Linearly polarized electric field

In summary: The equation for a harmonic wave is: ##E(\vec r,t) =\cdots##. When dealing with the z-direction, you need to use the equation for a vector: ##\vec k = k_x\hat x + k_y\hat y +...##. This will give you the direction of the electric field.
  • #1
rogeralms
19
0

Homework Statement


Consider a harmonic, electromagnetic plane wave traveling along a line from the origin to the point (3,5,6). It is linearly polarized and its electric field lies in a plane perpendicular to the direction of travel of the wave. The wavelength of the wave is 2.0mm and it has a frequency of 100 GHz.

a) Write an expression for the magnitude (i.e. don't worry about the orientation) of the electric field of this plane wave if the amplitude of the electric field is 10V/m.

b) What is the index of refraction of the medium in which the wave is traveling?

c) Compute the irradiance of the wave.

d) Determine a unit vector for a possible direction that would define the direction of the electric field.


Homework Equations



k = 3i + 5j + 6k

|k| = (32 + 52 + 62)(1/2)

unit vector in k direction = k / (70)(1/2)

n= v/c

E (dot) k = 0

I = (c * epsilonzero /2) * (E0)2

The Attempt at a Solution



a) I am really confused as to how to find the magnitude of the electric field

As you see, I have the magnitude of the direction vector k.

I know that the divergence of the E and k vectors is zero. But I don't know how the amplitude is used in the magnitude. Please I do not want the final answer, just a reference as to how to find the magnitude.

b) v = lambda * nu = 2*10-3 * 100*109 = 2*108 m/s

n = c/v = 3*108 / 2*108 = 1.5

c) I = (c*ep0 / 2)* E0 2 =
150 ep0 *108 V/m

d) I'm not sure on this one

It has something to do with divergence of E k = 0

Would it be something like

(3-x)Ex + (5-y)Ey + (6-z)Ez = 0?

Thanks for any help. I just need some hints.
 
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  • #2
I'll just start with this bit:
rogeralms said:
a) I am really confused as to how to find the magnitude of the electric field
You are not asked to find the magnitude - you are asked to write an expression for it.

You are told that the wave is "harmonic" and "polarized".
Do you understand what these terms mean?

You are given the amplitude, the frequency, and the wavelength ... you know the wave speed.
Do you know the equation for a harmonic wave?

I'll start you off: ##E(\vec r,t) =\cdots##

Note: you have used k to mean two different things.
Please makes sure that each symbol has only one meaning.
 
Last edited:
  • #3
Simon,

Thank you so much for answering and sorry for the double use of k. That was the convention used in my textbook and in my professor's lectures.

E(r,t) = E0 cos([itex]\omega[/itex] (t - r/c) + [itex]\epsilon[/itex])

where k = ω / c

Since cos varies between -1 and 1 and the amplitude is given as 10 V/m, then would the magnitude just be the expression

10 * cos (ω *t - k ° r) ?

Then would r = (x-3)i + (y-5)j + (z-6)k ?

I could not find a definition for magnitude of an electric field in the book, the posted lecture notes, or the internet. Again sorry for the double use of k as the direction of motion vector and the unit vector in the z direction, but that is the way my book and the professor use it.
 
  • #4
rogeralms said:
Simon,

Thank you so much for answering and sorry for the double use of k. That was the convention used in my textbook and in my professor's lectures.
Such sloppiness is quite common - but you have to be careful about it or you'll get confused.

If you have to use the wave vector as ##\vec k## then you will need an alternate form for the z-direction unit vector. ##\hat z## or ##\hat e_z## are common, with similar ones for x and y unit vectors.

E(r,t) = E0 cos([itex]\omega[/itex] (t - r/c) + [itex]\epsilon[/itex])

where k = ω / c
Can a vector be equal to a scalar?
Be careful here - it is usual to use c as the speed of light in a vacuum, the problem statement implies that this wave is traveling through a medium with a different speed.
(Mention of "refractive index"?)

Since cos varies between -1 and 1 and the amplitude is given as 10 V/m, then would the magnitude just be the expression

10 * cos (ω *t - k ° r) ?
I see you are struggling with the symbol chart as well as the mathematical description. Suggest you try LaTeX. If you use the "quote" button (below this post) you will get to see how I get the symbols to come out nicely.

i.e. ##\vec E(\vec r , t)=\hat n E_0\cos(\omega t - \vec k \cdot\vec r) = ## ... where ##\hat n## is the polarization direction.

Then would r = (x-3)i + (y-5)j + (z-6)k ?
##\vec r## usually just points to some location in space.
This may help... p7-8
... how does ##\vec k = k_x\hat x + k_y\hat y + k_z\hat z## come into this?

I could not find a definition for magnitude of an electric field in the book, the posted lecture notes, or the internet.
The magnitude of the electric field is defined the same way as the magnitude of any vector.

Basically the exercise is trying to get you to learn how to describe waves in 3D.
 
  • #5
Thanks

I finally understand part a)

k=3i+5j+6z so unit k vector = k/(70)^(1/2)

E(t) = E0 cos(k(dot)r - ωt)

E(t) = 10 V/m * cos ( (3x + 5y + 6z)/ 70(1/2) - ωt)

part d)

k(dot)<x,y,z> = 0 then 3x + 5y + 6z = 0. So choose y = 0 gives 3x + 6z = 0.

This makes x = -2z, so choose z = 1, then x = -2

A perpendicular vector to <3,5,6> is <-2,0,1>. To get the unit vector just divide by 5(1/2)

Thanks so much for your patience and gentle nudging!
 
  • #6
... well done - I think you are on your way.
 

Related to Linearly polarized electric field

1. What is a linearly polarized electric field?

A linearly polarized electric field is a type of electromagnetic wave that oscillates in a single plane. This means that the electric field vector, which describes the direction and magnitude of the electric field, only changes in one direction as the wave propagates through space.

2. How is a linearly polarized electric field created?

A linearly polarized electric field can be created by passing an electromagnetic wave through a polarizing filter, which only allows waves with a certain orientation of the electric field vector to pass through. It can also be created by using specialized antennas or by reflecting light off of certain surfaces.

3. What is the significance of a linearly polarized electric field?

The linear polarization of an electric field is important because it allows for the manipulation and control of electromagnetic waves. This is crucial in many modern technologies, such as telecommunications, radar, and satellite communication.

4. Can a linearly polarized electric field be converted into a circularly polarized one?

Yes, a linearly polarized electric field can be converted into a circularly polarized one by using a device called a waveplate. This device changes the phase difference between the two components of the electric field, resulting in circular polarization.

5. How does the orientation of a linearly polarized electric field affect its interaction with matter?

The orientation of a linearly polarized electric field can determine how it interacts with matter. For example, certain materials may absorb or reflect the wave more effectively depending on the orientation of the electric field vector. This is the principle behind polarized sunglasses, which are designed to block horizontally polarized light from entering the eyes.

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