# Electromagnetism and Polarisation question

1. Jan 26, 2009

### jazznaz

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

When a uniform electric field is applied to diamond, the induced dipole moment per unit volume (ie. polarisation) is $$P_{0} = 1.5 \times 10^{-7} \text{C m}^{-2}$$. Given that the density of diamond is $$3500 \text{kg m}^{-3}$$, 1 kmole of diamond has a mass of 12kg. Avogadro's constant is $$6 \times 10^{26} \text{kmol}^{-1}$$, the atomic number of carbon is 6.

How many carbon atoms are there per cubic metre?
What is the average induced dipole moment?
Estimate the average separation of the +ve and -ve charges.
If $$P = P_{0} \sin \omega t$$ as a result of the alternating electric field, evaluate the peak value of the resultant polarisation current density, J, at a frequency of $$10^{12} \text{Hz}$$
3. The attempt at a solution

First part,

$$\rho = \frac{m}{v} = 3500$$

$$m = 3500$$

$$m/0.012 = 291666.67 \text{moles}$$

$$291666.67 \times 6 \times 10^{23} = 1.8 \times 10^{29} \text{atoms m}^{-3}$$

Second part,

Average induced dipole moment is total polarisation divided by number of atoms,

$$= \frac{1.5 \times 10^{7}}{1.8 \times 10^{29}}= 8.3 \times 10^{-23} \text{C m}$$

Third part,

Dipole moment is $$\vec{p} = q\vec{L}$$

So $$L = \frac{p}{q} = \frac{8.3 \times 10^{-23}}{6 \times 1.6 \times 10^{-19}} = 8.6 \times 10^{-5} \text{m}$$

Which strikes me as far too small to be an estimate of the atomic radius of a carbon atom in the diamond lattice. I can't see where I've gone wrong, maybe I've missed a subtlety in the question?

Any comments would be great, thanks...

2. Jan 26, 2009

### chrisk

The avergare dipole moment is not the atomic radius of the atom. The value you obtained on the order of 10 microns is much larger than then atomic radii on the order of nanometers.

3. Jan 26, 2009

### jazznaz

In this case, the dipole is between the carbon nuclei and the surrounding electron cloud, so I expected the dipole moment to be acting over a length similar to the atomic radius.

4. Jan 26, 2009

### chrisk

Keep in mind an external electric field is applied thereby applying a force to the charges.