Torque on electric dipole placed in non-uniform electric field

When electric dipole placed in non-uniform electric field, what is the approach to calculate torque acting on it? Can it be zero?

• Calculate force on each charge separately.Then calculate torque about the point you want.
– user74370
Commented Jun 25, 2016 at 15:38

The torque $$\tau$$ on an electric dipole with dipole moment p in a uniform electric field E is given by $$\tau = p \times E$$ where the "X" refers to the vector cross product.

I will demonstrate that the torque on an ideal (point) dipole on a non-uniform field is given by the same expression.

I use bold to denote vectors.

Let us begin with an electric dipole of finite dimension, calculate the torque and then finally let the charge separation d go to zero with the product of charge q and d being constant.

We take the origin of the coordinate system to be the midpoint of the dipole, equidistant from each charge. The position of the positive charge is denoted by $$\mathbf r_+$$ and the associated electric field and force by $$\mathbf E_+$$ and $$\mathbf F_+$$, respectively. The notation for these same quantities for the negative charge are similarly denoted with a - sign replacing the + sign.

The torque about the midpoint of the dipole from the positive charge is given by

$$\mathbf \tau_+ = \mathbf r_+ \times \mathbf F_+$$

where

$$\mathbf F_+ = q \mathbf E_+(\mathbf r+)$$

Similarly for the negative charge contribution

$$\mathbf \tau_- = \mathbf r_- \times \mathbf F_-$$

where

$$\mathbf F_- = -q\mathbf E_-(\mathbf r-)$$

Note that

$$\mathbf r_- = -\mathbf r_+$$

We can now write the total torque as

$$\mathbf \tau_{tot} = \mathbf \tau_- + \mathbf \tau_+ =q\mathbf r_+ \times (\mathbf E(\mathbf r_+)+\mathbf E(\mathbf r_-))$$

It is clear that in taking the limit as the charge separation d goes to zero, the sum of electric fields will only contain terms of even order in d.

Noting that $$\mathbf |r_+| = \frac{d}{2}$$

and defining in the usual way $$\mathbf p = q\mathbf d = q(\mathbf r_+ - \mathbf r_- )$$

We can write that $$\tau_{tot} = \mathbf p \times \mathbf E(0) + \ second \ order \ in \ d$$

As we take the limit in which d goes to zero and the product qd is constant, the second order term vanishes.

Thus, for an ideal (point) dipole in a non-uniform electric field, the torque is given by the same formula as that of a uniform field.

Note that it is not correct to start with the expression for a force on an ideal/point dipole in a non-uniform field and then calculate torque from this force. To derive this expression one ends up first taking the limit of a point dipole (on which there is zero force in a uniform field) and then one finds a torque of zero, which is incorrect. One must start with the case of a finite dipole, calculate torque and only then pass to the limit.

When p and E are parallel and anti-parallel, the torque is zero, so yes zero is possible. But the case in which p and E are anti-parallel is one of an unstable equilibrium, and a small angular perturbation will cause the dipole to experience a torque which attempts to align the dipole with the electric field.

• with the proviso that the electric field is uniform
– jim
Commented Jun 26, 2016 at 16:13
• actually it is correct for both the uniform and non-uniform cases - I edited my response to demonstrate this. I have not seen this calculation anywhere else, but it seems a bit obvious now in retrospect. Commented Jun 27, 2016 at 8:33
• Yes, I agree with the result for a point dipole.
– jim
Commented Jun 27, 2016 at 8:44
• You've used Taylor expansion in this proof? Commented Apr 18, 2021 at 8:55

If the dipole is small enough, then the force on dipole would be:

$$\vec{F}=\nabla(\vec{p}.\vec{E})$$

and consequently the torque would be:

$$\vec{F} \times \vec{r}=\nabla(\vec{p}.\vec{E}) \times \vec{r}$$

where r is the length of the dipole

• At first glance this looks right. But for a uniform electric field, the gradient of p dot E is zero, no? So torque would be zero in a uniform field? Torque should be p cross E for a uniform field, right? Commented Jun 26, 2016 at 17:55