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

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    $\begingroup$ Calculate force on each charge separately.Then calculate torque about the point you want. $\endgroup$ – 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.

Ref: Wikipedia article on electric dipole moment.

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

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


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

Similarly for the negative charge contribution

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


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

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.

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  • $\begingroup$ with the proviso that the electric field is uniform $\endgroup$ – jim Jun 26 '16 at 16:13
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    $\begingroup$ 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. $\endgroup$ – RiskyScientist Jun 27 '16 at 8:33
  • $\begingroup$ Yes, I agree with the result for a point dipole. $\endgroup$ – jim Jun 27 '16 at 8:44

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


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

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  • $\begingroup$ 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? $\endgroup$ – RiskyScientist Jun 26 '16 at 17:55

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