While deriving the formula for potential energy of electric dipole I almost every time see that while the torque was demanding the dipole to rotate in one direction , we let the dipole to rotate in opposite direction by some other force .Find work done by integral and then find potential energy by $W=-∆U$

Why don't we let the dipole rotate in direction of torque caused by electric field Find work and then potential energy.

I tried this but got $U= pE\cos\theta$ but it is wrong.

  • $\begingroup$ There is no zero of potential energy defined so all you can find is the change in potential energy. There is no information about the initial orientation and/or the final orientation of the dipole relative to the electric field so finding the change in potential energy is not possible. $\endgroup$
    – Farcher
    Commented Aug 7, 2019 at 14:31
  • $\begingroup$ I'm just saying to derive the expression of change in potential energy of dipole in uniform electric field $\endgroup$ Commented Aug 7, 2019 at 14:55
  • $\begingroup$ If that is the case then you would have a dependence on the initial angle and the final angle that the dipole makes with the electric field. $\endgroup$
    – Farcher
    Commented Aug 7, 2019 at 15:05

1 Answer 1


It looks like you are just missing the negative sign in front of the potential energy expression. Let me explain.

Let us imagine the dipole to be at some angle in space with an electric field pointing horizontally to the right for simplicity.

In this scenario, the torque $\vec\tau$ = $\vec r_1$ $\times$ $q\vec E$ + $\vec r_2$ $\times$ $-q\vec E$ where $\vec r_1$ is the moment arm of the dipole towards the positive charge, $\vec r_2$ is the moment arm of the dipole towards the negative charge (both are of same magnitude in this case) and $\vec E$ is the electric field.

Evaluating the cross product, $\vec\tau$ = $-2rqE\sin\theta$ $\hat k$ as the electric field causes both $q$ and $-q$ to rotate in the clockwise direction.

Now, since the force caused by the electric field is conservative, we can say that: $$\Delta U = -\Delta K = -W_E = -\int\vec\tau\cdot d\vec\theta = 2rqE\int\sin\theta d\theta = -2rqE\cos\theta$$

If we define the dipole moment $|\vec p| = 2rq$, then $\Delta U = -pE\cos\theta$ or $-\vec p\cdot\vec E$.

  • $\begingroup$ See here that the direction of d(theta) vector must also be -k(hat) because it is rotating in clockwise direction and thus the dot product must be positive therefore. -(integral)of torque • d(theta). Is. -2rqE(integral)sin(theta)d(theta). Giving the result as 2rqEcos(theta) $\endgroup$ Commented Aug 8, 2019 at 3:50
  • $\begingroup$ From what I understand, the d(theta) vector direction is irrespective of the direction the torque acts in. I didn't state it clearly but in the beginning, we assume the positive theta direction to be in the counterclockwise direction, so that is the direction it takes when evaluating the integral, hence the negative sign for delta(U). $\endgroup$
    – smedepal
    Commented Aug 8, 2019 at 16:22
  • $\begingroup$ Yes the theta at the beginning is positive but due to the torque the angle is decreasing thus d(theta) is clockwise and hence it is negative . Direction of d(theta ) vector is -k hat or inside the device we are using similar to the direction of torque $\endgroup$ Commented Aug 8, 2019 at 16:37
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    $\begingroup$ For instance, considering the dipole to start in a vertical orientation with the electric field pointing to the right, if, as you say, the d(theta) points in the same direction as the torque by the electric field on the dipole, the delta(U) should be positive as it moves from the vertical orientation to a horizontal orientation. However, this cannot be correct since in order to align the dipole with the electric field (with the positive charge to the right), the electric field must do positive work on the dipole to rotate it, hence the delta(U) should be negative. $\endgroup$
    – smedepal
    Commented Aug 8, 2019 at 16:38
  • $\begingroup$ That's what my confusion is while the potential energy should be negative it is coming out positive by equations looks like we are missing a negative in the integral $\endgroup$ Commented Aug 9, 2019 at 3:12

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