So in the derivation of the Potential energy of a dipole due to an external field, we consider a dipole with charges $q_1= +q$ and $q_2= –q$ placed in a uniform external electric field. We know that in a uniform electric field, the dipole experiences no net force but it experiences a torque $τ$ given by

$$\vecτ = \vec p×\vec E$$

which will tend to rotate it. Now here's the tricky part. Suppose an external torque $τ _{ext}$ is applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle $θ_0$ to angle $θ_1$ at an infinitesimal angular speed and without angular acceleration. The amount of work done by the external torque

$$W= \int \:τ\left(\theta \right)d\theta =\int \:\left(pEsin\theta \right)d\theta $$

$$W=pE\left(cos\theta_0 -cos\theta_1 \right)$$

if $\theta_0=\pi/2$ then and $\theta_1=\theta$


My concern is if the torque due to an external electric field (the conservative force) is given by $\:pEsin\theta \:$, and if an external torque is applied to neutralise this torque, shouldn't its value be $-\:pEsin\theta \:$ in the derivation. I know the final answer is correct but I am also sure my 2 assumptions that (1) the external torque is negative of the conservative force and that (2) potential energy is work done by the external force in setting up a configuration (in this case rotating the dipole) are both correct. I think rotational dynamics is where I got something wrong


1 Answer 1


Yes you are absolutely right in stating that the external torque $τ_{ext}$ will have a sign opposite to that of the torque exerted by the electric field. However note that the expresion $$τ = 2aqEsin\theta = pEsin\theta$$ is only indicative of the magnitude of the torque applied by the field on the dipole. Consider the conventional diagram accompanying such derivations (for simplicity we take d = 2a):

enter image description here

It is clear that the torque due to the field causes a clockwise rotation of the dipole, which by definition is taken as negative. Alternatively you can use the standard definition for external torque: $$ \vec τ = \vec r× \vec F $$ taking the origin as the midpoint of the line joining the two charges to obtain $$ \vec τ = 2(-asin\theta .qE)\hat k$$

From here on as $τ_{ext} = -τ$, the negative signs cancel and the derivation follows suit.

Also as a sidenote, it is never a good idea to mention the absolute value of the potential energy of any system. The absolute value is physically meaningless, as only the change in potential energy is physically meaningful. Hence, try not to use the assumption that the intial angle is at 90 degrees and instead just use: $$ΔU_{E} = pE\left(cos\theta_0 -cos\theta_1 \right)$$

Hope this helps.

  • $\begingroup$ i am quite sure what you meant by " the negative signs cancel and the derivation follows suit..." but I hit one jack pot.... $A.B=\left|A\ \right|\left|B\right|\cos\theta$ the modulus means we ignore the negative right!! $\endgroup$ May 22, 2021 at 12:14
  • $\begingroup$ (1) No, the moduli in the expression for a dot product does not necessarily signify positive quantities of A and B. A and B are vectors and |A| represents the magnitude of the vector that can be positive or negative (it is the numeric value or scalar equivalent, which can be both positive or negative). If anything, the cosine term can invoke a negative sign. (2) It does not alter the derivation in any way does it? $\endgroup$
    – Cross
    May 22, 2021 at 12:18
  • $\begingroup$ (2) Aah i now get it. basically according to you, the formula for both the torque is the same but torque applied is negative as it in the anti-clockwise, then we apply negative once again as it is negative of torque... but then wouldn't the applied torque and the torque of the field be along the same direction (as opposed to the initial condition I wrote)? i thought that the sign of applied torque was opposite to the other torque as a consequence of the fact that one was clockwise and the other anti-clockwise. $\endgroup$ May 22, 2021 at 12:44
  • $\begingroup$ (1) I think if you take the modulus of vectors it is always going to be positive. for eg in my coordinate system, let the left side be negative. so if I go 5 units left my displacement is -5 units. But the modulus of displacement is the distance, which always positive. Another way to think is if my vector A=ai+bj+ck, where I,j,k are vectors along x,y,z axes. then the modulus is sqrt(a^2+b^2+c^2) which is always positive $\endgroup$ May 22, 2021 at 12:51
  • $\begingroup$ The torque applied by the field is negative as I wrote in my proof, because it causes a clockwise rotation. When you put in the external torque in the integral, you will have to invert the sign because the external torque causes an anticlockwise rotation which is taken to be positive (corresponds to $+\hat k$ ), and hence $pEsin\theta$ is used instead of $-pEsin\theta$ $\endgroup$
    – Cross
    May 22, 2021 at 12:53

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