Правка разделов «Statement», «Solution», «Answer»

Alexphysics правка от 17 июн. 2026 г., 05:34

правка #18968 предыдущая #18967 ← раньше позже →

en/12.1.21.md +10 −4

		@@ -1,19 +1,23 @@
			### Statement

		−	$12.1.21.$ ~~[Insert the problem statement]~~
		+	$12.1.21.$

		+	A plane electromagnetic wave falls on a metal wall perpendicular to its sur-
		+	face. Electric field strength of wave E. Determine in SI and GHS the linear
		+	current density in the wall and the wave pressure on it.
		+
			### Solution

		−	1. Configuration and fields
		+	Configuration and fields

			We take the z-axis perpendicular to the wall, with the metal occupying z < 0 and the vacuum in z > 0. The wave is incident from the right.

			Incident wave (traveling in the -z direction):

			$\mathbf{E}_i = E \cos(\omega t + kz)\,\hat{\mathbf{x}}, \qquad
			\mathbf{B}_i = \frac{E}{c} \cos(\omega t + kz)\,\hat{\mathbf{y}} \quad (\text{SI})$

		−	In CGS: $\mathbf{B}_i = E \cos(\omega t + kz)\,\hat{\mathbf{y}} ~~(same units as E)$~~.
		+	In CGS: $\mathbf{B}_i = E \cos(\omega t + kz)\,\hat{\mathbf{y}}$ .

			Reflected wave (from a perfect conductor):

			$\mathbf{E}_r = -E \cos(\omega t - kz)\,\hat{\mathbf{x}}, \qquad
			\mathbf{B}_r = \frac{E}{c} \cos(\omega t - kz)\,\hat{\mathbf{y}} \quad (\text{SI})$.

			In CGS:$ \mathbf{B}_r = E \cos(\omega t - kz)\,\hat{\mathbf{y}}$.

			At the surface (z = 0):

			$\mathbf{E}_{\text{total}} = E \cos\omega t - E \cos\omega t = 0 \quad (\text{node, electric field is zero})$.

			$\mathbf{B}_{\text{total}} = \frac{E}{c}\cos\omega t + \frac{E}{c}\cos\omega t = \frac{2E}{c}\cos\omega t\,\hat{\mathbf{y}} \quad (\text{SI}), \qquad
			\mathbf{B}_{\text{total}} = 2E\cos\omega t\,\hat{\mathbf{y}} \quad (\text{CGS})$

			At the surface of a perfect conductor, the magnetic field has a discontinuity related to the surface current $\mathbf{K}$

			In SI:

			$\mathbf{n} \times (\mathbf{H}_{\text{vacuum}} - \mathbf{H}_{\text{metal}}) = \mathbf{K}$

			The normal vector \mathbf{n} points from the metal into the vacuum:
			$\mathbf{n} = \hat{\mathbf{z}}$.

			Inside the metal

			$\mathbf{H} = 0$.

			In vacuum

			$\mathbf{H} = \mathbf{B}/\mu_0$.

			$\mathbf{H}_{\text{vacuum}} = \frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{y}}$.

			Then:

			$\mathbf{K} = \hat{\mathbf{z}} \times \left( \frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{y}} \right) = -\frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{x}}.$

			The amplitude (linear current density) is:

			$\boxed{i = \frac{2E}{\mu_0 c} = 2\varepsilon_0 c E \quad (\text{SI}).}$

			In CGS:

			$\mathbf{n} \times (\mathbf{B}_{\text{vacuum}} - \mathbf{B}_{\text{metal}}) = \frac{4\pi}{c}\mathbf{K}$

			With
			$\mathbf{B}_{\text{vacuum}} = 2E\cos\omega t\,\hat{\mathbf{y}}, \mathbf{B}_{\text{metal}} = 0, \mathbf{n} = \hat{\mathbf{z}}$:

			$\hat{\mathbf{z}} \times (2E\cos\omega t\,\hat{\mathbf{y}}) = -\frac{4\pi}{c}\mathbf{K} \quad\Rightarrow\quad \mathbf{K} = -\frac{cE}{2\pi}\cos\omega t\,\hat{\mathbf{x}}$

			Amplitude:

			$\boxed{i = \frac{cE}{2\pi} \quad (\text{CGS}).}$

			The average pressure on a perfect reflector is$ p = 2I/c$ where I is the average incident intensity.

			In SI:
			Average intensity: $I = \frac{1}{2}\varepsilon_0 c E^2$

			$p = \frac{2I}{c} = \varepsilon_0 E^2$

			$\boxed{p = \varepsilon_0 E^2 \quad (\text{SI}).}$

			In CGS:
			Average intensity:
			$ I = \frac{c}{8\pi} E^2$

			p = \frac{2I}{c} = \frac{E^2}{4\pi}.

			$\boxed{p = \frac{E^2}{4\pi} \quad (\text{CGS}).}$

		@@ -89,4 +93,6 @@Solution

			#### Answer

		−	[Insert a concise answer or boxed result]
		+	$\boxed{p = \varepsilon_0 E^2 \quad (\text{SI}).}$
		+
		+	$\boxed{p = \frac{E^2}{4\pi} \quad (\text{CGS}).}$

	@@ -1,19 +1,23 @@
	### Statement		### Statement

	$12.1.21.$ ~~[Insert the problem statement]~~		$12.1.21.$

			A plane electromagnetic wave falls on a metal wall perpendicular to its sur-
			face. Electric field strength of wave E. Determine in SI and GHS the linear
			current density in the wall and the wave pressure on it.

	### Solution		### Solution

	1. Configuration and fields		Configuration and fields

	We take the z-axis perpendicular to the wall, with the metal occupying z < 0 and the vacuum in z > 0. The wave is incident from the right.		We take the z-axis perpendicular to the wall, with the metal occupying z < 0 and the vacuum in z > 0. The wave is incident from the right.

	Incident wave (traveling in the -z direction):		Incident wave (traveling in the -z direction):

	$\mathbf{E}_i = E \cos(\omega t + kz)\,\hat{\mathbf{x}}, \qquad		$\mathbf{E}_i = E \cos(\omega t + kz)\,\hat{\mathbf{x}}, \qquad
	\mathbf{B}_i = \frac{E}{c} \cos(\omega t + kz)\,\hat{\mathbf{y}} \quad (\text{SI})$		\mathbf{B}_i = \frac{E}{c} \cos(\omega t + kz)\,\hat{\mathbf{y}} \quad (\text{SI})$

	In CGS: $\mathbf{B}_i = E \cos(\omega t + kz)\,\hat{\mathbf{y}} ~~(same units as E)$~~.		In CGS: $\mathbf{B}_i = E \cos(\omega t + kz)\,\hat{\mathbf{y}}$ .

	Reflected wave (from a perfect conductor):		Reflected wave (from a perfect conductor):

	$\mathbf{E}_r = -E \cos(\omega t - kz)\,\hat{\mathbf{x}}, \qquad		$\mathbf{E}_r = -E \cos(\omega t - kz)\,\hat{\mathbf{x}}, \qquad
	\mathbf{B}_r = \frac{E}{c} \cos(\omega t - kz)\,\hat{\mathbf{y}} \quad (\text{SI})$.		\mathbf{B}_r = \frac{E}{c} \cos(\omega t - kz)\,\hat{\mathbf{y}} \quad (\text{SI})$.

	In CGS:$ \mathbf{B}_r = E \cos(\omega t - kz)\,\hat{\mathbf{y}}$.		In CGS:$ \mathbf{B}_r = E \cos(\omega t - kz)\,\hat{\mathbf{y}}$.

	At the surface (z = 0):		At the surface (z = 0):

	$\mathbf{E}_{\text{total}} = E \cos\omega t - E \cos\omega t = 0 \quad (\text{node, electric field is zero})$.		$\mathbf{E}_{\text{total}} = E \cos\omega t - E \cos\omega t = 0 \quad (\text{node, electric field is zero})$.

	$\mathbf{B}_{\text{total}} = \frac{E}{c}\cos\omega t + \frac{E}{c}\cos\omega t = \frac{2E}{c}\cos\omega t\,\hat{\mathbf{y}} \quad (\text{SI}), \qquad		$\mathbf{B}_{\text{total}} = \frac{E}{c}\cos\omega t + \frac{E}{c}\cos\omega t = \frac{2E}{c}\cos\omega t\,\hat{\mathbf{y}} \quad (\text{SI}), \qquad
	\mathbf{B}_{\text{total}} = 2E\cos\omega t\,\hat{\mathbf{y}} \quad (\text{CGS})$		\mathbf{B}_{\text{total}} = 2E\cos\omega t\,\hat{\mathbf{y}} \quad (\text{CGS})$

	At the surface of a perfect conductor, the magnetic field has a discontinuity related to the surface current $\mathbf{K}$		At the surface of a perfect conductor, the magnetic field has a discontinuity related to the surface current $\mathbf{K}$

	In SI:		In SI:

	$\mathbf{n} \times (\mathbf{H}_{\text{vacuum}} - \mathbf{H}_{\text{metal}}) = \mathbf{K}$		$\mathbf{n} \times (\mathbf{H}_{\text{vacuum}} - \mathbf{H}_{\text{metal}}) = \mathbf{K}$

	The normal vector \mathbf{n} points from the metal into the vacuum:		The normal vector \mathbf{n} points from the metal into the vacuum:
	$\mathbf{n} = \hat{\mathbf{z}}$.		$\mathbf{n} = \hat{\mathbf{z}}$.

	Inside the metal		Inside the metal

	$\mathbf{H} = 0$.		$\mathbf{H} = 0$.

	In vacuum		In vacuum

	$\mathbf{H} = \mathbf{B}/\mu_0$.		$\mathbf{H} = \mathbf{B}/\mu_0$.

	$\mathbf{H}_{\text{vacuum}} = \frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{y}}$.		$\mathbf{H}_{\text{vacuum}} = \frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{y}}$.

	Then:		Then:

	$\mathbf{K} = \hat{\mathbf{z}} \times \left( \frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{y}} \right) = -\frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{x}}.$		$\mathbf{K} = \hat{\mathbf{z}} \times \left( \frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{y}} \right) = -\frac{2E}{\mu_0 c}\cos\omega t\,\hat{\mathbf{x}}.$

	The amplitude (linear current density) is:		The amplitude (linear current density) is:

	$\boxed{i = \frac{2E}{\mu_0 c} = 2\varepsilon_0 c E \quad (\text{SI}).}$		$\boxed{i = \frac{2E}{\mu_0 c} = 2\varepsilon_0 c E \quad (\text{SI}).}$

	In CGS:		In CGS:

	$\mathbf{n} \times (\mathbf{B}_{\text{vacuum}} - \mathbf{B}_{\text{metal}}) = \frac{4\pi}{c}\mathbf{K}$		$\mathbf{n} \times (\mathbf{B}_{\text{vacuum}} - \mathbf{B}_{\text{metal}}) = \frac{4\pi}{c}\mathbf{K}$

	With		With
	$\mathbf{B}_{\text{vacuum}} = 2E\cos\omega t\,\hat{\mathbf{y}}, \mathbf{B}_{\text{metal}} = 0, \mathbf{n} = \hat{\mathbf{z}}$:		$\mathbf{B}_{\text{vacuum}} = 2E\cos\omega t\,\hat{\mathbf{y}}, \mathbf{B}_{\text{metal}} = 0, \mathbf{n} = \hat{\mathbf{z}}$:

	$\hat{\mathbf{z}} \times (2E\cos\omega t\,\hat{\mathbf{y}}) = -\frac{4\pi}{c}\mathbf{K} \quad\Rightarrow\quad \mathbf{K} = -\frac{cE}{2\pi}\cos\omega t\,\hat{\mathbf{x}}$		$\hat{\mathbf{z}} \times (2E\cos\omega t\,\hat{\mathbf{y}}) = -\frac{4\pi}{c}\mathbf{K} \quad\Rightarrow\quad \mathbf{K} = -\frac{cE}{2\pi}\cos\omega t\,\hat{\mathbf{x}}$

	Amplitude:		Amplitude:

	$\boxed{i = \frac{cE}{2\pi} \quad (\text{CGS}).}$		$\boxed{i = \frac{cE}{2\pi} \quad (\text{CGS}).}$

	The average pressure on a perfect reflector is$ p = 2I/c$ where I is the average incident intensity.		The average pressure on a perfect reflector is$ p = 2I/c$ where I is the average incident intensity.

	In SI:		In SI:
	Average intensity: $I = \frac{1}{2}\varepsilon_0 c E^2$		Average intensity: $I = \frac{1}{2}\varepsilon_0 c E^2$

	$p = \frac{2I}{c} = \varepsilon_0 E^2$		$p = \frac{2I}{c} = \varepsilon_0 E^2$

	$\boxed{p = \varepsilon_0 E^2 \quad (\text{SI}).}$		$\boxed{p = \varepsilon_0 E^2 \quad (\text{SI}).}$

	In CGS:		In CGS:
	Average intensity:		Average intensity:
	$ I = \frac{c}{8\pi} E^2$		$ I = \frac{c}{8\pi} E^2$

	p = \frac{2I}{c} = \frac{E^2}{4\pi}.		p = \frac{2I}{c} = \frac{E^2}{4\pi}.

	$\boxed{p = \frac{E^2}{4\pi} \quad (\text{CGS}).}$		$\boxed{p = \frac{E^2}{4\pi} \quad (\text{CGS}).}$

	@@ -89,4 +93,6 @@Solution

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		$\boxed{p = \varepsilon_0 E^2 \quad (\text{SI}).}$

			$\boxed{p = \frac{E^2}{4\pi} \quad (\text{CGS}).}$