Новое решение

Alexphysics правка от 14 июн. 2026 г., 23:42

правка #18786 позже →

en/6.3.17.md +42 −0

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		+	### Statement
		+
		+	$6.3.17.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Given that L \gg R , we neglect edge effects in the middle region. The system behaves as infinitely long. Thus
		+
		+	$E= E(x)\hat{x}
		+
		+	The "cylindrical shield" of radius R acts as an outer conductor that shields the field.
		+
		+	The charge +q in the cavity induces a charge –q on the inner surface (ρ = r).
		+
		+	Since the conductor is neutral, a charge +q appears on its outer surface.
		+
		+	"We choose a cylindrical Gaussian surface of radius \rho and length L, coaxial with the system.
		+
		+	\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}
		+
		+	E \cdot 2\pi \rho L = \frac{Q_{\text{enc}}}{\varepsilon_0}
		+
		+	E(\rho) = \frac{Q_{\text{enc}}}{2\pi \varepsilon_0 L \rho}
		+	\]"
		+
		+	For r < \rho < R :
		+
		+	Q_{\text{enc}} = q.
		+
		+	E(\rho) = \frac{q}{2\pi \varepsilon_0 L \rho}
		+
		+	For \rho > R :
		+
		+	Q_{\text{enc}} = 0 \quad \rightarrow \quad E = 0.
		+
		+	If the shield were isolated and uncharged, the enclosed charge would be q and the field would remain as E = \frac{q}{2\pi \varepsilon_0 L \rho} .
		+
		+	E = 0 for \rho > R .
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,42 @@
			### Statement

			$6.3.17.$ [Insert the problem statement]

			### Solution

			Given that L \gg R , we neglect edge effects in the middle region. The system behaves as infinitely long. Thus

			$E= E(x)\hat{x}

			The "cylindrical shield" of radius R acts as an outer conductor that shields the field.

			The charge +q in the cavity induces a charge –q on the inner surface (ρ = r).

			Since the conductor is neutral, a charge +q appears on its outer surface.

			"We choose a cylindrical Gaussian surface of radius \rho and length L, coaxial with the system.

			\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}

			E \cdot 2\pi \rho L = \frac{Q_{\text{enc}}}{\varepsilon_0}

			E(\rho) = \frac{Q_{\text{enc}}}{2\pi \varepsilon_0 L \rho}
			\]"

			For r < \rho < R :

			Q_{\text{enc}} = q.

			E(\rho) = \frac{q}{2\pi \varepsilon_0 L \rho}

			For \rho > R :

			Q_{\text{enc}} = 0 \quad \rightarrow \quad E = 0.

			If the shield were isolated and uncharged, the enclosed charge would be q and the field would remain as E = \frac{q}{2\pi \varepsilon_0 L \rho} .

			E = 0 for \rho > R .

			#### Answer

			[Insert a concise answer or boxed result]