Edit to “Statement”

Alexphysics edited Jun 17, 2026, 5:34 AM

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			### 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
		+	face. Electric field strength of wave E. Determine in SI and CGS the linear
			current density in the wall and the wave pressure on it.

			### Solution

			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: $\mathbf{B}_i = E \cos(\omega t + kz)\,\hat{\mathbf{y}}$ .

			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}.

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


			#### Answer

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

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