Uniform plane wave (help with a short walkthrough for a worked problem)

[FrogPad]

I don't understand something on this problem:

Q: A uniform plane wave with $\vec E = \hat x E_x$ propagates in a lossless simple medium ($\epsilon_r = 4 \,\,\,\, \mu_r = 1 \,\,\,\, \sigma = 0$) in the +z-direction. Assume that $E_x$ is sinusoidal with a frequency of 100 (MHz) and has a maximum value of $+10^{-4}$ (V/m) at $t=0$ and $z = \frac{1}{8}$ (m)

Write the instantaneous expression for $\vec E$ for any t and z.

A:
Using $\cos \omega t$ as a reference, we find the instantaneous expression for $\vec E$ to be:

$$\vec E(z,t) = \hat x E_x = \hat x 10^{-4} \cos ( 2 \pi 10^8 t - kz + \psi)$$

Since $E_x$ equals $+10^{-4}$ when the argument of the cosine function equals zero, that is when:

$$2 \pi 10^8 - kz + \psi = 0$$

we have at $t = 0$, $z = \frac{1}{8}$

$$\psi = kz = \frac{4 \pi}{3} \frac{1}{8} = \frac{\pi}{6}$$

My question follows from this:
Why do we introduce the phase unknown phase $\psi$ and set $E_x$ according to this. Why couldn't we do this instead,

A:
Using $\cos \omega t$ as a reference, we find the instantaneous expression for $\vec E$ to be:

$$\vec E(z,t) = \hat x E_x = \hat x E_0 \cos ( 2 \pi 10^8 t - kz)$$

Set the according t=0, z=1/8 so that,
$$\hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4}$$ and meet the initial condition of a max of 10^-4 with $E_0$ instead of setting E_0 originally and changing the phase.

 

[berkeman]

I forget -- what is k in terms of lambda? It seems like the phi and the kz can influence the final equation under different constraints -- I could be wrong of course.

 

[OlderDan]

I don't understand something on this problem:

Q: A uniform plane wave with $\vec E = \hat x E_x$ propagates in a lossless simple medium ($\epsilon_r = 4 \,\,\,\, \mu_r = 1 \,\,\,\, \sigma = 0$) in the +z-direction. Assume that $E_x$ is sinusoidal with a frequency of 100 (MHz) and has a maximum value of $+10^{-4}$ (V/m) at $t=0$ and $z = \frac{1}{8}$ (m)

Write the instantaneous expression for $\vec E$ for any t and z.

A:
Using $\cos \omega t$ as a reference, we find the instantaneous expression for $\vec E$ to be:

$$\vec E(z,t) = \hat x E_x = \hat x 10^{-4} \cos ( 2 \pi 10^8 t - kz + \psi)$$

Since $E_x$ equals $+10^{-4}$ when the argument of the cosine function equals zero, that is when:

$$2 \pi 10^8 - kz + \psi = 0$$

we have at $t = 0$, $z = \frac{1}{8}$

$$\psi = kz = \frac{4 \pi}{3} \frac{1}{8} = \frac{\pi}{6}$$

My question follows from this:
Why do we introduce the phase unknown phase $\psi$ and set $E_x$ according to this. Why couldn't we do this instead,

A:
Using $\cos \omega t$ as a reference, we find the instantaneous expression for $\vec E$ to be:

$$\vec E(z,t) = \hat x E_x = \hat x E_0 \cos ( 2 \pi 10^8 t - kz)$$

Set the according t=0, z=1/8 so that,
$$\hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4}$$ and meet the initial condition of a max of 10^-4 with $E_0$ instead of setting E_0 originally and changing the phase.

If you used

$$\hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4}$$

the maximum would not be 10^-4 and at t = 0 the maximum would be at z = 0.

 

[FrogPad]

If you used

$$\hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4}$$

the maximum would not be 10^-4 and at t = 0 the maximum would be at z = 0.

oooooooooooooooooohhhh :)

Thanks man, that definitely makes sense! I really appreciate it.

berkeman:

$$k = \omega \sqrt{\mu \epsilon}$$