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Electric dipole in Uniform Electric Field (3D)

  1. Mar 2, 2017 #1
    1. The problem statement, all variables and given/known data
    An electric dipole with magnitude ##p = 0.2Cm## is placed inside a uniform electric field of ##\vec{E} = 100\vec{i} + 70\vec{j} + 40\vec{k} \frac {N} {C}##. The dipole was initially pointing along the +x direction. You then start to rotate it first on xz-plane towards the z-axis. After it is aligned with the z-axis, you rotate it on the yz-plane towards the y-axis. At the final moment, the dipole is aligned parallel to the +y axis.
    (a) What is the total work done by you during this process?
    (b)In the final orientation, the electric field applies a torque ~ τ on the dipole. As you want to keep the dipole in this orientation, you apply another torque ~ τhand to prevent it from rotating. Find both torques.

    2. Relevant equations
    ##\vec {τ}=\vec {p}x\vec {E}##
    ##W=τΔθ##


    3. The attempt at a solution
    Our dipole moment is ##\vec {p}=(0,2\vec{i})## and ##\vec{E} = 100\vec{i} + 70\vec{j} + 40\vec{k} \frac {N} {C}## is so
    ##\vec{τ}=\vec {p} x \vec {E}=(-8\vec{j}+14\vec{k})## and the magnitude is ##16.1Nm##
    Its uniform electric field so I think we dont need an integral.Or the integral becomes
    ##W=τΔθ##
    In the first rotation The electric field wlll gonna rotate the dipole.But I think we need an extra torque to make it turn just 90 degree so I am little bit confused after here.I think I dont have enough info to calculate the torque done by me or work ?
     
  2. jcsd
  3. Mar 2, 2017 #2

    kuruman

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    Your use of the work equation is incorrect. The torque ##\vec{p} \times \vec{E}## is not constant because the dipole moment changes direction relative to the field. Remember that the electric force is conservative, so consider using the change in potential energy.
     
  4. Mar 2, 2017 #3
    But we define potantial energy over work right ? Yeah you are right ##\veac {p}## will change with time.
     
  5. Mar 2, 2017 #4

    kuruman

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  6. Mar 2, 2017 #5
  7. Mar 2, 2017 #6

    kuruman

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    That does not look correct. Can you show me how you got that number?
     
  8. Mar 2, 2017 #7
    Yeah I saw that...I am tired I did a lot of physics today..I ll reply when I feel I can conti.
     
  9. Mar 2, 2017 #8

    kuruman

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    No problem. Get some rest and clear your head.
     
  10. Mar 3, 2017 #9
    12 j ?
     
  11. Mar 3, 2017 #10

    kuruman

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    Better, but still not correct. How did you get this number?
     
  12. Mar 3, 2017 #11
    ##U=\vec {-p}⋅\vec{E}## First position gave me ##U_i=-20J## and ##U_f=-8## so ##ΔU=12J##
     
  13. Mar 3, 2017 #12

    kuruman

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    I agree with the Ui but not with Uf. It looks like you assumed that the dipole ends up aligned parallel to the +z-axis instead of the +y axis as the problem statement specifies.
     
  14. Mar 3, 2017 #13
    I was trying to calculate just firstt rotation.I never thought second rotation
     
  15. Mar 3, 2017 #14
    ##ΔU=6j##
     
  16. Mar 3, 2017 #15

    kuruman

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    Correct. Now what is total work done by the external agent (you) in this case?
     
  17. Mar 3, 2017 #16
  18. Mar 3, 2017 #17
    Wait its 6 j again
     
  19. Mar 3, 2017 #18

    kuruman

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    Yes, - 6 J (the negative of ΔU) is the work done by the electric force. The work done by you is the negative of that, therefore equal to ΔU.
    Your answer to part (b) is not OK. In the final orientation the dipole moment is in the y-direction. Your torque shows a component in the y-direction and that cannot be correct. You need to redo the cross product.
     
  20. Mar 3, 2017 #19
    Then its ##0.2jxE=τ## ? İts the applied torque by electrric field.The work done by hand will be equal to The rotation energy minusThe work done by electrci field ?
     
  21. Mar 3, 2017 #20

    kuruman

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    There is no work involved in part (b). You are holding the dipole with your hand fixed in space (in equilibrium). There is a ##\vec{\tau}_{elec.}=\vec{p} \times \vec{E}##, the torque exerted by the electric field. That's one torque you have to find. The other torque you have to find is ##\vec{\tau}_{hand}##, the torque exerted by your hand.
     
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