# 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 Jun 25 '16 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

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

Similarly for the negative charge contribution

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

where

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

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 Jun 26 '16 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. – RiskyScientist Jun 27 '16 at 8:33
• Yes, I agree with the result for a point dipole. – jim Jun 27 '16 at 8:44

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? – RiskyScientist Jun 26 '16 at 17:55