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A $H_2$ molecule with charges $\pm p$ separated by $d=10^{-10}$m and masses $1.66\times 10^{-27}$kg is put into an electric field with uniform strength $E$, making an angle $\theta$ with this field. Show that the torque on the dipole is $pE\sin\theta$.

I assume that the dipole will rotate about its centre of mass. Thus, since $$\tau=|\bf r||F|\sin\theta$$ we have that $$\tau = 2 (d/2)(pE)\sin\theta$$

I don't know what I'm doing wrong and I just want to confirm the result. Thanks! Also I don't know why I'm given the masses. Do I have to treat the problem like a rigid body?:

$$\tau = I\alpha, ~~~~~~~I=\frac 12 m d^2,~~~~\alpha = \frac ar, ~~~a=\frac Fm,~~~~F=pE.$$

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    $\begingroup$ Something is wrong in the setup of the problem. If $p$ is charge, then the torque you're trying to obtain doesn't have torque units; it has force units. I'm thinking that the charges are $\pm q$. Then the dipole moment would be $pd$. $\endgroup$ – Bill N Mar 5 '15 at 4:33
  • $\begingroup$ @BillN - completely agree with you. I wrote an answer - but if you decide to convert your comment to an answer I will happily delete mine so you can get the credit. $\endgroup$ – Floris Mar 5 '15 at 4:44
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Something is wrong in the setup of the problem. If $p$ is charge, then the torque you're trying to obtain doesn't have torque units; it has force units. I'm thinking that the charges are $\pm q$. Then the dipole moment would be $qd$.

The torque on the dipole is $$\tau = 2qE\sin\theta\left(\frac{d}{2}\right) = qdE\sin\theta = pE\sin\theta $$

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Agreeing with Bill N 's comment - but since comments tend to vanish on this site, I'm writing this as an answer.

It is unusual for the letter $p$ to be used for charge, but common for it to be the dipole moment, $ p = qd $.

Making that change in the original problem statement, your analysis is correct: a dipole of dipole moment $p$ at an angle $\theta$ to an electric field $E$ experiences a torque

$$\Gamma = pE\sin\theta = qdE\sin\theta$$

It appears the wording in the problem was wrong.

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