I am trying to derive the potential energy because of torque in a dipole placed in a uniform electric field. But the answer I am getting is different from the answer I saw everywhere. So could someone tell me, where in the integral or where have I made a mistake? In the given diagram the total torque is given by $p\times E$. Now the torque is in the clockwise direction. Lets consider $U$ at $90^{\circ}$ to be $0$. Hence if we move the dipole by an angle $d\theta$ the work done is given by $$W=\int_{90^{\circ}}^{90^{\circ}+\theta} pE\sin\theta\, d\theta.$$ Now in the derivations that I have seen, I see everyone integrating from limits $90^{\circ}$ to $\theta$ but does not the angle change from $90^{\circ}$ to $90^{\circ}+\theta$. As the $\theta$ in $pE\sin\theta$ is the angle between vector p and E. So if we move it anticlockwise the angle between them changes from $90$ to $90 +\theta$.

Now doing the integral. we get $-pE\cos(90+\theta$)

This is also the answer given online and it makes sense. But lets suppose that we move the dipole in the opposite direction. Then the Work $$W=\int_{90^{\circ}}^{\theta} -pE\sin\theta\, d\theta.$$ In this case we get the integral as $pE\cos\theta$ Now in the earlier case we got that if the angle is greater than $90$ then the Potential is greater than $0$ as it becomes $pE\sin\theta$. But in this case even though $\theta$ is less than $90$ the answer is positive. Now both the cases cannot be true.enter image description here

  • $\begingroup$ The angle in the limit should not be indicated as being the same as the angle in the function. In the first integral, the angle in the function is measured from where the two vectors are parallel. The angle in the limit is measured from where they are perpendicular. $\endgroup$ – R.W. Bird Jul 4 at 17:03

You have to integrate from 90 to θ as you are measuring angle from horizontal in your diagram. If you are measuring any other angle as θ,obviously result would differ.

First of all, Understand that you are taking angle between P and E as θ( By bold i meant vectors).And integrating doesn't mean that you are moving it by θ in some way.Suppose if torque is not varying with θ you have to just multiply i.e. $\tau$ ×($\Delta$$\theta$ ).But now you have to integrate $\tau$ dθ. By integrate i mean limit of sum of work in infinitesimally small intervals of θ (i.e. dθ). Now $$Work\ done\ by\ external\ agent = change\ in\ internal\ eneregy\ $$

Also $τ_{ext}$ is what i meant above as $\tau$. It is equal and opposite to torque due to electric field so that process is quasistatic (no kinetic energy is observed at any instant).

$$\int_{\theta_1}^{\theta_2}\tau\ d\theta$$

Now the above expression implies that you are taking limit of sum of $d\ W$ from position $\theta_1$to $\theta_2$ which gives you total work done in that process.And remember that $d\theta$ should be taken as an increment in $\theta$ which means you are rotating dipole in the increasing direction of $\theta$ by small $d\theta$.

So finally you have to set $\theta_1$=$90^{\circ}$ because it is an assumption that P.E is zero at that position and $\theta_2$=$\theta$ which can be anything from 0 to 360 degrees . $${U(\theta)}=-PE{\cos\theta}$$

The above one can represent P.E.

There is nothing wrong with your first integral in which you are finding energy as function of angle between P and a perpendicular to electric field(Taking it as $\alpha$) .But the second integral is faulty, you should not put negative sign even if you feel you are moving in reverse direction and you need not change your upper limit of integral to $\theta$. You should understand that even in second case your integral is same as first one . $${U(\alpha)}= PE{\sin\alpha}$$

This also is correct.

Hope you understand.

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  • $\begingroup$ Mate if you move it bu theta the angle between p and E will be 90+theta not theta $\endgroup$ – BlackSusanoo Jul 4 at 4:46
  • $\begingroup$ your torque expression is having 'theta' measured from horizontal, so you set limits in integral as θ=90 to θ=θ. I am not moving it by θ , i am telling it is at angle θ from horizontal $\endgroup$ – sanjay Jul 4 at 5:03
  • $\begingroup$ But in the cross product which yeilds pEsintheta over there theta is the angle between them. You cannot take any other theta $\endgroup$ – BlackSusanoo Jul 4 at 5:38
  • $\begingroup$ yeah, i am taking that only as theta as in diagram. $\endgroup$ – sanjay Jul 4 at 6:40
  • $\begingroup$ You are having many misconceptions(even in integration) which can't be clarified here(i don't have patience to type that much) .You should meet your physics teacher. $\endgroup$ – sanjay Jul 6 at 5:16

Think of changing the value of $\theta$ (denoted in the figure) from $\theta_1$ to an angle $\theta_2$. Thus your integral becomes

\begin{align} \Delta U&=\int_{\boldsymbol{\theta_1}}^{\boldsymbol{\theta_2}}(\mathbf p\times\mathbf E)\cdot\mathrm d\boldsymbol{\theta}\\ &=\int_{\theta_1}^{\theta_2}(pE\sin\theta)\:\mathrm d\theta\\ \Delta U=U_2-U_1&=(-pE\cos\theta_2)-(-pE\cos\theta_1) \end{align} Now, it isn't hard to see that the expression of $U$ at any angle $\theta$ should be of the form

$$U=-pE\cos\theta+\rm constant$$

The constant appears because the value of potential energy can be arbitrarily shifted by a constant without really changing anything else. For simplicity's sake, we take the $\rm constant$ to be $0$. Thus the final value of the potential energy becomes


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You have found the work done by the force. According to the work-energy theorem, the work done is equal to the increase in the kinetic energy. So, in a system like this in which the total (kinetic plus potential) energy is conserved, the work done is the decrease in the potential energy.

This is just like the potential energy of a conservative force $F$ that is a function of position is $$U_{2}-U_{1}=-\int_{x_{1}}^{x_{2}}F\, dx,$$ with a minus sign in front. The equivalent with a conservative torque $N$ is $$U_{2}-U_{1}=-\int_{\theta_{1}}^{\theta_{2}}N\, d\theta.$$ Choosing $\theta_{1}$ (where $U=0$) to be at the the angle $\theta=\frac{\pi}{2}=90^{\circ}$, the minus sign in the formula for the work gives $$U(\theta)=-\int_{\frac{\pi}{2}}^{\theta}pE\sin\theta\, d\theta=pE\cos\theta.$$

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  • $\begingroup$ Mate the work done by the force is the energy is stored. You could do negative of the torque due ti electric field it will give same result also the theta here is 90+ theta as if you move it by a theta angle between p and E will be 90 +theta $\endgroup$ – BlackSusanoo Jul 4 at 4:45

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