An infinite slab of charge of thickness 2𝒵₀ lies in the xy-plane between 𝒵 = -𝒵₀ and 𝒵 = +𝒵₀ . The volume charge density p (C/m3) is a constant. Find an expression for the electric field strength above the slab (𝒵 ≥ 𝒵₀).

Ch 24: Gauss' Law

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 24: Gauss' Law

Problem 45b

Chapter 24, Problem 45b

An infinite slab of charge of thickness 2𝒵₀ lies in the xy-plane between 𝒵 = -𝒵₀ and 𝒵 = +𝒵₀ . The volume charge density p (C/m³) is a constant. Find an expression for the electric field strength above the slab (𝒵 ≥ 𝒵₀).

Verified step by step guidance

Start by recognizing that the problem involves a uniformly charged infinite slab, and we need to find the electric field above the slab (𝒵 ≥ 𝒵₀). Use Gauss's law, which states: ∮𝐄·d𝐀 = Q_enclosed/ε₀, where Q_enclosed is the charge enclosed by the Gaussian surface, and ε₀ is the permittivity of free space.

Choose an appropriate Gaussian surface. For this problem, a cylindrical Gaussian surface (pillbox) is ideal. Place the pillbox such that it extends above the slab into the region 𝒵 ≥ 𝒵₀ and partially inside the slab. The flat surfaces of the pillbox are parallel to the xy-plane, and the curved surface does not contribute to the flux because the electric field is perpendicular to it.

Calculate the charge enclosed by the Gaussian surface. The volume of the slab enclosed by the pillbox is the cross-sectional area A of the pillbox multiplied by the thickness of the slab within the pillbox (from -𝒵₀ to +𝒵₀). The charge enclosed is Q_enclosed = p × A × (𝒵 - 𝒵₀), where p is the volume charge density and (𝒵 - 𝒵₀) is the thickness of the slab within the pillbox.

Apply Gauss's law. The electric flux through the pillbox is given by ∮𝐄·d𝐀 = E × A (since the electric field is uniform and perpendicular to the surface). Substituting this and the expression for Q_enclosed into Gauss's law gives: E × A = (p × A × (𝒵 - 𝒵₀))/ε₀.

Simplify the expression to solve for the electric field strength E. Cancel out the area A from both sides of the equation, leaving E = (p × (𝒵 - 𝒵₀))/ε₀. This is the expression for the electric field strength above the slab (𝒵 ≥ 𝒵₀).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field Due to a Continuous Charge Distribution

The electric field generated by a continuous charge distribution, such as an infinite slab, can be calculated using Gauss's law. This law relates the electric flux through a closed surface to the charge enclosed by that surface. For an infinite slab, the electric field is uniform and directed perpendicular to the surface, depending on the charge density and the distance from the slab.

Gauss's Law

Gauss's law states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. This principle is particularly useful for calculating electric fields in symmetrical charge distributions, such as planes, cylinders, and spheres. By choosing an appropriate Gaussian surface, one can simplify the calculations significantly.

Superposition Principle

The superposition principle states that the total electric field created by multiple charge distributions is the vector sum of the electric fields produced by each distribution individually. In the case of an infinite slab, the electric field above the slab can be determined by considering the contributions from both the positive and negative sides of the slab, allowing for a straightforward calculation of the resultant field.

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Textbook Question

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Textbook Question

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Textbook Question

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Textbook Question

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Textbook Question

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