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Torque and Dipoles

  1. Oct 28, 2013 #1
    "Problem 4.5 In Fig. 4.6 ##p_{1}## and ##p_{2}## are (perfect) dipoles a distance r apart. What is the torque on ##p_{1}## due to ##p_{2}##? What is the torque on ##p_{2}## due to ##p_{1}##? [In each case I want the toruqe on the dipole about its own centre. If it bothers you that the answers are not equal and opposite, see Prob. 4.29.]" (Introduction to Electrodynamics, 3rd edition by David J. Griffiths; pg. 165)

    Because I have not included Fig. 4.6 I will attempt to describe it (I also think it's safe to guess that most people own a copy of this text aha...).

    ##p_{1}## is on the left pointing upwards and ##p_{2}## is to its right pointing right; they are separated by a distance r.



    Here's what I've done so far:

    ##\tau## = ##p## x ##E##

    ##E_{dip}(r)## = ##\frac{p}{4\pi \epsilon_{o} r^{3}} (2 cos\theta \hat{r} + sin\theta \hat{\theta})##


    To find the toque on 2 from 1 I first find the electric field produced by 1 using the above formula.

    ##E_{2}## = ##\frac{p_{1}}{4 \pi \epsilon_{o} r^{3}} \hat{\theta}## (since the angle between these two vectors is ##\frac{\pi}{2}##.

    Then I can find the torque by apply the above formula and voila!


    The part I'm having difficulty with is that when I have to find the torque on the second dipole to the first. The question asks for the torque about its own centre...and I'm not sure what to do with that; but what I first tried was to "move" the dipoles in vector space and rotate them so that the second was pointing upwards at the origin and then did the same thing I did above. No matter how I think about it the angle ##\theta## in the formula for the electric field is ##\frac{\pi}{2}## (or some integer multiple of it).

    However, the solution manual uses a value of ##\pi## for ##\theta## in calculating this second electric field and I really don't understand why. It certainly results in the torques not being equal and opposite - but I can't grasp it. Any help understanding this would be much appreciated. Thanks a bunch in advance!
     
  2. jcsd
  3. Oct 29, 2013 #2
    It looks like theta measures the angle between the dipole moment (of the source) and the direction of the electric field. So when you go horizontally (perpendicular to the line connecting opposite charges), you use theta = pi/2. But when going parallel, you use 0 or pi, depending on which side you are looking at.
     
  4. Oct 29, 2013 #3
    Ahh, if that's true (that ##\theta## is a measure of the angle between the dipole moment and the electric field I can finally see how this works out. Thanks so much for clearing that up for me!
     
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