Ch 24: Gauss' Law

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 24: Gauss' Law

Problem 48

Chapter 24, Problem 48

An infinite slab of charge is centered in the xy-plane. It has charge density $ρ = ρ_{0} e^{- ∣ z ∣ / z_{e}}$ , where ρ₀ and z₀ are constants. This is a charge density that decreases exponentially as you move away from z = 0 in either the positive or negative direction. Find the electric field strength at distance z from the center of the slab.

Verified step by step guidance

Understand the problem: The charge density ρ = ρ₀e−|z|/z₀ varies with the distance z from the center of the slab. The goal is to find the electric field strength at a distance z from the center. This requires using Gauss's law and integrating the charge density over the slab.

Set up Gauss's law: Gauss's law states that the electric flux through a Gaussian surface is proportional to the enclosed charge. Mathematically, it is expressed as ∫E·dA = Q_enclosed/ε₀, where E is the electric field, dA is the area element, Q_enclosed is the enclosed charge, and ε₀ is the permittivity of free space.

Choose a Gaussian surface: Since the slab is infinite and symmetric about the xy-plane, choose a rectangular Gaussian surface that extends symmetrically above and below the slab, with its faces parallel to the xy-plane. This simplifies the problem because the electric field will only depend on z and point perpendicular to the slab (along the z-axis).

Calculate the enclosed charge: The enclosed charge Q_enclosed is obtained by integrating the charge density ρ over the volume enclosed by the Gaussian surface. The charge density is given as ρ = ρ₀e−|z|/z₀. For a slab of thickness 2z, the integral becomes Q_enclosed = ∫ρ₀e−|z|/z₀ dV, where dV = A dz (A is the area of the Gaussian surface). Split the integral into two parts for z > 0 and z < 0 to account for the absolute value in the exponential.

Solve for the electric field: Once Q_enclosed is determined, substitute it into Gauss's law. The electric field strength E will be proportional to Q_enclosed and inversely proportional to the area of the Gaussian surface. Use symmetry to argue that the electric field points along the z-axis and has the same magnitude above and below the slab. Simplify the expression to find E as a function of z.

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Key Concepts

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

Charge Density

Charge density is a measure of electric charge per unit volume, represented by the symbol ρ. In this case, the charge density varies with the distance from the center of the slab, following an exponential decay function. Understanding how charge density affects the electric field is crucial for solving problems involving electric fields generated by continuous charge distributions.

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 influenced by the distribution of charge in space. For an infinite slab of charge, the electric field can be derived using Gauss's law, which relates the electric flux through a closed surface to the charge enclosed by that surface.

Gauss's Law

Gauss's Law states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. This law is particularly useful for calculating electric fields in symmetric charge distributions, such as infinite slabs. By applying Gauss's Law to the given charge density, one can derive the electric field strength at a distance z from the center of the slab.

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Gauss' Law

Related Practice

Textbook Question

The three parallel planes of charge shown in FIGURE P24.44 have surface charge densities ─ ½ η , η , and ─ ½ η. Find the electric fields $\vec{E_{A}}$ to $\vec{E_{D}}$ in regions A to D. The upward direction is the + y-direction.

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Textbook Question

All examples of Gauss’s law have used highly symmetric surfaces where the flux integral is either zero or EA. Yet we’ve claimed that the net Φₑ = Qᵢₙ / ϵ₀ is independent of the surface. This is worth checking. FIGURE CP24.57 shows a cube of edge length L centered on a long thin wire with linear charge density λ. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral. Now integrate dΦ to find the total flux through this face.

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Textbook Question

FIGURE P24.46 shows an infinitely wide conductor parallel to and distance d from an infinitely wide plane of charge with surface charge density η. What are the electric fields $\vec{E_{A}}$ to $\vec{E_{D}}$ in regions A to D?

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Textbook Question

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).

Textbook Question

All examples of Gauss’s law have used highly symmetric surfaces where the flux integral is either zero or EA. Yet we’ve claimed that the net $Φ_{e} = Q_{i n} / ϵ_{0}$ is independent of the surface. This is worth checking. FIGURE CP24.57 shows a cube of edge length L centered on a long thin wire with linear charge density λ. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral. Consider the face parallel to the yz-plane. Define area $d \vec{A}$ as a strip of width dy and height L with the vector pointing in the 𝓍-direction. One such strip is located at position y. Use the known electric field of a wire to calculate the electric flux dΦ through this little area. Your expression should be written in terms of y, which is a variable, and various constants. It should not explicitly contain any angles.

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Textbook Question

An early model of the atom, proposed by Rutherford after his discovery of the atomic nucleus, had a positive point charge + Ze (the nucleus) at the center of a sphere of radius R with uniformly distributed negative charge -Ze. Z is the atomic number, the number of protons in the nucleus and the number of electrons in the negative sphere. A uranium atom has Z = 92 and 𝑅 = 0.10nm . What is the electric field strength at r = ½𝑅?