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FrogPad

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

Write the instantaneous expression for [itex] \vec E [/itex] for any t and z.

A:

Using [itex] \cos \omega t [/itex] as a reference, we find the instantaneous expression for [itex] \vec E [/itex] to be:

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

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

[tex] 2 \pi 10^8 - kz + \psi = 0 [/tex]

we have at [itex] t = 0 [/itex], [itex] z = \frac{1}{8} [/itex]

[tex] \psi = kz = \frac{4 \pi}{3} \frac{1}{8} = \frac{\pi}{6} [/tex]

**My question**follows from this:

Why do we introduce the phase unknown phase [itex] \psi [/itex] and set [itex] E_x [/itex] according to this. Why couldn't we do this instead,

A:

Using [itex] \cos \omega t [/itex] as a reference, we find the instantaneous expression for [itex] \vec E [/itex] to be:

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

Set the according t=0, z=1/8 so that,

[tex] \hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4}[/tex] and meet the initial condition of a max of 10^-4 with [itex] E_0 [/itex] instead of setting E_0 originally and changing the phase.