An infinitely long cylindrical conductor has radius r rr and uniform surface charge density σσ. In terms of σσ, what is the magnitude of the electric field produced by the charged cylinder at a distance R"\u003er\u003eRr \u003e R from its axis? Then, express the result in terms of λλ and show that the electric field outside the cylinder is the same as if all the charge were on the axis.

Ch 22: Gauss' Law

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 22: Gauss' Law

문제 29b

22장, 문제 29b

An infinitely long cylindrical conductor has radius $r$ and uniform surface charge density $σ$ . In terms of $σ$ , what is the magnitude of the electric field produced by the charged cylinder at a distance $r > R$ from its axis? Then, express the result in terms of $λ$ and show that the electric field outside the cylinder is the same as if all the charge were on the axis.

검증된 단계별 안내

To find the electric field at a distance r > R from the axis of the cylinder, we use Gauss's Law. Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space, $ \varepsilon_0 $.

Consider a cylindrical Gaussian surface of radius r and length L, coaxial with the charged cylinder. The electric field $ E $ is radial and constant over this surface, and the surface area of the cylindrical Gaussian surface is $ 2\pi r L $.

According to Gauss's Law, $ \Phi = E \cdot 2\pi r L = \frac{Q_{\text{enclosed}}}{\varepsilon_0} $. The charge enclosed by the Gaussian surface is $ Q_{\text{enclosed}} = \sigma \cdot 2\pi R L $, where $ \sigma $ is the surface charge density and $ R $ is the radius of the cylinder.

Substitute $ Q_{\text{enclosed}} $ into Gauss's Law: $ E \cdot 2\pi r L = \frac{\sigma \cdot 2\pi R L}{\varepsilon_0} $. Simplify to find the electric field: $ E = \frac{\sigma R}{\varepsilon_0 r} $.

To express the electric field in terms of linear charge density $ \lambda $, note that $ \lambda = \sigma \cdot 2\pi R $. Substitute $ \lambda $ into the expression for $ E $: $ E = \frac{\lambda}{2\pi \varepsilon_0 r} $. This shows that the electric field outside the cylinder is the same as if all the charge were concentrated along the axis.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. For a cylindrical conductor, it helps determine the electric field by considering a Gaussian surface coaxial with the cylinder. The symmetry of the problem simplifies calculations, as the electric field is uniform over the surface.

Surface Charge Density

Surface charge density, denoted by σ, is the amount of charge per unit area on the surface of a conductor. It is crucial for calculating the total charge on the cylindrical surface, which is used in Gauss's Law to find the electric field. Understanding σ allows us to relate it to linear charge density λ for further analysis.

Linear Charge Density

Linear charge density, λ, is the charge per unit length along a line, such as the axis of a cylinder. In this problem, it is used to express the electric field outside the cylinder, showing that the field behaves as if all the charge were concentrated along the axis. This simplification is possible due to the symmetry and infinite length of the cylinder.