Electromagnetisim exam question (can someone please check)

[vorcil]

Homework Statement

A dielectric material has the following properties:
electrical conductivity $$\sigma = 0$$
relative permittivity $$\epsilon_r = 3$$
relative permeability $$\mu_r =1$$

The electric field in the dielectric is given by
$$\mathbf{E} = E_0 cos(kz-\omega t)\hat{x}$$

There are no time independent magnetic fields in the dielectric:

i) Write down maxwell's equation in matter in differential form
ii) find an expression for the polarization vector $$\mathbf{P}$$
iii) find an expression for the volume density of bound charge
iv) find an expression for the volume density of free charge
v) find an expression for $$\frac{\parital \mathbf{E}}{\partial \mathbf{t}}$$
iv) find an expression for $$\nabla \times \mathbf{E}$$, and hence deduce an expression for the magnetic field $$\mathbf{B}$$, in the dielectric.
vii) find $$\nabla . \mathbf{B}$$ and explain what this means physically
viii) What is the magnetization vector M in the dielectric?
ix) Find an expression for the phase speed $$\frac{\omega}{k}$$

Homework Equations

The Attempt at a Solution

i) Maxwells equations in differential form (THOUGH I'm not sure what they are in matter?)

curl/divergence of both magnetic and electric fields:

$$\nabla . \mathbf{E} = \frac{\rho}{\epsilon_0}$$ gauss's law
$$\nabla . \mathbf{B} = 0$$
$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$ faraday's law
$$\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}$$ maxwells fixed ampere's law

the question does say it wants the equations in matter,
so should I be using the auxhillary magnetic field and electric displacement field vectors H and D?

or would the equations in the form I gave be enough?

ii) (writing it up now)

 

[vorcil]

ii)
Find an expression for the polarization vector P:


There are two ways I could think of to approach this question,

using $$\mathbf{P} =\epsilon_0\chi_e\mathbf{E}$$

and the electrical displacement route, using $$\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P}$$

but in both these methods, I don't know either $$\chi_e$$ the electrical susceptibility, nor do I know the Displacement vector D,

-

if I were to attempt this in an exam I would have just put down,

$$\mathbf{P} = \epsilon_0 \chi_e \epsilon_0 cos(kz-\omega t) = \epsilon_0^2 \chi_e cos(kz-\omega t)$$

 

[vorcil]

iii) Find an expression for the volume density of bound charge:

Using the equation 4.12 from griffiths, $$\rho_b = -\nabla . \mathbf{P}$$

If the Polarization vector P, (I had $$\epsilon_0^2 \chi_e cos(kz-\omega t))$$is uniform, the volume density of bound charge, $$\rho_b$$ = 0,

but I'm not sure it is uniform,

does that mean I should take the divergence of the Polarization vector P?
I would use Cartesian coordinates,

[tex] (\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})(P)?

I'm not too sure

 

[vorcil]

:( ??

 

[paranormal]

ii) The polarization vector \mathbf{P} is defined as the dipole moment per unit volume of a dielectric material. In this case, it can be expressed as:

\mathbf{P} = \epsilon_0 \chi_e \mathbf{E}

Where \chi_e is the electric susceptibility of the material. Since the material has a relative permittivity of \epsilon_r = 3, we can rewrite this as:

\mathbf{P} = 3\epsilon_0 \mathbf{E}

iii) The volume density of bound charge is given by:

\rho_{bound} = -\nabla \cdot \mathbf{P}

Using the expression for \mathbf{P} from part ii), we can write this as:

\rho_{bound} = -3\epsilon_0 \nabla \cdot \mathbf{E}

iv) The volume density of free charge is given by:

\rho_{free} = \nabla \cdot \mathbf{D}

Since there are no time independent magnetic fields in the dielectric, \mathbf{D} is equal to \epsilon_0 \mathbf{E}. Therefore, we can write this as:

\rho_{free} = \epsilon_0 \nabla \cdot \mathbf{E}

v) Using the given electric field, \mathbf{E} = E_0 cos(kz-\omega t)\hat{x}, we can find the time derivative as:

\frac{\partial \mathbf{E}}{\partial t} = -E_0 \omega sin(kz-\omega t)\hat{x}

vi) To find an expression for \nabla \times \mathbf{E}, we can use the definition of the curl operator:

\nabla \times \mathbf{E} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ E_x & E_y & E_z\end{vmatrix}

Plugging in the components of \mathbf{E} from the given electric field, we get:

\nabla \times \mathbf{E} = -E_0 k sin(kz-\omega t