# What is the force on a dipole in a uniform electric field?

For an electric dipole of dipole moment $$\vec{p}$$ placed in an electric field $$\vec{E}$$ experiences a force given by $$\vec{F}=-\nabla U=-\nabla(\vec{p}\cdot\vec{E})=(\vec{p}\cdot\nabla)\vec{E}.$$ So $$\vec{F}=0$$ if $$\vec{E}$$ is uniform. But if we use $$\vec{F}=-\frac{1}{r}\frac{\partial U}{\partial\theta}\hat{\theta}$$ and $$U=-pE\cos\theta$$ we find $$\vec{F}\neq 0$$. What is the reason of this contradiction?

The angle $$\theta$$ which appears in the polar coordinates of the position of the dipole is not the same as the angle $$\theta'$$ between the direction of the dipole and the direction of the electric field. When you translate the dipole in space, $$\theta$$ changes, but $$\theta'$$ does not. You have conflated the two.
It may help to express the force and torque in vector form. The interaction energy is $$U=-\vec{p}\cdot\vec{E}$$ and we are taking $$\vec{E}$$ to be uniform in space. Therefore, this expression is independent of position $$\vec{r}$$. We can write the force as $$\vec{F} = -\vec{\nabla}_{\vec{r}} U$$ and express it, if we wish, in the polar coordinates defined via $$\vec{r}=(r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta)$$, but for a uniform field, we will get $$\vec{F}=\vec{0}$$ either way, because $$U$$ has no dependence on $$\vec{r}$$.
If we write the dipole in terms of its magnitude and a unit vector like this $$\vec{p}=p\vec{e}$$, the torque may be expressed $$\vec{\tau} = -\vec{e}\times \vec{\nabla}_{\vec{e}} U$$ where it should be noted that the gradient is with respect to the components of the vector $$\vec{e}$$, not $$\vec{r}$$. For the given form of the energy, $$U=-p\vec{e}\cdot\vec{E}$$ this becomes $$\vec{\tau} = p\vec{e}\times \vec{E} = \vec{p}\times \vec{E}.$$ Again, it is possible to write this in polar coordinates if we wish. We need to define $$\vec{e}=(\sin\theta'\cos\phi',\sin\theta'\sin\phi',\cos\theta')$$, where the angles $$\theta'$$ and $$\phi'$$ are the polar angles of the dipole, not the same as $$\theta$$ and $$\phi$$. If we choose $$\vec{E}$$ to point in the $$z$$ direction, then $$\vec{p}\cdot\vec{E}=pE\cos\theta'$$ so we can write $$U=-pE\cos\theta'$$. I won't write the derivation out, but it is not too complicated. It gives the same result.
There may be some superficial similarities between the expression for the torque on the dipole, acting about the centre of the dipole, and the commonly used formula for the torque about the origin produced by a force $$\vec{F}$$ acting at a point $$\vec{r}$$, namely $$\vec{r}\times\vec{F}$$. This last formula is not applicable here: the force $$\vec{F}$$ is zero, and in any case the expression $$\vec{r}\times\vec{F}$$ would involve the polar angles $$\theta$$ and $$\phi$$, not $$\theta'$$ and $$\phi'$$. One can derive an expression for the torque by writing the dipole as two charges at $$\pm d\vec{e}$$ (where $$d$$ is small), evaluating the force on each charge, and applying the formula $$\pm d\vec{e}\times\vec{F}$$ to both the charges. This gives a superficially similar-looking formula to $$\vec{r}\times\vec{F}$$, but involving the orientation of $$\vec{e}$$, not $$\vec{r}$$.
• But if we do not distinguish between $\theta$ and $\theta^\prime$, the torque expression is correctly obtained. Why? Note that by considering $\vec{r}\times\vec{F}$, I get the correct formula for the torque using the second set of equations. – mithusengupta123 Feb 25 at 8:55
• The torque is defined in terms of the gradient with respect to the dipole orientation, not position. Therefore, it is correctly expressed in terms of $\theta'$, not $\theta$, in my notation. – user197851 Feb 25 at 8:58
• I've extended my answer, hopefully to clarify further, in case anyone else looks at this. The expression $\vec{r}\times\vec{F}$ is not applicable here: it is not how one defines the torque on a dipole. One can derive an expression for the torque by writing the dipole as two charges at $\pm d\vec{e}$ (where $d$ is small), evaluating the force on each charge, and applying the formula $\pm d\vec{e}\times\vec{F}$ to both the charges. This is why the formulae look similar. Hopefully this helps. – user197851 Feb 25 at 10:00