equal and opposite charges separated by a small distance. It is a fundamental concept in electromagnetism and can be represented by two point charges or a distribution of charges.

2. How is an electric dipole affected by an electric field?

An electric dipole experiences a torque when placed in an electric field. This torque causes the dipole to align itself with the direction of the electric field. The magnitude of the torque is directly proportional to the strength of the electric field and the dipole moment.

3. How do you calculate the electric field at a point due to an electric dipole?

The electric field at a point due to an electric dipole can be calculated using the equation E = 1/(4πε0) * (3cos2θ - 1) * (p/r3), where ε0 is the permittivity of free space, θ is the angle between the dipole moment and the line connecting the point to the dipole, p is the dipole moment, and r is the distance between the point and the dipole.

4. Can an electric dipole be in equilibrium in an electric field?

Yes, an electric dipole can be in equilibrium in an electric field if the electric field is uniform and the dipole is aligned with the field. In this case, the torque on the dipole will be zero and it will not experience any rotational motion.

5. How is the potential energy of an electric dipole in an electric field calculated?

The potential energy of an electric dipole in an electric field can be calculated using the equation U = -pEcosθ, where p is the dipole moment, E is the electric field strength, and θ is the angle between the dipole moment and the electric field.

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