A hollow, conducting sphere with an outer radius of 0.2500.2500.250 m and an inner radius of 0.2000.2000.200 m has a uniform surface charge density of +6.37×10−6+6.37\$$times10^{-6}+6.37×10$$−6 C/m2. A charge of −0.500−0.500−0.500 μ\muμC is now introduced at the center of the cavity inside the sphere. Calculate the strength of the electric field just outside the sphere?

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 19b

22장, 문제 19b

A hollow, conducting sphere with an outer radius of $0.250$ m and an inner radius of $0.200$ m has a uniform surface charge density of $+6.37$\times$10^{-6}$ C/m². A charge of $−0.500$ $\mu$ C is now introduced at the center of the cavity inside the sphere. Calculate the strength of the electric field just outside the sphere?

검증된 단계별 안내

First, understand that the electric field just outside a conducting sphere is determined by the total charge enclosed within the sphere. The charge on the sphere itself and any charge inside the cavity contribute to this total charge.

Calculate the total charge on the surface of the sphere using the formula for surface charge density: $ Q_{surface} = \sigma \times A $, where $ \sigma = 6.37 \times 10^{-6} $ C/m² is the surface charge density and $ A = 4\pi r_{outer}^2 $ is the surface area of the sphere with $ r_{outer} = 0.250 $ m.

Add the charge introduced at the center of the cavity to the charge on the surface to find the total charge enclosed by the sphere. The charge introduced is $ -0.500 \mu C $ or $ -0.500 \times 10^{-6} $ C.

Use Gauss's Law to find the electric field just outside the sphere. Gauss's Law states that $ \Phi = \frac{Q_{enclosed}}{\varepsilon_0} $, where $ \Phi $ is the electric flux and $ \varepsilon_0 $ is the permittivity of free space ($ 8.85 \times 10^{-12} $ C²/N·m²). The electric field $ E $ is related to the flux by $ \Phi = E \times A $, where $ A $ is the area of the Gaussian surface.

Solve for the electric field $ E $ using the equation $ E = \frac{Q_{enclosed}}{\varepsilon_0 \times A} $, where $ A = 4\pi r_{outer}^2 $. This will give you the strength of the electric field just outside the sphere.

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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. It is essential for calculating electric fields in symmetric charge distributions, such as spherical conductors. The law states that the total electric flux is equal to the enclosed charge divided by the permittivity of free space.

Conductors in Electrostatic Equilibrium

In electrostatic equilibrium, conductors have no electric field inside them, and any excess charge resides on the surface. This principle helps determine the distribution of charge on the sphere and the resulting electric field outside it. The surface charge density affects the field strength just outside the conductor.

Superposition Principle

The superposition principle states that the total electric field due to multiple charges is the vector sum of the fields due to each charge individually. This concept is crucial for calculating the net electric field outside the sphere, considering both the surface charge and the introduced charge at the center.