Finding the dipole moment of a molecule in an electric field

lundi 1 septembre 2014

1. The problem statement, all variables and given/known data

"A molecule has its dipole moment aligned with a 1.8 kN/C electric field. If it takes 3.5x10^(-27) J to reverse the molecule's orientation, what is its dipole moment?"





2. Relevant equations

Potential energy of dipole in an electric field

U = -p*E = -p*E*cos(theta)



U = 0 corresponds to the dipole aligned at right angles to the field



Torque on a dipole in an electric field

tau = p X E



Dipole moment vector is the product of the equal and opposite charges separated by distance d

p = qd







3. The attempt at a solution

Greetings. My solution attempt is as follows:

U = -p*E

energy required to reverse dipole's orientation = potential energy of the dipole

From this I found:

-3.5x10^(-27) J /1.8 kN/C = -1.9x10^-30 C-m

But this is wrong and I don't know why. The dimensional analysis shows the correct units using this approach.

Then I tried to included the cos(theta) for the magnitude of the dot product, taking theta as pi because the dipole has to have its orientation reversed, and this produced the answer 1.9x10^-30 C-m, which is also incorrect.



I assume that the dipole has no kinetic energy since it is not described as moving, so the energy needed to reverse it's orientation must be large enough to equal its potential energy. This energy has to be applied by doing work on the dipole, which is given, but since there are no charges or distance given, I cannot use p = qd. Then because I don't have p, the torque equation does not help. I must be overlooking something really simple. There are no examples like this in my book. The ones I have found on other sites all show the approach that I tried, so they are no help. The professor did not discuss dipole moments in lecture and none of his slides have a calculation of it, but do show finding the electric field generated by a dipole, which does not work since the problem does not give a charge, q.



If anybody has any comments or ideas, they would be greatly appreciated. Cheers.





0 commentaires:

Enregistrer un commentaire