Polar molecule in electric field

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Main Question or Discussion Point

what happens to gas dipole molecule when electric field E is turn on ?
a) is going to rotate to lined up with vector E and stay like that ?
b) is going to oscilate, like pendulum, pointing on average toward E ?
c) none of the above.
d) is every collision going to flip molecule unpredictible, or it maintain it's orientation with respect to E ?

-ael
 

Answers and Replies

  • #2
olgranpappy
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what happens to gas dipole molecule when electric field E is turn on ?
a) is going to rotate to lined up with vector E and stay like that ?
b) is going to oscilate, like pendulum, pointing on average toward E ?
c) none of the above.
d) is every collision going to flip molecule unpredictible, or it maintain it's orientation with respect to E ?

-ael
since it's a gas you can think of one atom at a time, and there is a means of maintaining equilibrium apparently, so the dipole will line up with the field (there is some "damping" to get rid of mechanics oscillations such as in option "b" above). There should be thermal fluctuations of the dipole away from the field direction tho. Thus answer "a" is fairly correct (moreso than b or d).
 
  • #3
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what mechanism is providing a "dumping" ?
Is this EM radiation from oscilating charge ?

-ael
 
  • #4
olgranpappy
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probably something like that.
 
  • #5
Meir Achuz
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what happens to gas dipole molecule when electric field E is turn on ?
a) is going to rotate to lined up with vector E and stay like that ?
b) is going to oscilate, like pendulum, pointing on average toward E ?
c) none of the above.
d) is every collision going to flip molecule unpredictible, or it maintain it's orientation with respect to E ?

-ael
Due to collisions, the polarization will be thermalized.
The probability of polarization along the field will be ~exp(p.E/kT).
 
  • #6
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Thx Meir,

I already found article on "STARK EFFECTS ON RIGID ROTOR WAVEFUNCTIONS" which explained how probability of molecule orientation will depend on kinetic enrgy, dipole moment and strength of electric field. There are QM states to be considered, but simple classical approximation can be obtained based on:

<max rotation angle> ~= arccos( <1 axis thermal energy ~ 0.5kT> / <dipole moment> * E)

I also tried to find an apprximate strength of the E field to force a molecule of cyanamide (long polar molecule) to limit its rotation to +-15deg. I got something like 150V/um at room temp. This is 50 times more then voltage causing air breakage.
 
  • #7
The deciding factor will be the strength of the electric field .
If the potential energy of the dipole molecule in the electric field is very much grater than the thermal energy of the molecule,then option (a) is your answer.
If the potential energy of the dipole molecule in the electric field is comparable to the thermal energy of the molecule,then option (b) is your answer.
If the potential energy of the dipole molecule in the electric field is less than the thermal energy of the molecule,then option (d-unpredictability) is your answer.

mail me if anyone doesnot stand to this explanation.
 
  • #8
A followup question....

If one applies a sufficiently high voltage to change rotation to pendular motion about the electric field axis, what are the thermodynamic consequences? IE:

1) does the electric field do net work on the molecules?
2) if yes, a) how much and b) does the energy go end up in increased translational energy resulting in higher temperature/pressure?

Thanks
 

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