Electric dipole in an electric field

  1. 1. The problem statement, all variables and given/known data
    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.



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


    3. The attempt at a solution

    im not asking for anyone to do the problem, but I dont 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.
     
  2. jcsd
  3. Dick

    Dick 25,624
    Science Advisor
    Homework Helper

    You have the potential U. The force is the negative of the gradient of the potential. Is that a good starting point?
     
  4. 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?
     
    Last edited: Oct 5, 2007
  5. Dick

    Dick 25,624
    Science Advisor
    Homework Helper

    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?
     
  6. i just dont know. i think that it might be E=[tex]\delta[/tex]V/[tex]\delta[/tex]x
     
  7. Dick

    Dick 25,624
    Science Advisor
    Homework Helper

    The gradient of a function U is (dU/dx,dU/dy,dU/dz) (where the derivatives are partial derivatives).
     
  8. im sorry, i just dont 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 dont 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
     
  9. Dick

    Dick 25,624
    Science Advisor
    Homework Helper

    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.
     
  10. 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?
     
  11. Dick

    Dick 25,624
    Science Advisor
    Homework Helper

    I give up. Can somebody else take this post please? I really give up.
     
  12. sorry dont bother
     
  13. learningphysics

    learningphysics 4,124
    Homework Helper

    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
     
  14. thank you so much, im sorry i didnt get it before, now that you explain it, it seems so simple like i thought, i just dont know why i didnt get it before. sorry dick.
     
  15. Dick

    Dick 25,624
    Science Advisor
    Homework Helper

    Sometimes having somebody else say the same thing makes all the difference. Sorry, I lost patience as well.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook