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Torque on a dipole near a infinite conducting plane

  1. Aug 5, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-8-5_14-44-17.png upload_2017-8-5_14-44-32.png

    2. Relevant equations


    3. The attempt at a solution
    upload_2017-8-5_16-31-24.png

    I have drawn this diagram using MS paint, could you please tell me some other software in which I can draw and insert greek symbols, too ?


    Let me take the origin at O.
    ## \hat r \left(\alpha\right) ≡ \hat r\text{ at angle }\alpha ~with ~z - axis ##
    Considering the image problem,
    ## \vec E \left (O' \right) ## due to the image dipole ## = \frac { -kp} { {2 z}^3 } \left( 2 \cos \alpha ~ \hat r \left( \theta = 0 \right) + \sin \alpha ~ \hat \theta \left ( \theta =0 \right) \right )##
    ##
    \\ \hat r = \sin \theta ~ \hat s + \cos \theta ~ \hat z
    \\ \hat \theta = - \sin \theta ~ \hat z +\cos \theta ~ \hat s
    \\ \text {here } \hat s\text{ is the cylindrical coordinate unit vector, I don't know its name, please tell me this so if it has a name. }
    \\\text { at } \theta =0, ~ \hat r = \hat z ~ and ~ \hat \theta = \hat s

    \\
    \\ \vec p = p~ \hat r \left ( \alpha \right)##

    Now,
    torque ## \vec N = \vec p \times \vec E ##

    ## = p ~ \hat r ## ## \left( \alpha \right ) \times ## ## \frac { -kp} { {2z}^3 }
    \left( 2 \cos \alpha ~ \hat r \left( \theta = 0 \right) + \sin \alpha ~ \hat \theta \left ( \theta =0 \right) \right) ##

    ##\\ = p \hat r\left( \alpha \right ) \times

    \frac { -kp} {{2z}^3 } \left( 2 \cos \alpha ~\hat z + \sin \alpha ~ \hat s \right) ##

    ## = \frac {-kp^2}{ \left (2z\right) ^3 } \left[ - 2 \cos \alpha \sin \alpha ~\hat \phi + \cos \alpha \sin \alpha ~ \hat \phi \right ]
    ##

    ## \\ = \frac {kp^2}{16 z ^3 } \sin \alpha ~ \hat \phi ##

    To answer the final part,
    I guess the orientation of plane is x-y and ##\hat \phi ## is also in the x-y orientation. So, the dipole will come to rest in the orientation which is parallel to the plane .
    But why will it come to rest? Why won't it oscillate (as there are no other net force acting on the dipole)?
     

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    Last edited: Aug 5, 2017
  2. jcsd
  3. Aug 5, 2017 #2

    TSny

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    I had a difficult time following all of your unit vector notations. I believe your unit vector ##\hat{s}## could be called the "radial polar unit vector" in cylindrical coordinates.
    Are you sure both terms in the parentheses are positive? There should be a set of parentheses in the denominator (which you finally put in near the very end of your calculation.)

    Did you mean to type ##2 \alpha## for the argument of the sine function? I don't get the same numerical factor out front due to the possible sign error that I mentioned above.

    Yes, it would oscillate. I guess they want you to assume small damping so that it comes to rest at the equilibrium position. I agree that the equilibrium position would be parallel to the conducting plane.
     
  4. Aug 5, 2017 #3
    ##\vec E ## due to an ideal dipole \vec p at a point ## \hat r \left( \theta \right ) ## is given by

    upload_2017-8-5_23-4-9.png

    Here, r = 2z
    ## \theta = \alpha##
    The electric field at O' due to p ##= \frac { kp} { \{2 z\}^3 } \left( 2 \cos \alpha ~ \hat r \left( \theta = 0 \right) + \sin \alpha ~ \hat \theta \left ( \theta =0 \right) \right ) ##
    So, The electric field at O' due to -p ##= \frac { -kp} { \{2 z\}^3 } \left( 2 \cos \alpha ~ \hat r \left( \theta = 0 \right) + \sin \alpha ~ \hat \theta \left ( \theta =0 \right) \right ) ##

    Is there anything wrong with it?
     

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  5. Aug 5, 2017 #4

    TSny

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    Now you're using spherical coordinates.

    This is confusing. You say ##\theta = \alpha## but then you say ##\theta = 0## and ##\alpha \neq 0##.

    There is a nice way to write the electric field of a dipole that is independent of coordinate system. See equation 13 here:
    http://phys.columbia.edu/~nicolis/Dipole_electric_field.pdf

    You can apply this expression to any coordinate system that you find convenient. Thus, you can apply it to this picture to find E at O' in terms of unit vectors ##\hat y## and ##\hat z##.
    upload_2017-8-5_13-17-30.png

    You can then switch the sign of E to correspond to p pointing in the opposite direction to that shown in this picture.
     
  6. Aug 5, 2017 #5
    upload_2017-8-5_16-31-24-png.png ##
    \vec N = \vec p \left ( ~at ~ O' \right ) \times

    \{ ## ## {\vec E \left ( ~at ~ O' \right )} = { \frac k { \{ 2 z \}^3 } }## ## \left [ 3 { \left ( - \vec p ⋅ \hat z \right ) \hat z } +\vec p \right ] ## ## \}##
    ##\\ = \frac {- 3 k p^2} { \{ 2 z \}^3 } \cos \alpha \{ \vec p \times \hat z\}
    \\ = \frac {- 3 k p^2} {16 z^3} \sin 2 \alpha \{- \hat \phi \} ##

    The book's solution says that the equilibrium position would be perpendicular to the conducting plane or did I iterpret it wrong?



    upload_2017-8-6_8-53-59.png
    upload_2017-8-6_8-54-18.png
     
  7. Aug 6, 2017 #6

    TSny

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    Sorry, it was my mistake. I was not orienting the image dipole correctly which led to a sign error on my part. The image dipole is not oriented as shown in your picture (post #5) but as shown in part (a) of the solution of your last post. The expressions for E and N given in the solution look correct. And the direction of the torque is such as to make the equilibrium orientation of the dipole perpendicular to the conducting plane:

    equilibrium is ##\uparrow## if initially ##-\pi/2 < \theta < \pi/2##
    equilibrium is ##\downarrow## if initially ##\pi/2 < \theta < 3\pi/2##

    where ##\theta## is as shown in the figure for part (a) solution.

    Hope I didn't cause you too much of a headache.
     
    Last edited: Aug 6, 2017
  8. Aug 6, 2017 #7
    The image dipole given in the book's solution is not -p. I was taking -p as image dipole so that the potential due to p at the plane will get cancelled by potential due to -p and hence we will get the potential on the space containing the plane to be 0.
    What is wrong with this argument?
     
  9. Aug 6, 2017 #8

    TSny

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    In your diagram, the image dipole looks to me to be parallel to the real dipole. But they are not parallel (in general). See the figure in the solution part (a).
     
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