Правка разделов «Statement», «Solution», «Answer»
en/6.3.17.md
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| @@ -1,12 +1,16 @@ | |||
| ### Statement | |||
| − | $6.3.17.$ | ||
| + | $6.3.17.$ Inside the cavity of a long uncharged conductor, the radius of which is r, there | ||
| + | is a charge q. The conductor is surrounded by a cylindrical screen of radius | ||
| + | R. Conductor length L ≫ R. How does the electric field strength outside the | ||
| + | cavity in the middle part of the system depend on the distance to the axis of | ||
| + | this system? | ||
| ### Solution | |||
| − | Given that L | ||
| + | Given that L ≫ R , we neglect edge effects in the middle region. The system behaves as infinitely long. Thus | ||
| − | $E= E(x)\hat{x} | ||
| + | $E= E(x)\hat{x}$ | ||
| 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. | |||
| @@ -16,27 +20,40 @@Solution | |||
| "We choose a cylindrical Gaussian surface of radius \rho and length L, coaxial with the system. | |||
| − | |||
| + | ${E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$ | ||
| − | E \cdot 2\pi \rho L = \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} | ||
| − | \]" | ||
| + | $E(\rho) = \frac{Q_{\text{enc}}}{2\pi \varepsilon_0 L \rho}$ | ||
| − | For r < \rho < R : | ||
| + | $For \ r < \rho < R :$ | ||
| − | Q_{\text{enc}} = q | ||
| + | $Q_{\text{enc}} = q$ | ||
| − | E(\rho) = \frac{q}{2\pi \varepsilon_0 L \rho} | ||
| + | $E(\rho) = \frac{q}{2\pi \varepsilon_0 L \rho}$ | ||
| − | For | ||
| + | $For\ \rho > R :$ | ||
| − | Q_{\text{enc}} = 0 \quad \rightarrow \quad E = 0. | ||
| + | $Q_{\text{enc}} = 0 \quad \rightarrow \quad E = 0.$ | ||
| − | If the shield were isolated and uncharged, the enclosed charge would be q and the field would remain as | ||
| + | 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 | |||
| − | |||
| + | The final answer is . | ||
| + | |||
| + | $\boxed{E(\rho) = \frac{q}{2\pi \varepsilon_0 L}\,\frac{1}{\rho} \quad \text{for} \quad r < \rho < R}$ | ||
| + | |||
| + | $\boxed{E(\rho) = 0 \quad \text{for} \quad \rho > R \ \text{(shield grounded)}}$ | ||
| + | |||
| + | The direction is radial: outward if | ||
| + | $q > 0,$ | ||
| + | inward if | ||
| + | $ q < 0.$ | ||
| + | |||
| + | and Inside the conductor metal $(\rho = r) $ | ||
| + | the field is zero. | ||
| @@ -1,12 +1,16 @@ | |||
| ### Statement | ### Statement | ||
| $6.3.17.$ |
$6.3.17.$ Inside the cavity of a long uncharged conductor, the radius of which is r, there | ||
| is a charge q. The conductor is surrounded by a cylindrical screen of radius | |||
| R. Conductor length L ≫ R. How does the electric field strength outside the | |||
| cavity in the middle part of the system depend on the distance to the axis of | |||
| this system? | |||
| ### Solution | ### Solution | ||
| Given that L |
Given that L ≫ R , we neglect edge effects in the middle region. The system behaves as infinitely long. Thus | ||
| $E= E(x)\hat{x} | $E= E(x)\hat{x}$ | ||
| The "cylindrical shield" of radius R acts as an outer conductor that shields the field. | 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). | 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. | Since the conductor is neutral, a charge +q appears on its outer surface. | ||
| @@ -16,27 +20,40 @@Solution | |||
| "We choose a cylindrical Gaussian surface of radius \rho and length L, coaxial with the system. | "We choose a cylindrical Gaussian surface of radius \rho and length L, coaxial with the system. | ||
| ${E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$ | |||
| E \cdot 2\pi \rho L = \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} | $E(\rho) = \frac{Q_{\text{enc}}}{2\pi \varepsilon_0 L \rho}$ | ||
| \]" | |||
| For r < \rho < R : | $For \ r < \rho < R :$ | ||
| Q_{\text{enc}} = q |
$Q_{\text{enc}} = q$ | ||
| E(\rho) = \frac{q}{2\pi \varepsilon_0 L \rho} | $E(\rho) = \frac{q}{2\pi \varepsilon_0 L \rho}$ | ||
| For |
$For\ \rho > R :$ | ||
| Q_{\text{enc}} = 0 \quad \rightarrow \quad E = 0. | $Q_{\text{enc}} = 0 \quad \rightarrow \quad E = 0.$ | ||
| If the shield were isolated and uncharged, the enclosed charge would be q and the field would remain as |
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 | #### Answer | ||
| The final answer is . | |||
| $\boxed{E(\rho) = \frac{q}{2\pi \varepsilon_0 L}\,\frac{1}{\rho} \quad \text{for} \quad r < \rho < R}$ | |||
| $\boxed{E(\rho) = 0 \quad \text{for} \quad \rho > R \ \text{(shield grounded)}}$ | |||
| The direction is radial: outward if | |||
| $q > 0,$ | |||
| inward if | |||
| $ q < 0.$ | |||
| and Inside the conductor metal $(\rho = r) $ | |||
| the field is zero. | |||