Find ##E_0## and ##k## for ##E= E_0 \sin(k r -\omega t)## using Gauss

[happyparticle]

Homework Statement

Find $E_0$ and $k$, $E= E_0 \sin(k r -\omega t)$ using Gauss's equation.

Relevant Equations

$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$

$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$

$\vec{E}_0 k cos(kr -\omega t) = \frac{\rho}{\epsilon_0}$

$E_0 = \frac{\rho}{\epsilon_0} / k cos(kr -\omega t)$

and

$k^2 = (\arccos{\frac{\rho}{E_0 \epsilon_0}} + \omega t)/r$

I don't think it makes sense since I found $k = \pm \frac{\omega}{c}$ using

$\Delta \vec{E} = \frac{1}{c^2} \frac{\partial ^2 E}{\partial t^2}$

should I use Gauss's equation in vacuum?

Edit: Even with Gauss's equation in vacuum I don't get the same answer.

 

[Steve4Physics]

Homework Statement:: Find $E_0$ and $k$, $E= E_0 sin(k r -\omega t)$ using Gauss's equation.
Relevant Equations:: $\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$

Not sure I can help directly, but I can see some problems with the actual question…

“Find $E_0$ and $k$” - in terms of what other parameters?

The equation $E= E_0 sin(k r -\omega t)$ is for the electric field of a plane EM wave propagating in free space (no charges), so it is not clear (to me) how Gauss’s Law is relevant in this situation.

Have you stated the original question completely/accurately?

 

[Gordianus]

Homework Statement:: Find $E_0$ and $k$, $E= E_0 sin(k r -\omega t)$ using Gauss's equation.
Relevant Equations:: $\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$

$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$

$\vec{E}_0 k cos(kr -\omega t) = \frac{\rho}{\epsilon_0}$

$E_0 = \frac{\rho}{\epsilon_0} / k cos(kr -\omega t)$

and

$k^2 = (\arccos{\frac{\rho}{E_0 \epsilon_0}} + \omega t)/r$

I don't think it makes sense since I found $k = \pm \frac{\omega}{c}$ using

$\Delta \vec{E} = \frac{1}{c^2} \frac{\partial ^2 E}{\partial t^2}$

should I use Gauss's equation in vacuum?

Edit: Even with Gauss's equation in vacuum I don't get the same answer.

When we write $E=E_0\cos (kr-\omega t)$ we give the modulus of the electric field of a plane wave in a sourceless medium. In such a wave the electric field points in a direction (call it $x$ )orthogonal to the traveling direction (call it $z$). Thus, the electric field should be written as: $\vec{E}=E_0\cos(kz-\omega t)\hat{x}$. Its divergence is zero

 

[happyparticle]

Have you stated the original question completely/accurately?

Yeah I did. I don't really understand the question as well.

When we write $E=E_0\cos (kr-\omega t)$ we give the modulus of the electric field of a plane wave in a sourceless medium. In such a wave the electric field points in a direction (call it $x$ )orthogonal to the traveling direction (call it $z$). Thus, the electric field should be written as: $\vec{E}=E_0\cos(kz-\omega t)\hat{x}$. Its divergence is zero

How do you know this is the modulus of the electric field in a sourceless medium?

In this case $\nabla \cdot \vec{E} = 0$
$\frac{d}{dr}(E_0 sin (k r -\omega t)) =0$
$E_0 k cos(k r -\omega t) = 0$

$E_0 = 0$
or
$k = 0$

I don't think $r$ or $x$ change anything. IfI have to find the conditions on $E_0$ and $k$

 

[Gordianus]

Several authors show that the proposed electric field is the one associated with a plane wave (David Griffiths, "Introduction to Electrodynamics").
I'd also suggest reading about the divergence operator, you aren't using it the right way. Divergence operates on a vector, not a scalar.
Finally, $E_0$ can be any real number for it will certainly satisfy Gauss' Law.

 

[happyparticle]

I probably don't explain my question correctly.
I should deduce the conditions of $\vec{E}_0$ and $k$ by using gauss's law.
It should have a way to do it.

The only way I see is to plug $\vec{E} = \vec{E}_0 sin(\vec{k} \vec{r} - \omega t)$ to $\nabla \cdot \vec{E} = 0$

To be honest, I'm not sure to fully understand the question.

I have to do the same thing for $B_0$ and $k$ by using $\nabla \cdot \vec{B} = 0$ for
$\vec{B} = \vec{E}_0 sin(\vec{k} \vec{r} - \omega t)$

Then by using faraday's equation.

 

[Gordianus]

Let's say, for the sake of simplicity, that $\vec k$ points along $z$. You already know $\vec\nabla\cdot\vec E=0$ and this implies the direction $\vec E$ points to must bear a relationship with the direction of $\vec k$

 

[happyparticle]

So, I could just say, $\vec{E}_0$ must be perpendicular to $\vec{k}$ ?

 

[Gordianus]

Not just say, prove it.

 

[happyparticle]

I'm not sure if this is correct.

$\vec{k}\vec{r} -\omega t = \frac{\pi}{2}$

 

[Gordianus]

No, you've found the set of positions and time that result in zero field.
Try $\vec E=(a \hat x+b \hat y+ c \hat z) \cos(kz-\omega t)$
Can you find the valid values of $a, b, c$?

 

[happyparticle]

I thought I have to find values of $\vec{E}_0$ and $k$ where $\vec{E}_0 \vec{k} cos (\vec{k} \vec{r} -\omega t) = 0$
thus, if $\vec{k}\vec{r} - \omega t = \frac{\pi}{2}$
$\vec{E}_0\vec{k}$ are perpendicular, no?

$a,b,c$ should be 0?

I see 3 ways that $\vec{E}_0 \vec{k} cos (\vec{k} \vec{r} -\omega t) = 0$

$\vec{E}_0 = 0$ , $\vec{k} = 0$ or $\vec{E_0} \vec{k}$ must be perpendicular so the angle between must be 90 or $\frac{\pi}{2}$ rad, so $k =(\frac{\pi}{2} + \omega t)/r$

 

[happyparticle]

No, you've found the set of positions and time that result in zero field.
Try $\vec E=(a \hat x+b \hat y+ c \hat z) \cos(kz-\omega t)$
Can you find the valid values of $a, b, c$?

I'm not sure what "conditions" means. Does it means the direction of the vectors? If so, I don't know neither $\vec{E_0}$ nor $\vec{k}$.

Is $\vec{r}$ means radial or like $\vec{r} = xyz$?

If $\vec{r}$ means radial then I guess $\vec{k} = k\hat{r}$, thus $\vec{E_0}$ must be $E_0\hat{\theta}$

 

[Gordianus]

Let's use Cartesian coordinates, thus $\vec \nabla=(\frac{\partial}{\partial x}\hat x+\frac{\partial}{\partial y}\hat y+\frac{\partial}{\partial z}\hat z)$.
I already suggested the wave travels in the z direction:$\vec k=k \hat z$ and the electric field has the form: $\vec E=(a \hat x+b \hat y+ c \hat z)\cos(kz-\omega t)$.
Now, compute $\vec \nabla \cdot \vec E$ and check tha acceptable values of a, b and c.

 

[happyparticle]

What about $\vec{r}$?
$\vec{E}(\vec{r},t) = \vec{E_0} sin(\vec{k} \vec{r} - \omega t)$
then we should have
$\vec{E}(\vec{r},t) = \vec{E_0} sin(k\hat{z} \vec{r} - \omega t)$

$\vec{r} = x\hat{x} + y\hat{y} + z\hat{z}$

$\vec{E}(\vec{r},t) = \vec{E_0} sin(k\hat{z} (x\hat{x} + y\hat{y} + z\hat{z}) - \omega t)$$\vec{E}(\vec{z},t) = \vec{E_0} sin(k\hat{z} z\hat{z} - \omega t)$

$\vec{E}(z,t) = \vec{E_0} sin(k z - \omega t)$

I have to find the conditions of $\vec{k}$ and not $k$

However,I'm still not convinced about $\vec{k} = k\hat{z}$

http://www.physics.usu.edu/Wheeler/Waves/WaveSolutions.pdf
There we have $kx$

$\nabla \cdot \vec{E} = \frac{\partial}{\partial x}(a sin(kz-\omega t)) \hat{x} + \frac{\partial}{\partial y}(b sin(kz-\omega t)) \hat{y} + \frac{\partial}{\partial z}(c sin(kz-\omega t)) \hat{z}$

$= 0 + 0 + c k cos(kz- \omega t)$

So, a and b can be any values and c must be 0 ?

Does that means $\vec{E_0} = E_0 \hat{z}$?

 

[happyparticle]

Thus, the electric field should be written as: $\vec{E}=E_0\cos(kz-\omega t)\hat{x}$.

How do you know the electric field $\vec{E} = E\hat{x}$?

If my previous post is correct, that means $\vec{E_0}$ and $\vec{k}$ are in the same direction.

 

[Gordianus]

No, $\vec E_0$ and $\vec k$ don't point in the same direction.

 

[happyparticle]

If $\nabla \cdot \vec{E} = ck cos(kz-\omega t)$ and $\nabla \cdot \vec{E} = 0$
c must be 0. I don't see what that tells me about $\vec{E_0}$ and $\vec{k}$.

$E_z$ must be 0. Which means $\vec{E_0}$ must point in x or y direction.

 

[Gordianus]

If c=0 then $\vec E_0$ lies in the xy plane AND Is orthogonal to $\vec k$

 

[happyparticle]

Yeah, I had finally saw that. So is that what I was looking for ? because to be honest I'm still not sure what "the conditions on $\vec{E_0}$ and $\vec{k}$ means.Furthermore, how do you know that $\vec{E}$ points in the x direction?

 

[vela]

Homework Statement:: Find $E_0$ and $k$, $E= E_0 \sin(k r -\omega t)$ using Gauss's equation.

Yeah I did. I don't really understand the question as well.

I probably don't explain my question correctly.
I should deduce the conditions of $\vec{E}_0$ and $k$ by using gauss's law.
It should have a way to do it.

The only way I see is to plug $\vec{E} = \vec{E}_0 sin(\vec{k} \vec{r} - \omega t)$ to $\nabla \cdot \vec{E} = 0$

I'm going to point out that you did not, in fact, state the original problem accurately and correctly. There's a significant difference between $E_0 \sin(kr-\omega t)$ and $\vec E_0 \sin(\vec k \vec r - \omega t)$. I can't tell if you were just being sloppy or you really don't understand that $k$ and $\vec k$, for instance, aren't the same thing. If it's the latter, it may explain why you struggled so much with this question, and you'd be doing yourself a favor to clarify your understanding of the mathematics and notation.

By the way, you should have written $\vec E_0 \sin (\vec k \cdot \vec r - \omega t)$.

To be honest, I'm not sure to fully understand the question.

My guess is you're being asked to actually prove what you were earlier only told about electromagnetic waves in a vacuum: the electric field and magnetic field are perpendicular to the direction of propagation and to each other.

 

[happyparticle]

I'm going to point out that you did not, in fact, state the original problem accurately and correctly. There's a significant difference between $E_0 \sin(kr-\omega t)$ and $\vec E_0 \sin(\vec k \vec r - \omega t)$. I can't tell if you were just being sloppy or you really don't understand that $k$ and $\vec k$

I saw that sorry, but it was too late to edit the post.
It's $\vec{E} = \vec E_0 \sin(\vec k \cdot \vec r - \omega t)$.

However, is the post #18 correct? If so, is there a way to know if $\vec{E}$ is pointing toward x or y.
I can prove that $\vec{B}$ is perpendicular to $\vec{k}$ by using $\nabla \cdot \vec{B} = 0$

the electric field and magnetic field are perpendicular to the direction of propagation and to each other.

I think I can prove that using faraday's law, but I have to find the direction of $\vec{E}$

 

[vela]

I saw that sorry, but it was too late to edit the post.
It's $\vec{E} = \vec E_0 \sin(\vec k \vec r - \omega t)$.

That's still not really correct. It's the dot product of the two vectors that appears in the argument of sine.

However, is the post #18 correct? If so, is there a way to know if $\vec{E}$ is pointing toward x or y.
ultimately, I have to find the pointing vector $\frac{\vec{E} \times \vec{B}}{\mu_0}$, so I think I have to find the direction of $\vec{E}$ and \vec{B}.

You don't need to make any assumptions about the directions of $\vec E_0$ or $\vec k$. You want to show that Gauss's law leads to the condition $\vec k \cdot \vec E_0 = 0$.

 

[happyparticle]

That's still not really correct. It's the dot product of the two vectors that appears in the argument of sine.

I fixed it a second before your post.

You don't need to make any assumptions about the directions of $\vec E_0$ or $\vec k$. You want to show that Gauss's law leads to the condition $\vec k \cdot \vec E_0 = 0$.

Alright, but I have to show the same thing using Faraday's law. However, I don't know if I can do it without knowing the direction of $\vec{E}$

Maybe what I said is not clear. I'll type what I did.

$\vec{\nabla} \times \vec{E} = - \frac{d\vec{B}}{dt}$
$\vec{\nabla} \times \vec{E} = - \frac{dE_y}{dz}\hat{x} + \frac{dE_x}{dz}\hat{y} + \frac{dE_y}{dx}\hat{z} - \frac{dE_x}{dy}\hat{z}$

Because I know $\vec{E}$ is in the plane xy
$\vec{\nabla} \times \vec{E} = - \frac{dE_y}{dz}\hat{x} + \frac{dE_x}{dz}\hat{y}$

$-\frac{d\vec{B}}{dt} = - \frac{dE_y}{dz}\hat{x} + \frac{dE_x}{dz}\hat{y}$Thus, I have 2 coordinates for $\vec{B}$ I don't think I can show that $\vec{E}$ and $\vec{B}$
are perpendicular.

 

[Delta2]

Take the most general case for the vector $\vec{k}$ that is that $\vec{k}=k_x\hat x+k_y\hat y+k_z\hat z$
Also $\vec{E_0}=E_{0x}\hat x+E_{0y}\hat y+E_{0z}\hat z$. Also $\vec{r}=x\hat x+y\hat y+z\hat z$.

Then you can prove by using the definition of divergence in cartesian coordinates that $$\nabla\cdot\vec{E}=\vec{E_0}\cdot\vec{k}\cos(\vec{k}\cdot\vec{r}-\omega t)$$ where the $\cdot$ is the dot product of the vectors $\vec{E_0},\vec{k}$.

Also assuming that $\vec{E}$ propagates in free space where $\rho=0$ you 'll have that $$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}=0$$. With the last two equations for divergence you can prove that $\vec{E_0},\vec{k}$ are orthogonal.

 

[vela]

You have $\sin(\vec k \cdot \vec r - \omega t) = \sin(k_x x + k_y y + k_z z - \omega t)$. If you differentiate it with respect to $x$, you get $k_x \cos(\vec k \cdot \vec r - \omega t)$. The differentiation has the effect of pulling out a factor of $k_x$. Similarly, differentiating wrt $y$ and $z$ pulls out factors of $k_y$ and $k_z$ respectively.

You should be able to use that idea to show
$$\nabla \cdot \vec E = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} = (\vec k \cdot \vec E_0) \cos(\vec k \cdot \vec r - \omega t).$$ Similarly, if you calculate $\nabla \times \vec E$, you can show it's equal to $(\vec k \times \vec E_0) \cos(\vec k \cdot \vec r - \omega t).$

There's really no need to make any assumptions about the directions of $\vec k$, $\vec E_0$, and $\vec B_0$.

 

[happyparticle]

Alright,

$(\vec{k} \times \vec{E_0}) cos (\vec{k} \cdot \vec{r} - \omega t) = -\frac{\partial \vec{B}}{\partial t}$

$\vec{B} = \frac{(\vec{k} \times \vec{E_0}) sin(\vec{k} \cdot \vec{r} - \omega t)}{\omega}$

Because $\vec{k} \times \vec{E_0}$ is not equal to 0, $\vec{k}$ and $\vec{E_0}$ are orthogonal.
Since $\vec{B}$ is equal to the cross product between $\vec{k}$ and $\vec{E_0}$, $\vec{B}$ is orthogonal to $\vec{k}$ and $\vec{E_0}$ which means $\vec{B_0}$ is also orthogonal to $\vec{k}$ and $\vec{E_0}$

There's still one more thing.
$\vec{S} = \frac{\vec{E} \times \vec{B}}{\mu_0}$

I have to deduce $\vec{S} \times \vec{k}$ and $\vec{S} \cdot \vec{k}$
I know from faraday's law above that $\vec{S} \times \vec{k} = 0$ because $\vec{E} \times \vec{B}$ are orthogonal to each other and orthogonal to $\vec{k}$

And for the same reason $\vec{S} \cdot \vec{k} \neq 0$

However, I can't show that $\vec{S} = \frac{\vec{E} \times \vec{B}}{\mu_0}$, that $\vec{S}$ is orthogonal to $\vec{E}$ and $\vec{B}$. Where $\vec{B} = \vec{B_0}sin(\vec{k} \cdot \vec{r} - \omega t)$ from the question

I have hard time to figure this out without direction for my vectors.

For instance, if I had $E\hat{x}$,$B\hat{y}$and $k\hat{z}$, then $E\hat{x} \times B\hat{y} = (E\cdot B)\hat{z}$. Then it's easy to show $\vec{S} \times \vec{k}$ and $\vec{S} \cdot \vec{k}$

 

[happyparticle]

I was thinking if $\vec{E} \times \vec{B} = \vec{E_0} sin(\vec{k} \cdot \vec{r} - \omega t) \times \vec{B_0} sin(\vec{k} \cdot \vec{r} - \omega t) = \vec{E_0} \times \vec{B_0} sin (\vec{k} \cdot \vec{r} - \omega t)$

Thus, $\vec{S} = \frac{\vec{E} \times \vec{B}}{\mu_0} = \frac{\vec{E_0} \times \vec{B_0} sin (\vec{k} \cdot \vec{r} - \omega t)}{\mu_0}$

which is perpendicular to $\vec{B_0}$ and $\vec{E_0}$, then $\vec{S} \cdot \vec{k} \neq 0$

 

[Gordianus]

I was thinking if $\vec{E} \times \vec{B} = \vec{E_0} sin(\vec{k} \cdot \vec{r} - \omega t) \times \vec{B_0} sin(\vec{k} \cdot \vec{r} - \omega t) = \vec{E_0} \times \vec{B_0} sin (\vec{k} \cdot \vec{r} - \omega t)$

Thus, $\vec{S} = \frac{\vec{E} \times \vec{B}}{\mu_0} = \frac{\vec{E_0} \times \vec{B_0} sin (\vec{k} \cdot \vec{r} - \omega t)}{\mu_0}$

which is perpendicular to $\vec{B_0}$ and $\vec{E_0}$, then $\vec{S} \cdot \vec{k} \neq 0$

The term $\sin (\vec k \cdot \vec r-\omega t)$ should be squared. This ensures $\vec S$ never reverses direction.

 

[happyparticle]

The term $\sin (\vec k \cdot \vec r-\omega t)$ should be squared. This ensures $\vec S$ never reverses direction.

You are absolutely right, I make so much mistakes.
otherwise is it correct?