Calculating Average Power Density of a Uniform Plane Wave

[cathode-ray]

Homework Statement

A plane monocromatic electromagnetic wave propagates in the air hitting a cristal plate with an incident angle of $$60$$ degrees. The cristal plate has an area $$A=0.5m^{2}$$, and is fully illuminated by the wave. The average power density carried by the wave is $$I=10^{-4}\sqrt{\frac{\varepsilon_{0}}{\mu_{0}}}W.m^{-2}$$ and his electric field is given by:

$$\overrightarrow{E}=E_{x}\overrightarrow{u}_{x}+E_{z}\overrightarrow{u}_{z}$$
$$E_{x}=E_{0}cos(\omega t-ky)$$
$$E_{z}=E_{0}sen(\omega t-ky)$$

c)Calculate $$E_{0}$$

Homework Equations

$$I=\left\langle \overrightarrow{S}\right\rangle$$ S is the Poynting vector
$$\overrightarrow{H}=\frac{\overrightarrow{B}}{\mu_{0}}$$
$$v=\frac{E}{B}$$

The Attempt at a Solution

This problem came with a solution but I don't understand one of the steps:

$$I=\left\langle \overrightarrow{S}\right\rangle =\left\langle \overrightarrow{E}\times\overrightarrow{H}\right\rangle =\left\langle \overrightarrow{E}\times\frac{\overrightarrow{B}}{\mu_{0}}\right\rangle =\frac{E_{0}^{2}}{v}$$

How is $$\left\langle \overrightarrow{E}\times\frac{\overrightarrow{B}}{\mu_{0}}\right\rangle =\frac{E_{0}^{2}}{v}$$ ?

 

[graphene]

Hello cathode ray

The E and B fields in an EM wave are related as $$|\vec B| = \frac{|\vec E|}{c}$$
This relation is a consequance of Maxwell's curl E equation.
You could check Griffiths sec.9.2.2.

 

[cathode-ray]

Hi!
Sorry for the time I took to answer.
I use that relation to get:

$$\left\langle \overrightarrow{E}\times\frac{\overrightarrow{B}}{\mu_{0}}\right\rangle \Rightarrow\left\langle \frac{|\overrightarrow{E}|^{2}}{c\mu_{0}}\right\rangle$$

But I still get stuck. Why is $$\left\langle \frac{|\overrightarrow{E}|^{2}}{c\mu_{0}}\right\rangle =\frac{E_{0}^{2}}{v}$$? And shouldn't we get a factor of 1/2 in the final expression because of the average?

 

[graphene]

Hey,

The field are being considered in a medium.
So $$|\vec B| = \frac{k}{\omega}|\vec E| = \frac{|\vec E|}{v}$$.

 

[cathode-ray]

Then in that case we get:

$$\left\langle \frac{|\overrightarrow{E}|^{2}}{v\mu_{0}}\right\rangle$$

Sorry, but I still don't understand the equality:

$$\left\langle \frac{E_{o}^{2}}{v\mu_{0}}\right\rangle =\frac{E_{0}^{2}}{v}$$