# Homework Help: Average Effective Dipole moment

1. Apr 2, 2013

### xz4chx

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

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)?

Then

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.

2. Relevant equations
X = eps - 1 and P=eps0*X*E
P=Nv(p) or P=1/ΔVƩ(p)

3. The attempt at a solution
X=81-1 = 80
And so P=eps080x
From that I would plug into the other equation correct, but i am not sure what to do once i have that plugged in for P and what to plug into for the rest

2. Apr 2, 2013

### xz4chx

So I found out the average dipole moment by doing

80eph0/(Density of water)(# molecules of water) and got 2.119e-38 [C*m]

Now it ask for the displacement angle (the angle between the x axis and the angle of the each dipole inside the water) when the E field is 1.0x. Since the E field is in the x direction only would the displacement angle be zero since the E field runs along the x axis

3. Apr 3, 2013

### xz4chx

I need to be able to prove this if this is true and i don't really have any formula for finding a displacement angle. I assume it has something to do with using trig but I am not 100%

P=ε080 x
Average dipole moment is 2.119e-38 [C*m]
E = 1.0 x

And I am assuming the values are correct.