Question

2. b) A surface wave with frequency w = 3 × 10−5 rad s−1 is propagating at a metal-free
space boundary. The plasma frequency of metal is 9 × 1015 rad and er = 7. Estimate
k. If the amplitude of the wave is 105 Vm−1 at the surface, what will be its value at
the depth of 0.2 μm in the metal? Ignore the collision effect.

Answer 1

Answer on Question #80087, Physics / Other

2. b) A surface wave with frequency $w = 3 \times 10^{-5}$ rad s$^{-1}$ is propagating at a metal-free space boundary. The plasma frequency of metal is $9 \times 10^{15}$ rad and $\varepsilon_{r} = 7$. Estimate k. If the amplitude of the wave is $10^{5} \, \text{Vm}^{-1}$ at the surface, what will be its value at the depth of $0.2 \, \mu \text{m}$ in the metal? Ignore the collision effect.

Solution:

Dispersion relation:

$$\omega = \sqrt{\omega_{P}^{2} + c^{2} k^{2}}$$

So,

$$k = \sqrt{\frac{\omega^{2} - \omega_{P}^{2}}{c^{2}}} = \sqrt{\frac{(3 \times 10^{-5})^{2} - (9 \times 10^{15})^{2}}{(3 \times 10^{8})^{2}}} = 3 \times 10^{7} \, \text{i} \, \text{m}^{-1}$$

K is imaginary and the wave decays exponentially with distance inside the metal.

$$E(x) = E_{0} e^{ikx}$$

So, for $x = 0.2 \, \mu \text{m}$

$$E(x) = 10^{5} e^{-3 \times 10^{7} \times 0.2 \times 10^{-6}} = 247.9 \, \text{V/m}$$

Answer: 247.9 V/m

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