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Potential Energy for dipole in Electric Field

  1. May 28, 2016 #1
    1. The problem statement, all variables and given/known data

    To derive Potential Energy for dipole p in Electric Field E.

    2. Relevant equations


    Potential Energy is the work done by the external agent in turning the angle of the dipole from the U=0 position to another position against the influence of the electric field applied right?

    3. The attempt at a solution

    So if Torque exerted by field for a particular θ is given by $$ \tau = pE\sin\theta $$
    then when working out potential energy, should we not take the following:
    τapp will act in same sense as dθ and opposite sense of τfield right?
    So $$ \tau_{app} = - pE\sin\theta $$
    And the potential energy is just
    $$ U = \int_{\theta_1}^{\theta_2}\tau_{app}\,d\theta $$
    $$ U = \int_{\theta_1}^{\theta_2}-pE\sin\theta\, d\theta $$
    $$U=-pE(-\cos(\theta_2)+\cos(\theta_1))$$
    Now if θ1 =π/2 and θ2
    $$U=pE\cos(\theta)$$
    $$U=p\cdot E$$
    But the traditional derivation outputs $$-p\cdot E$$ and takes τfield and not τapp in the first step. Why is this the case?
     
  2. jcsd
  3. May 29, 2016 #2

    ehild

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    The sign error is caused by the controversial signs of the angle and torque.
    upload_2016-5-29_8-44-54.png

    You see that the torque of the electric field causes decreasing angle, clockwise (negative) direction of rotation of the dipole. So τ(field) = -p x E.
     
  4. May 29, 2016 #3
    But isnt it only in the work integral that it becomes a - wrt +dθ? because Wfield=∫τfielddθcos π?
     
  5. May 29, 2016 #4

    ehild

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    What do you mean?
    The electric field tends to align the dipole, decreasing the angle θ. The "natural" processes tend to make the potential energy minimum. If we set the zero of the potential energy at angle θ=pi/2, the potential energy is negative at θ=0.
     
  6. May 29, 2016 #5
    0Hq2tdM.png
    τapp is -p x E will mean that it points out of the plane in this figure right? But how can that be? How do I derive potential energy in this scenario? the violet arrow is the dipole moment vector p. :cry: I know im missing something big but i am not able to figure out why..
     
  7. May 29, 2016 #6

    ehild

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    The electric field turns the dipole in the opposite direction as in your figure, clockwise. So the torque points into the plane. Calculate the potential energy as you did in the first post, only the sign of torque was wrong.
    upload_2016-5-29_16-16-54.png
     
  8. May 29, 2016 #7
    Then i'd be getting the work done by the field right? Im trying to get work done by external agent in turning the dipole as in the figure in my previous post.. The turning i indicated was that produced by external torque..this work is stored in the system right?
     
  9. May 29, 2016 #8

    ehild

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    You need a positive external torque to turn the dipole by an angle dθ.
     
  10. May 29, 2016 #9
    since τfield= p x E is intrinsically negative(?) i put the negative sign in front of it to make it positive in τapp..
     
  11. May 30, 2016 #10

    ehild

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    upload_2016-5-30_7-15-59.png
    The electric field tends to rotate the dipole in the direction of the green arrow, clockwise, to make it aligned with the field. To move the dipole by the red angle dθ, the external agent has to exert a torque, which would rotate the dipole in anti-clockwise direction, according to the thin blue arrows. The initial angle of the dipole with respect to the electric field is shown by the blue angle.
    The elementary work of the external agent is dW=PEsin(θ)dθ, positive, as the displacement and force have the same sign. ( Both the angular displacement and the torque are positive). The work between θ1 and θ2 is equal to the potential difference U(θ2)-U(θ1) = -PE(cos(θ2)-cos(θ1)). If θ1=0 and θ2=pi/2, U(π/2)-U(0) = -PE(cos(π/2)-cos(0)). What is U(0) if U(π/2)=0?
     
  12. May 30, 2016 #11
    Ah so I have been confusing angles and magnitudes.. τapp.dθ has to be positive yes! :O
    U(θ2)-U(θ1)=-pE(cos θ2-cosθ1)
    so if we take U(π/2)=0, then U(0)=-pE.
    Is this correct?
     
  13. May 30, 2016 #12

    ehild

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    yes.
     
  14. May 30, 2016 #13
    Thanks very much sir! :D
     
  15. May 30, 2016 #14

    ehild

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    You are welcome.:smile:
     
  16. May 30, 2016 #15
    This is where you are getting it wrong .

    The confusion you are having is that even though $$ \vec{\tau}_{app} = - \vec{p} \times \vec{E} $$ , $$W_{app} = U = \int (pEsin\theta) d\theta $$ .
     
  17. May 30, 2016 #16
    There should be no minus sign . Except this , I agree with everything you have written :smile: .
     
  18. May 30, 2016 #17

    ehild

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    Yes, τ(field) = p x E, but with the angle shown, it is -pEsin(θ).
    The vector product uses the sin of the angle the second vector encloses with the first, anti-clockwise. In this problem, we used the angle the first vector (p) made with second (E). In this way, the field produced negative torque.
     
  19. May 30, 2016 #18
    If by putting a minus sign you are representing a clockwise torque ,then it is alright . But if the - sign has something to do with the way angle is measured , then I do not understand .

    o_O Why complicate things ?

    The cross product uses the smaller of the two angles between the vectors when joined from tail to tail .
     
  20. May 30, 2016 #19

    ehild

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    Yes, calculating the magnitude of the torque. But from τ = pxE, it does not follow that τ=pEsinθ, where 0 ≤ θ ≤ π. The torque has sign, in addition that its direction is perpendicular to both p and E, which has to be taken into account when calculating work.
     
    Last edited: May 30, 2016
  21. May 30, 2016 #20
    I believe that the sign of torque is not required for calculating work because work ( dot product ) requires magnitude of torque . All that matters is that whether the torque tries to increase or decrease the angle . When calculating work , the minus sign is placed when torque tries to decrease the angle between the vectors .
     
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