Solve Media Homework: Relative Permittivity of Non-dissipative Medium

[Maat]

Homework Statement

A uniform plane wave propagates in a non-dissipative medium in the positive z direction. The frequency of the wave is 20 MHz. A probe located at z = 0 measures the phase of the wave to be 98 degrees. An identical probe located at z = 2 m measures the phase to be -15 degrees. What is the relative permittivity (e/e0) of the medium.

Homework Equations

k = w*sqrt(u*e)

w = 2*pi*f

wt - kz = a constant

w is angle velocity, u is permeability, e is permittivity

The Attempt at a Solution

wt = phase angle

Phase angle = kz

K2 = -15/2 = -7.5

w = 2*pi*(20*10^6)

K2 = w*sqrt(u*e), solve for e (e2)

e2 = 2.83 * 10^-9 F/m

Now I'm stuck on how to get the other e since z = 0. I think this problem should be easy, but I'm not seeing something.

 

[rude man]

Why is your topic "conducting" media? It's not conducting. Conducting media dissipate. Yours is stated to be "non-dissipative".

Homework Statement

k = w*sqrt(u*e)
Phase angle = kz
w = 2*pi*(20*10^6)

All correct. (Some of the other stuff you posted wasn't, but never mind).

So how about equating delta degrees to delta kz?

 

[Maat]

If I equated the deltas, how would I get two different e's? My thought was to get two different k's and from that find the two e's to compare.

 

[rude man]

If I equated the deltas, how would I get two different e's? My thought was to get two different k's and from that find the two e's to compare.

Why are you looking for two different epsilons? there is only one epsilon.
There is also only one k.
There are however two z's and two phase angles.

 

[Maat]

Ah is that because the other epsilon is just e0?

 

[Maat]

Ok I see now. Thank you very much.