An infinite cylinder of radius R has a linear charge density λ. The volume charge density (C/m3) within the cylinder (r ≤ R) is p(r)=rp0/Rp(r) = $$rp_0 / R$$, where p₀ is a constant to be determined. Use Gauss’s law to find an expression for the electric field strength E inside the cylinder, r ≤ R, in terms of λ and R.

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 58c

Chapter 24, Problem 58c

An infinite cylinder of radius R has a linear charge density λ. The volume charge density (C/m³) within the cylinder (r ≤ R) is $p of r equals r p sub 0 of slash R$ , where p₀ is a constant to be determined. Use Gauss’s law to find an expression for the electric field strength E inside the cylinder, r ≤ R, in terms of λ and R.

Verified step by step guidance

Start by recalling Gauss's law, which states that the electric flux through a closed surface is proportional to the total charge enclosed by that surface. Mathematically, it is expressed as: ∮E·dA = Q_enclosed / ε₀, where E is the electric field, dA is the infinitesimal area element, Q_enclosed is the total charge enclosed by the Gaussian surface, and ε₀ is the permittivity of free space.

Choose a cylindrical Gaussian surface of radius r (where r ≤ R) and length L, coaxial with the charged cylinder. The symmetry of the problem ensures that the electric field E is radial and constant in magnitude over the curved surface of the Gaussian cylinder.

Calculate the total charge enclosed within the Gaussian surface. The volume charge density is given as p(r) = (r * p₀) / R. The infinitesimal charge element in a cylindrical shell of radius r and thickness dr is dQ = p(r) * (2πrL * dr). Integrate this expression from 0 to r to find the total charge enclosed: Q_enclosed = ∫[0 to r] [(r * p₀) / R] * (2πrL) dr.

Perform the integration: Q_enclosed = (2πL * p₀ / R) ∫[0 to r] r² dr. The integral of r² is (r³ / 3), so Q_enclosed = (2πL * p₀ / R) * (r³ / 3). Simplify this to Q_enclosed = (2πL * p₀ * r³) / (3R).

Apply Gauss's law: ∮E·dA = E * (2πrL) = Q_enclosed / ε₀. Substitute Q_enclosed into this equation: E * (2πrL) = [(2πL * p₀ * r³) / (3R)] / ε₀. Simplify to find the electric field: E = (p₀ * r²) / (3R * ε₀). Finally, relate p₀ to λ using the total charge of the cylinder (λ = ∫[0 to R] p(r) * 2πr dr), and substitute this into the expression for E.

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Key Concepts

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

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. Mathematically, it states that the total electric flux is equal to the enclosed charge divided by the permittivity of free space. This principle is fundamental in electrostatics, allowing for the calculation of electric fields in symmetric charge distributions, such as cylinders.

Linear Charge Density

Linear charge density (λ) is defined as the amount of electric charge per unit length along a line, typically measured in coulombs per meter (C/m). In the context of the infinite cylinder, λ represents the charge distributed uniformly along the length of the cylinder, influencing the electric field generated around and within the cylinder.

Volume Charge Density

Volume charge density (ρ) describes the distribution of electric charge per unit volume, expressed in coulombs per cubic meter (C/m³). In this problem, the volume charge density varies with the radial distance (r) from the center of the cylinder, which affects the electric field inside the cylinder. Understanding how ρ is defined and varies is crucial for applying Gauss's Law effectively.

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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

An infinite cylinder of radius R has a linear charge density λ . The volume charge density (C/m³) within the cylinder (r ≤ R ) is p (r) = rp₀ / R, where p₀ is a constant to be determined. The charge within a small volume dV is dq = pdV. The integral of pdV over a cylinder of length L is the total charge Q = λL within the cylinder. Use this fact to show that p₀ = 3λ / 2πR² Hint: Let dV be a cylindrical shell of length L, radius r, and thickness dr. What is the volume of such a shell?

2017

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Textbook Question

A spherical ball of charge has radius R and total charge Q. The electric field strength inside the ball (r ≤ R ) is $E (r) = r^{4} E_{m a x} / R^{4}$ . Find an expression for the volume charge density ρ(r) inside the ball as a function of r.

1867

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Textbook Question

A sphere of radius R has total charge Q. The volume charge density (C/m³) within the sphere is p(r) = C/r², where C is a constant to be determined. Use Gauss’s law to find an expression for the electric field strength E inside the sphere, r ≤ R, in terms of Q and R.

124

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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 $Φ_{e} = Q_{i n} / ϵ_{0}$ 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. Show that the net flux through the cube is $Φ_{e} = Q_{i n} / ϵ_{0}$ .

192

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