What is the Relationship Between EM Wave Velocity and Electric Field in Space?

[kasse]

Homework Statement

Find the velocity of EM waves as a function of $$\epsilon_{0}$$ and $$\mu_{0}$$

2. The attempt at a solution

$$E = E_{0}cos(kx-\omega t)$$

Using $$v= \frac{\omega}{k}$$

 

[gabbagabbahey]

When you used Maxwell's equations to derive the wave equation, you should have ended up with an answer to this

 

[kasse]

No, I didn't. But I can substitute my expression for E into the wave equation. What is $$\vec{\nabla}^{2}E$$?

$$\frac{\partial^{2}E}{\partial x^{2}} + \frac{\partial^{2}E}{\partial t^{2}}$$?

 

[gabbagabbahey]

No, I didn't.

Last time I checked, Maxwell's equations were in terms of [itex]\epsilon_0[/tex] and $\mu_0$ not $c$; so you should have ended up with a wave equation where the propagation speed is in terms of [itex]\epsilon_0[/tex] and $\mu_0$...if you didn't, then you did something wrong...I think you should go back to that problem and show me your work.[/itex][/itex]

 

[kasse]

double

 

[kasse]

Of course...

So $$\vec{\nabla}^{2}E$$ = $$\frac{\partial^{2}E}{\partial x^{2}} + \frac{\partial^{2}E}{\partial y^{2}} + \frac{\partial^{2}E}{\partial z^{2}}$$ (only spatial dimension, not time)?

 

[gabbagabbahey]

Did you even read my last post?

 

[kasse]

Yes.$$\frac{1}{v^{2}} = \mu_{0}\epsilon_{0}$$, so $$\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}} = v$$. That's what you meant, right?

That would mean that (if I substitute my expression for E into the wave equation) $$\vec{\nabla}^{2}E = \frac{\partial^{2}E}{\partial x^{2}} + \frac{\partial^{2}E}{\partial y^{2}} + \frac{\partial^{2}E}{\partial z^{2}}$$.

Can I also write $$\vec{\nabla}^{2}E = \frac{\partial^{2}E}{\partial \vec{r}^{2}}$$?

 

[gabbagabbahey]

Yes.$$\frac{1}{v^{2}} = \mu_{0}\epsilon_{0}$$, so $$\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}} = v$$. That's what you meant, right?

Yes.

That would mean that (if I substitute my expression for E into the wave equation) $$\vec{\nabla}^{2}E = \frac{\partial^{2}E}{\partial x^{2}} + \frac{\partial^{2}E}{\partial y^{2}} + \frac{\partial^{2}E}{\partial z^{2}}$$.

Can I also write $$\vec{\nabla}^{2}E = \frac{\partial^{2}E}{\partial \vec{r}^{2}}$$?

First, the electric field is vector, not a scalar so this relation is incorrect...second what does this have to do with finding v...or anything else for that matter?