# Electromagnetism and Polarisation question

• jazznaz
In summary, the induced dipole moment per unit volume in diamond is P_{0} = 1.5 \times 10^{-7} \text{C m}^{-2}. With a density of 3500 \text{kg m}^{-3} and Avogadro's constant of 6 \times 10^{26} \text{kmol}^{-1}, there are 1.8 \times 10^{29} \text{atoms m}^{-3} in diamond. The average induced dipole moment is 8.3 \times 10^{-23} \text{C m} and the estimated separation of the charges is 8.6 \times 10^{-5} \
jazznaz

## Homework Statement

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 $$10^{12} \text{Hz}$$

## 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...

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.

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.

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

## 1. What is electromagnetism?

Electromagnetism is a branch of physics that deals with the study of electric and magnetic fields and their interactions with charged particles. It explains the relationship between electricity and magnetism, and how they are both different manifestations of the same fundamental force.

## 2. What is a polarised electromagnetic wave?

A polarised electromagnetic wave is a type of electromagnetic radiation in which the electric and magnetic fields oscillate in a specific direction, known as the polarization direction. This means that the wave oscillates in a single plane, rather than in all directions like an unpolarised wave.

## 3. How are electromagnetic waves polarised?

Electromagnetic waves can be polarised through various methods, such as passing them through a polarising filter or by reflection, refraction, or scattering. These methods cause the electric and magnetic fields to align in a specific direction, resulting in polarisation.

## 4. What are the applications of polarised electromagnetic waves?

Polarised electromagnetic waves have various applications in different fields. In the telecommunications industry, polarisation is used to transmit and receive signals in specific directions. In photography, polarising filters are used to reduce glare and enhance colors. In medicine, polarised light is used in imaging techniques such as polarised microscopy to study biological samples.

## 5. How is electromagnetism related to other fundamental forces?

Electromagnetism is one of the four fundamental forces of nature, along with gravity, strong nuclear force, and weak nuclear force. It is related to the other forces through the theory of quantum electrodynamics, which explains the interactions between charged particles and electromagnetic fields. This theory also provides a unified understanding of electromagnetism and quantum mechanics.

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