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Potential energy of a dipole

  1. Aug 29, 2015 #1
    Hi, friends! I read that the torque exerced by a uniform electric field ##\mathbf{E}## on a dipole with moment ##\mathbf{p}## is ##\boldsymbol{\tau}=\mathbf{p}\times\mathbf{E}##. Then the book, Gettys' Physics 2, explain that the work made to rotate the dipole around a fixed point is

    ##\int_{\theta_1}^{\theta_2}\tau d\theta=\int_{\theta_1}^{\theta_2}pE\sin\theta d\theta=-pE\cos\theta_2-(-pE\cos\theta_1)=\Delta U##
    and therefore potential energy can be defined as ##U=-\mathbf{p}\cdot\mathbf{E}##.

    Well, there is a major obstacle to my comprehension of that: I think that ##\int_{\theta_1}^{\theta_2}\tau d\theta=\int_{\theta_1}^{\theta_2}pE\sin\theta d\theta## (where I think ##\theta## to be the oriented angle from ##\mathbf{p}## to ##\mathbf{E}##) is the work done by the electric fied while the dipole rotates from an angle ##\theta_1## with the direction of ##\mathbf{E}## to an angle ##\theta_2## with ##\mathbf{E}##, but I know the definition of the variation of potential energy ##\Delta U## as the opposite of the work done by a conservative force field: ##\Delta U=-W_{\text{conservative}}##.

    Or isn't ##\int_{\theta_1}^{\theta_2}\tau d\theta## the work done by the electric field? I would think that it is, because the work done by a force ##\mathbf{F}## to move along the circumference ##\gamma## parametrized by ##\mathbf{r}:[0,2\pi]\to\mathbb{R}^3## is ## \int_{\theta_1}^{\theta_2} \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(\theta)d\theta##. Here ##\mathbf{F}\cdot \mathbf{r}'=F_t R## where ##F_t## is the tangential component of the force (positive if and only if counterclockwisely oriented) and ##R## the distance of the moved object from the centre of the circle and, if I correctly understand, ##RF_t## precisely is the torque ##\tau_z## with respect to the centre of rotation of a rotating body, therefore the work done by the forces acting on the rotating body are ##W_{\text{tot}}=\int_{\theta_1}^{\theta_2} \sum\tau_z d\theta##. Or am I wrong?

    I heatily thank you for any answer!​
     
  2. jcsd
  3. Aug 29, 2015 #2
    I have understood: in the formula for ##\Delta U## the angle ##\theta## is from ##\mathbf{E}## to ##\mathbf{p}##, i.e ##\tau## is the opposite of the torque ##\tau_z##.
     
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