Question

Show that silver behaves as a good conductor at a frequency of $10^8 s^{-1}$. Also, calculate penetration depth at this frequency given that $\epsilon = \epsilon_0, \mu = \mu_0, \sigma = 3 \times 10^7 mhos/m$.

Answer 1

Silver behaves as a good conductor at $10^8 s^{-1}$ because its conductivity ($\sigma = 3 \times 10^7 \, S/m$) is much greater than $\omega\epsilon_0 \approx 5.56 \times 10^{-3} \, S/m$. The penetration depth is approximately $9.19 \, \mu m$.

Answer 2

A material is a good conductor if its conductivity $\sigma$ is much greater than the product of angular frequency $\omega$ and permittivity $\epsilon$ ($\sigma \gg \omega \epsilon$).

Given frequency $f = 10^8 \, s^{-1}$, so angular frequency $\omega = 2\pi f = 2\pi \times 10^8 \, s^{-1}$. Conductivity $\sigma = 3 \times 10^7 \, S/m$. Permittivity $\epsilon = \epsilon_0 \approx 8.85 \times 10^{-12} \, F/m$.

Calculate $\omega \epsilon_0$: $\omega \epsilon_0 = (2\pi \times 10^8) \times (8.85 \times 10^{-12}) \approx 5.56 \times 10^{-3} \, S/m$ Since $\sigma = 3 \times 10^7 \, S/m \gg 5.56 \times 10^{-3} \, S/m$, silver is a good conductor.

The penetration depth $\delta$ is given by: $\delta = \sqrt{\frac{2}{\omega \mu \sigma}}$ Using $\mu = \mu_0 = 4\pi \times 10^{-7} \, H/m$: $\omega \mu_0 \sigma = (2\pi \times 10^8) \times (4\pi \times 10^{-7}) \times (3 \times 10^7) \approx 2.37 \times 10^{10} \, m^{-2}$ $\delta = \sqrt{\frac{2}{2.37 \times 10^{10}}} \approx \sqrt{0.844 \times 10^{-10}} \approx 0.919 \times 10^{-5} \, m = 9.19 \, \mu m$