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Homework Help: Deriving the potential energy of a dipole

  1. Aug 5, 2017 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    In case of uniform E, torque ##\vec N = \vec p \times \vec E
    \\ U = \int_0 ^ \theta N d \theta = pE \int_0 ^ \theta \sin \theta d \theta = - pE \cos \theta
    \text { Here, I assumed that the direction of } \vec N ~ and ~ \vec d \theta ~ are ~ same .
    \\ U = - \vec p \dot \vec E ##

    How to show the same for non - uniform E ?
  2. jcsd
  3. Aug 5, 2017 #2


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    What about the lower limit of the integration?

    An "ideal" dipole has negligible length.
  4. Aug 6, 2017 #3
    Sorry, I took the wrong lower limit of the integration.

    If the dipole moment and the applied electric field are co - linear initially , then the torque on it will be zero. So, the potential energy will not change.

    Actually I got confused by the wordings of the problem.

    Change in the potential energy of a dipole in the presence of external electric field is given as
    ## U_f - U_i = \int_{\theta _i } ^ {\theta_f} N d \theta = pE \int_{\theta _i } ^ {\theta_f} \sin \theta d \theta = - pE\{ \cos \theta_f - \cos \theta _i \} ##

    I guess when it is said potential energy of the dipole, this potential energy corresponds to U_f with U_i = 0 J.
    For U_i to be 0 , I have to take ## \theta _i = 90° ##.

    Hence, the potential energy of the dipole ## U = - \vec p⋅ \vec E ##

    If I just come and see a dipole whose p makes an angle θ with E , I will say that the energy of the dipole is -pE cosθ. While saying this,it is assumed that initially the dipole's orientation was such that p made an angle 90° with E; and due to the torque due to E, the dipole got rotated in the final position.

    Now consider the following case,
    Initially , there was no applied E. Later I switch on the E such that E and p are co-linear . Now, using the formula ## U = - \vec p⋅ \vec E ##, I will calculate its energy to be -pE, while in reality, it is 0. So, my calculation will give the wrong result, Right?

    Does it imply that what we calculate for uniform field could be applied to non - uniform field in case of ideal dipole?
    Then, in case of an ideal dipole in the presence of non -uniform field , the net force on it be 0. Isn't it?
  5. Aug 6, 2017 #4


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    Yes, that's the convention. You could choose some other angle to correspond to zero U, but then you would need to carry around an added constant in the formula for U. But it would not be wrong to do this.

    I wouldn't say that you must conclude that in reality it is 0. After the field is turned on, you have the freedom to define which orientation corresponds to zero potential energy. The potential energy just corresponds to work required to rotate the dipole from some arbitrarily chosen zero-potential-energy orientation.

    An ideal dipole has infinitesimal length (dr) but "almost infinite" values of the charges (±q) so that the product qdr is some finite value.

    When calculating the torque, you are dealing with force time lever arm. So, the torque is of the order qEdr. Note, qE is huge while dr is infinitesimal. But the product (qE)(dr) can turn out to be some reasonable finite number. If you worried about the variation of E in a nonuniform field, then that would add a correction of order q(∂E/∂r dr) dr. The quantity in parentheses gives the variation of E over the length of the dipole. But now you have two infinitesimals dr appearing in the expression, so the overall quantity is negligible. Thus, you can neglect the non-uniformity of E when finding the torque.

    I'll leave it to you to do a similar argument for the net force on the ideal dipole in a non-uniform E. Show that the effect of the non-uniformity of E is not negligible now.
  6. Aug 7, 2017 #5
    To illustrate this, let's follow this convention in which the P.E. of the dipole is 0 at ## \theta = 45 \degree ##
    Then, for ## \theta_i = 90 \degree ##

    ## U_f - U_i = \int_{\theta _i } ^ {\theta_f} N d \theta = pE \int_{\theta _i } ^ {\theta_f} \sin \theta d \theta = - pE\{ \cos \theta_f - \cos \theta _i \} = - \vec p⋅ \vec E##
    But this doesn't give ## U = - \vec p⋅ \vec E##
    ## U_f - U_{45\degree } = - pE\cos {\theta _f} + pE{ \cos 45\degree }##
    Now taking ## U_{45\degree } = 0 ## gives,

    ## U = - \vec p⋅ \vec E + pE{ \cos 45 \degree }##

    How to write degree in Latex?
  7. Aug 7, 2017 #6
    O.K. Now, I understood the point.
    Let's say that the arbitrarily chosen zero potential energy orientation is ## \theta_0 ##.
    Now, P.E. of the dipole is given by ## U = - \vec p⋅ \vec E + pE \cos{\theta_0 } ##
    While the first term corresponds to that orientation of the dipole in which its potential energy is calculated.
    It doesn't matter whether the dipole has got rotated in the presence of the applied field or not.
    The work done by the electric field on the dipole is 0, not the potential energy of the dipole.

    While calculating P.E ( taking ## \theta_0 = 90\degree ##), what I did was

    ##U_f - U_ i = \int_{\theta _i } ^ {\theta_f} N d \theta = pE \int_{\theta _i } ^ {\theta_f} \sin \theta d \theta = - pE\{ \cos \theta_f - \cos \theta _i \} = - \vec p⋅ \vec E ##
    The right side is the work done by the torque due to the Electric field. While the change in potential energy is defined as the negative of the work done by the conservative force ( in general).

    So, I think I have done some mistake in calculating the R.H.S. Will you please point it out?
    Last edited: Aug 7, 2017
  8. Aug 7, 2017 #7


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    I just use ^o. Thus, ##45^o##.
    With a quick search, I found ^{\circ}. Thus, ##45^{\circ}##.
  9. Aug 7, 2017 #8


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    Is it the sign that you are worrying about? It can be tricky.


    In the figure above, the positive direction for θ is counterclockwise from the electric field direction.
    So, positive torque would be counterclockwise. You can see that the torque on the dipole is clockwise. So, the torque N is negative and should be written N = -pEsinθ. The work done for rotation by dθ is Ndθ = -pEsinθdθ.

    dU is the negative of the work done. So, dU = -(-PEsinθ)dθ = +pEsinθdθ. So, your calcuation for Uf - Ui was correct!

    Attached Files:

  10. Aug 7, 2017 #9
    I was not clear about the direction of ## \theta ##. Thanks for pointing it out.
    Work done by the electric force is defined as W = ##\int_{\theta _i}^ {\theta _f} \vec N⋅ d \vec \theta ##
    Now, ## \vec N = \vec p \times \vec E = pE \sin \theta ~ \left (- \hat n \right)##
    ## d\vec \theta = d \theta \left ( \hat n \right),~~~ d \theta = - |d \theta| ##
    The clockwise direction of ##d\vec\theta ## is taken care of by the limits of the integration.
    So, W = ##\int_{\theta _i}^ {\theta _f} -pE \sin \theta d \theta ##
    U_f - U_i = - W = ##\int_{\theta _i}^ {\theta _f} pE \sin \theta d \theta = ## ##\int_{\theta _i}^ {\theta _f} N d \theta ##

    So, ##\int_{\theta _i}^ {\theta _f} N d \theta ## is the work done against the electrostatic torque, not work done due to the electrostatic torque.
  11. Aug 8, 2017 #10


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    I believe I follow up to here.
    This would be true for clockwise rotation.
    OK. In the last equality you are apparently taking the symbol ##N## to equal ##pE \sin \theta##. I was taking ##N## to represent the amount of counterclockwise torque. So, since the torque is clockwise, I took ##N## to be ##-pE \sin \theta##. As long as we know how we're interpreting the symbols, we're OK.

    Yes, I agree (if ##N## is as you've defined it).
  12. Aug 8, 2017 #11
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