Average effective dipole moment?

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SUMMARY

The discussion focuses on calculating the average effective dipole moment of water molecules under an applied electric field. Given the density of water at 1000 kg/m³ and the number of molecules per kg as 3.34 × 10²⁵, participants clarify that the average dipole moment is zero due to random molecular orientation. However, the effective dipole moment, induced by an electric field, can be determined using the dielectric constant of water, approximately 81. The relevant equations include the relationship between relative permittivity and susceptibility, as well as the formula for dipole polarization per unit volume.

PREREQUISITES
  • Understanding of dipole moments and molecular orientation
  • Knowledge of dielectric constants and their significance
  • Familiarity with the concept of polarization per unit volume
  • Basic grasp of relevant equations in electrostatics, including Er=1+X
NEXT STEPS
  • Research the calculation of dipole moments in polar molecules
  • Learn about the relationship between dielectric constant and electric field effects
  • Explore the concept of molecular susceptibility in materials
  • Study the implications of dipole polarization in various substances
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Students and researchers in physical chemistry, materials science, and electrical engineering who are interested in the behavior of polar molecules in electric fields.

brad sue
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Hi ,
I crashes myself with this problem about diploe.
Please can I have some suggestions.

Assume that the density of water is 1000 [kg/m3]
and that there are 3.34 × 1025 molecules per kg of water. (One mole of water weighs 18.0[g] since atomic weight of water is 18.0. A mole contains a number of molecules that is equal to Avogadro number (6.02 × 10^23)).

Determine an average effective dipole moment of each water molecule when E= x [V/m].(x is unit vector)

Note: each molecule of water has dipole moment. The average dipole moment,
however, is zero since these dipole moments are randomly oriented. The average effectivedipole moment px is induced by the applied electric filed because the molecules are preferentially oreinted along the field.


they also talk about average effective dipole moment, I don't know what they mean by that!

thank you
B
 
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You also need the dielectric constant of water. Then you can work out the polarization per unit volume.
 
You are correct in fact it was the preceding problem.

the question was :
water relative permitivity is about 81. If the electric field E=x[V/m] (x unit vector) is inside the water, what is the vector P(dipole polarization per unit volume)?

I found P=7.08x10^(-10)x

I don't know if I am right.
Then can you please give me some guidance for the second part(original post)?

Thank you
 
The dipole per unit volume should be in units of charge*length/length^3 = charge/length^2

The relevant eqns are:

Er=1+X
where Er is relative permittivity, X is susceptibility. Both are dimensionless.

X=<P>/E eps0

where E is applied field, eps0 is perm. of free-space.
 
Average effective dipole moment is presumably just the dipole per unit volume / Number of molecules per unit vol. i.e., the dipole per molecule.

It's an 'effective' dipole, because, as the question says- for zero field, the ave. net dipole per unit volume is zero- but it doesn't follow that the dipole per molecule is zero (even though the 'effective' dipole per molecule is zero).

That's because though each molecule has a dipole in zero ext. field, the molecules are all pointing in random orientations- so the addition of all those dipole vectors comes to zero (per unit volume).

So, the effective dipole is not the same as the real dipole moment per molecule.
 
Thank you very much christianjb :smile:
 

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