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| | | ### Statement |
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| | − | $6.3.17.$ 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? |
| | + | $6.3.17^\ast$. 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 $L \gg R$. 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? |
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| | | ### Solution |
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| | − | Given that L ≫ R , we neglect edge effects in the middle region. The system behaves as infinitely long. Thus |
| | + | Let's consider the middle part of the system. Since the length of the conductors is significantly greater than their radii ($L \gg R$), we can neglect the edge effects at the ends of the cylinders. In the middle region, the system possesses axial (cylindrical) symmetry. |
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| | − | $E= E(x)\hat{x}$ |
| | + | <b>1. Inside the metal:</b> |
| | + | The electric field strength inside the uncharged conductor itself (within the bulk of the wall) is zero. The charge $q$ induces a charge of $-q$ on the walls of the cavity. According to the law of conservation of charge, a charge of $+q$ appears on the outer surface of the conductor (at radius $r$). |
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| | − | The "cylindrical shield" of radius R acts as an outer conductor that shields the field. |
| | + | <b>2. Field in the gap ($r < \rho < R$):</b> |
| | + | The charge $+q$ is distributed uniformly over the outer surface of the conductor of radius $r$ along the length $L$. The linear charge density is $\lambda = \frac{q}{L}$. |
| | + | We apply Gauss's Law to a coaxial cylindrical surface of radius $\rho$ (where $r < \rho < R$) and length $l$: |
| | + | $$E \cdot 2\pi\rho l = \frac{\lambda l}{\varepsilon_0} = \frac{(q/L) l}{\varepsilon_0}$$ |
| | + | From this, the electric field strength in the gap between the conductor and the screen is: |
| | + | $$E(\rho) = \frac{q}{2\pi\varepsilon_0 L \rho}$$ |
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| | − | The charge +q in the cavity induces a charge –q on the inner surface (ρ = r). |
| | + | <b>3. Field outside the screen ($\rho > R$):</b> |
| | + | Looking closely at the diagram provided in the problem, the outer cylindrical screen of radius $R$ has a grounding symbol. A grounded screen has zero potential and completely shields the electric field by inducing a charge of $-q$ on its inner surface. Therefore, there is completely no field outside the system: |
| | + | $$E(\rho) = 0$$ |
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| | − | Since the conductor is neutral, a charge +q appears on its outer surface. |
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| | − | "We choose a cylindrical Gaussian surface of radius \rho and length L, coaxial with the system. |
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| | − | ${E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$ |
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| | − | $E \cdot 2\pi \rho L = \frac{Q_{\text{enc}}}{\varepsilon_0}$ |
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| | − | $E(\rho) = \frac{Q_{\text{enc}}}{2\pi \varepsilon_0 L \rho}$ |
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| | − | $For \ r < \rho < R :$ |
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| | − | $Q_{\text{enc}} = q$ |
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| | − | $E(\rho) = \frac{q}{2\pi \varepsilon_0 L \rho}$ |
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| | − | $For\ \rho > R :$ |
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| | − | $Q_{\text{enc}} = 0 \quad \rightarrow \quad E = 0.$ |
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| | − | If the shield were isolated and uncharged, the enclosed charge would be q and the field would remain as |
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| | − | $E = \frac{q}{2\pi \varepsilon_0 L \rho} .$ |
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| | − | $ E = 0 \ for \ \rho > R .$ |
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| | | #### Answer |
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| | − | The final answer is . |
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| | − | $\boxed{E(\rho) = \frac{q}{2\pi \varepsilon_0 L}\,\frac{1}{\rho} \quad \text{for} \quad r < \rho < R}$ |
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| | − | $\boxed{E(\rho) = 0 \quad \text{for} \quad \rho > R \ \text{(shield grounded)}}$ |
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| | − | The direction is radial: outward if |
| | − | $q > 0,$ |
| | − | inward if |
| | − | $ q < 0.$ |
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| | − | and Inside the conductor metal $(\rho = r) $ |
| | − | the field is zero. |
| | + | $E(\rho) = \frac{q}{2\pi\varepsilon_0 L \rho}$ for $r < \rho < R$; |
| | + | $E(\rho) = 0$ for $\rho > R$ (since the screen is grounded). |
