Average effective dipole moment?

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Homework Help Overview

The discussion revolves around calculating the average effective dipole moment of water molecules in the presence of an electric field. The context includes the density of water and the number of molecules per kilogram, as well as the concept of dipole moments and their orientation in an electric field.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between dipole moment, electric field, and polarization. Questions arise regarding the definition of average effective dipole moment and its calculation based on dipole per unit volume and number of molecules.

Discussion Status

Some participants have provided insights into the necessary parameters, such as the dielectric constant of water and the relevant equations for calculating dipole polarization. There is an ongoing exploration of how to relate these concepts to find the average effective dipole moment.

Contextual Notes

Participants note the importance of understanding the orientation of dipole moments in the absence of an electric field and how this affects the calculation of effective dipole moments. The discussion also references the relationship between relative permittivity and susceptibility.

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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