Edits to “Statement”, “Answer”
en/12.1.18.md
+6 −2
| @@ -1,7 +1,11 @@ | |||
| ### Statement | |||
| − | $12.1.18.$ | ||
| + | $12.1.18.$ | ||
| + | Estimate the depth of penetration of an electromagnetic wave perpendicular | ||
| + | to its surface into the conductor. The wave frequency v = 1015Hz, the number | ||
| + | of conduction electrons per unit volume ne= 1022cm−3. | ||
| + | |||
| ### Solution | |||
| To estimate the penetration depth of an electromagnetic wave, the free electron plasma model is used. | |||
| At optical frequencies, the permittivity of the conductor is | |||
| $\varepsilon(\omega) = 1 - \frac{\omega_p^2}{\omega^2}, | |||
| \qquad \omega = 2\pi\nu$, | |||
| where the plasma frequency in SI is | |||
| $\omega_p = \sqrt{\frac{n_e e^2}{\varepsilon_0 m_e}}$. | |||
| With e = $1.6\times10^{-19}\,\text{C}, \varepsilon_0 = 8.85\times10^{-12}\,\text{F/m}, m_e = 9.1\times10^{-31}\,\text{kg}$: | |||
| $\omega_p^2 = \frac{10^{28}(1.6\times10^{-19})^2}{8.85\times10^{-12}\times9.1\times10^{-31}} \approx 3.2\times10^{31}\,\text{rad}^2/\text{s}^2, | |||
| \quad \omega_p \approx 5.6\times10^{15}\,\text{rad/s}$ | |||
| $Since \omega = 2\pi\times10^{15} \approx 6.3\times10^{15}\,\text{rad/s} > \omega_p$ | |||
| the wave propagates with attenuation. The penetration depth \delta in a conductor is given by the damping length of the field intensity: | |||
| $\delta = \frac{c}{\sqrt{\omega_p^2 - \omega^2}} \approx \frac{c}{\omega_p} | |||
| \quad (\text{since } \omega \sim \omega_p)$ | |||
| Numerically, | |||
| $\delta \approx \frac{3\times10^{10}\,\text{cm/s}}{5.6\times10^{15}\,\text{s}^{-1}} \approx 5.4\times10^{-6}\,\text{cm}$ | |||
| $\boxed{\delta \approx 5\times10^{-6}\,\text{cm} \; (= 50\,\text{nm})}$. | |||
| @@ -34,4 +38,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{\delta \approx 5\times10^{-6}\,\text{cm} \; (= 50\,\text{nm})}$. | ||
| @@ -1,7 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $12.1.18.$ |
$12.1.18.$ | ||
| Estimate the depth of penetration of an electromagnetic wave perpendicular | |||
| to its surface into the conductor. The wave frequency v = 1015Hz, the number | |||
| of conduction electrons per unit volume ne= 1022cm−3. | |||
| ### Solution | ### Solution | ||
| To estimate the penetration depth of an electromagnetic wave, the free electron plasma model is used. | To estimate the penetration depth of an electromagnetic wave, the free electron plasma model is used. | ||
| At optical frequencies, the permittivity of the conductor is | At optical frequencies, the permittivity of the conductor is | ||
| $\varepsilon(\omega) = 1 - \frac{\omega_p^2}{\omega^2}, | $\varepsilon(\omega) = 1 - \frac{\omega_p^2}{\omega^2}, | ||
| \qquad \omega = 2\pi\nu$, | \qquad \omega = 2\pi\nu$, | ||
| where the plasma frequency in SI is | where the plasma frequency in SI is | ||
| $\omega_p = \sqrt{\frac{n_e e^2}{\varepsilon_0 m_e}}$. | $\omega_p = \sqrt{\frac{n_e e^2}{\varepsilon_0 m_e}}$. | ||
| With e = $1.6\times10^{-19}\,\text{C}, \varepsilon_0 = 8.85\times10^{-12}\,\text{F/m}, m_e = 9.1\times10^{-31}\,\text{kg}$: | With e = $1.6\times10^{-19}\,\text{C}, \varepsilon_0 = 8.85\times10^{-12}\,\text{F/m}, m_e = 9.1\times10^{-31}\,\text{kg}$: | ||
| $\omega_p^2 = \frac{10^{28}(1.6\times10^{-19})^2}{8.85\times10^{-12}\times9.1\times10^{-31}} \approx 3.2\times10^{31}\,\text{rad}^2/\text{s}^2, | $\omega_p^2 = \frac{10^{28}(1.6\times10^{-19})^2}{8.85\times10^{-12}\times9.1\times10^{-31}} \approx 3.2\times10^{31}\,\text{rad}^2/\text{s}^2, | ||
| \quad \omega_p \approx 5.6\times10^{15}\,\text{rad/s}$ | \quad \omega_p \approx 5.6\times10^{15}\,\text{rad/s}$ | ||
| $Since \omega = 2\pi\times10^{15} \approx 6.3\times10^{15}\,\text{rad/s} > \omega_p$ | $Since \omega = 2\pi\times10^{15} \approx 6.3\times10^{15}\,\text{rad/s} > \omega_p$ | ||
| the wave propagates with attenuation. The penetration depth \delta in a conductor is given by the damping length of the field intensity: | the wave propagates with attenuation. The penetration depth \delta in a conductor is given by the damping length of the field intensity: | ||
| $\delta = \frac{c}{\sqrt{\omega_p^2 - \omega^2}} \approx \frac{c}{\omega_p} | $\delta = \frac{c}{\sqrt{\omega_p^2 - \omega^2}} \approx \frac{c}{\omega_p} | ||
| \quad (\text{since } \omega \sim \omega_p)$ | \quad (\text{since } \omega \sim \omega_p)$ | ||
| Numerically, | Numerically, | ||
| $\delta \approx \frac{3\times10^{10}\,\text{cm/s}}{5.6\times10^{15}\,\text{s}^{-1}} \approx 5.4\times10^{-6}\,\text{cm}$ | $\delta \approx \frac{3\times10^{10}\,\text{cm/s}}{5.6\times10^{15}\,\text{s}^{-1}} \approx 5.4\times10^{-6}\,\text{cm}$ | ||
| $\boxed{\delta \approx 5\times10^{-6}\,\text{cm} \; (= 50\,\text{nm})}$. | $\boxed{\delta \approx 5\times10^{-6}\,\text{cm} \; (= 50\,\text{nm})}$. | ||
| @@ -34,4 +38,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{\delta \approx 5\times10^{-6}\,\text{cm} \; (= 50\,\text{nm})}$. | ||