Ch 27: Current and Resistance

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo

Knight Calc 5th Edition

Ch 27: Current and Resistance

Problema 23

Capitolo 27, Problema 23

A hollow copper sphere has inner radius 1.0 cm and outer radius 2.5 cm. A 5.0 A current flows radially outward from the inner surface to the outer surface. What is the electric field strength at r=2.0 cm?

Guida verificata passo dopo passo

Step 1: Understand the problem. The electric field strength at a distance r from the center of the hollow sphere can be calculated using the relationship between current density, charge flow, and electric field. The current flows radially outward, and the geometry of the sphere suggests that the electric field depends on the radial distance r.

Step 2: Calculate the current density J. The current density is defined as the current per unit area. For a spherical shell, the area at a radial distance r is given by the surface area of a sphere: \( A = 4 \pi r^2 \). Use \( J = \frac{I}{A} \), where \( I \) is the current and \( A \) is the area.

Step 3: Relate the current density to the electric field. In a conductor, the current density \( J \) is related to the electric field \( E \) by the equation \( J = \sigma E \), where \( \sigma \) is the conductivity of the material. Rearrange this equation to solve for \( E \): \( E = \frac{J}{\sigma} \).

Step 4: Determine the conductivity \( \sigma \) of copper. Look up the conductivity of copper (a known material property) or use the value provided in the problem if specified. Substitute \( \sigma \) and \( J \) into the equation for \( E \).

Step 5: Evaluate \( E \) at \( r = 2.0 \) cm. Substitute \( r = 2.0 \) cm into the expressions for \( A \) and \( J \), and then calculate \( E \) using the formula derived in the previous steps. Ensure all units are consistent (e.g., convert cm to meters).

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Electric Field

The electric field is a vector field that represents the force exerted by an electric charge on other charges in its vicinity. It is defined as the force per unit charge and is measured in volts per meter (V/m). In this scenario, the electric field strength at a specific radius within the hollow sphere is crucial for understanding how the current affects the surrounding space.

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Percorso guidato

03:16

Intro to Electric Fields

Ohm's Law

Ohm's Law states that the current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. This relationship is expressed as V = IR. In the context of the hollow sphere, understanding how current relates to electric field and resistance is essential for calculating the electric field strength.

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Percorso guidato

03:07

Resistance and Ohm's Law

Gauss's Law

Gauss's Law relates the electric flux passing through a closed surface to the charge enclosed by that surface. It is mathematically expressed as Φ_E = Q_enc/ε_0, where Φ_E is the electric flux, Q_enc is the enclosed charge, and ε_0 is the permittivity of free space. This law is particularly useful in this problem to determine the electric field at a point within the hollow sphere by considering the symmetry and the distribution of the current.

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Percorso guidato

10:20

Gauss' Law

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