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Finding torque on a dipole

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    a perfect dipole (p) is situated a distance z above an infinite grounded plane. The dipole is at an angle theta with the perpendicular to the plane. find the torque on p.

    if you have griffiths text, its problem 4.6
    2. Relevant equations

    E(dipole)= (p/4pi*epsilon.*r^3)*(2cos(theta) r^ +sin(theta) theta^), ^refers to unit vector

    torque = PxE


    3. The attempt at a solution


    Well i cant seem to figure out how to find E.

    The above equation for E doesnt take into consideration the grounded plate, so i thought maybe i need to solve for the potential first. But the potential on the plate would be zero since it is grounded, so I dont really know what to do with that.

    Im not really sure how to approach this honestly, a nudge in the right direction would be appreciated.
     
  2. jcsd
  3. Nov 30, 2011 #2
    Here's hint number one: where is the image dipole? And did you account for it in your equation of E?

    You need to use the method of image charges for this problem, Griffiths explains the topic sufficiently if you need any reference.
     
  4. Nov 30, 2011 #3
    well,

    i think the image problem is as follows.


    a dipole p at z = +z, and a dipole p' at z=-z, where p' makes an angle - theta with the perpendicular to the plane

    To find the torque (N) on p, 1st find V as a function of z, then take the divergence to obtain E, then take the cross product P X E.

    V(z) = (1/4*pi*eps.) * [ ( p' . z^ ) / (2*z)^2 ] , where 2*z is the distance from p' to p, and p'. z^ is the dot product of p' and the z unit vector.

    ill wont bother posting the rest of the calculation, its pretty straight forward from here on.


    However, I'm not fully convinced my image problem is set up correctly.

    My reasoning is that, in a simple scenario, the dipole (a distance z above a grounded conducting plane) is just a + and - charge separated by some distance d.

    To solve for V with an equivalent mirror image distribution, a + charge near this grounded plane would require a - charge located at an equal distance "behind" this plane. Similarly, a - charge would require a + charge at an equal distance behind the plane.

    I sketched this and its not exactly a "mirror image"

    Also, the electric field I obtained is oriented in the +z direction. Since the + charge is closer to the grounded plane, I would expect the electric field to be oriented in the -z direction.



    Im pretty sure my method for finding the torque is correct, my calculated potential must be wrong.
     
  5. Dec 1, 2011 #4
    I have attached a rough (very ugly) sketch of what the mirror charge configuration should look like.

    E=[p/((4pi*e0)*(2z)^3)]*(2costheta(rhat)+sintheta(thetahat) now your p is going to equal what?
     

    Attached Files:

  6. Dec 2, 2011 #5
    the p term in E is the image dipole,



    E= p/(32*pi*z^3*eps.)*(2cos(theta) rhat +sin(theta) thetahat)

    then the torque is N = [sin(2theta)p^2 / (32*pi*eps.*z^3)] xhat and thats the answer

    That diagram helped alot thank you. I previously made the mistake of thinking p points from + to - .


    The question also asks what direction the dipole will to rest.

    The dipole would rotate into the direction of the electric field. But I am not sure if the dipole would ever actually come to rest.

    The dipole will have angular momentum when it lines up with E, so it will continue rotating, stop, and rotate back into the direction of E. I believe this would repeat indefinitely in the absence of some dampening force.

    so i guess p never comes to a complete rest, rather oscillates across the direction of the electric field
     
  7. Dec 3, 2011 #6
    Well first define p - I'll give ya that p=pcostheta(rhat)+psintheta(thetahat)

    So then, N=p x E=((p^2)/(4pie0(2z)^3)*[(costheta(rhat))+(sintheta(thetahat))] X[(2costheta(rhat)+(sintheta(thetahat))]

    The N that I gave you can be simplified further - in its simplified form you will get the direction of N. From there you can figure where how N tends to rotate p.
     
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