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JAMF правка от 16 июн. 2026 г., 01:07

правка #18873 предыдущая #18871 ← раньше позже →

en/12.1.22.md +36 −0

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		+	### Statement
		+
		+	$12.1.22.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	$12.1.22$ The amplitude of the electric field strength of a plane sinusoidal wave is equal
		+	to $E_0$. What is the average pressure exerted by this wave on a flat metal wall
		+	when it is normally incident on it?
		+
		+	The energy per unit area per unit time striking the metal wall is equal to the Poynting vector:
		+
		+	\begin{equation}
		+	\vec{S} = c^2 \epsilon_0 (\vec{E} \times \vec{B})
		+	\end{equation}
		+
		+	For this case:
		+
		+	\begin{equation}
		+	\vec{E} = \hat{y} E_0 \cos(\omega t) \quad \text{and} \quad \vec{B} = \hat{z} \frac{E_0}{c} \cos(\omega t)
		+	\end{equation}
		+	The pressure exerted by the incoming flow on the wall is:
		+
		+	\begin{equation}
		+	P = \frac{\left\langle S\right\rangle_T }{c} = c \epsilon_0 \left\langle \vec{E} \times \vec{B}\right\rangle_T = \frac{\epsilon_0 E_0^2}{2}
		+	\end{equation}
		+	In this equation we have taken the time average of the cross product.
		+	This is half of the answer, because there is a pressure exerted by the incoming flow and another equal pressure from the outgoing flow; our answer must be twice that for this reason:
		+
		+	\begin{equation}
		+	P_{\text{total}} = \epsilon_0 E_0^2
		+	\end{equation}
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,36 @@
			### Statement

			$12.1.22.$ [Insert the problem statement]

			### Solution

			$12.1.22$ The amplitude of the electric field strength of a plane sinusoidal wave is equal
			to $E_0$. What is the average pressure exerted by this wave on a flat metal wall
			when it is normally incident on it?

			The energy per unit area per unit time striking the metal wall is equal to the Poynting vector:

			\begin{equation}
			\vec{S} = c^2 \epsilon_0 (\vec{E} \times \vec{B})
			\end{equation}

			For this case:

			\begin{equation}
			\vec{E} = \hat{y} E_0 \cos(\omega t) \quad \text{and} \quad \vec{B} = \hat{z} \frac{E_0}{c} \cos(\omega t)
			\end{equation}
			The pressure exerted by the incoming flow on the wall is:

			\begin{equation}
			P = \frac{\left\langle S\right\rangle_T }{c} = c \epsilon_0 \left\langle \vec{E} \times \vec{B}\right\rangle_T = \frac{\epsilon_0 E_0^2}{2}
			\end{equation}
			In this equation we have taken the time average of the cross product.
			This is half of the answer, because there is a pressure exerted by the incoming flow and another equal pressure from the outgoing flow; our answer must be twice that for this reason:

			\begin{equation}
			P_{\text{total}} = \epsilon_0 E_0^2
			\end{equation}

			#### Answer

			[Insert a concise answer or boxed result]