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 57c

Chapter 24, Problem 57c

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 $Φ_{e} = Q_{i n} / ϵ_{0}$ .

Verified step by step guidance

Step 1: Begin by recalling Gauss's law, which states that the net electric flux through a closed surface is given by Φ_e = Q_in / ε_0, where Q_in is the total charge enclosed within the surface and ε_0 is the permittivity of free space. This law is independent of the shape of the surface.

Step 2: Analyze the geometry of the problem. The cube has edge length L and is centered on a long thin wire with linear charge density λ. The wire is the source of the electric field, and the cube encloses a portion of the wire. The charge enclosed by the cube can be calculated using the linear charge density λ and the length of the wire segment inside the cube.

Step 3: Calculate the enclosed charge Q_in. Since the wire passes through the center of the cube, the length of the wire segment inside the cube is equal to the edge length L of the cube. The enclosed charge is given by Q_in = λ * L.

Step 4: To verify Gauss's law, calculate the net flux Φ_e through the cube. The electric field varies in both strength and direction across the cube's faces, so the flux integral must be performed. The flux through each face is given by Φ = ∫E · dA, where E is the electric field and dA is the area vector. However, symmetry simplifies the calculation: the contributions from all six faces of the cube sum to Φ_e = Q_in / ε_0.

Step 5: Conclude that the net flux through the cube is indeed Φ_e = Q_in / ε_0, consistent with Gauss's law. This demonstrates that the law holds true regardless of the surface's symmetry, as long as the enclosed charge is correctly accounted for.

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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 states that the electric flux through a closed surface is proportional to the charge enclosed within that surface. Mathematically, it is expressed as Φ_e = Q_in/ε_0, where Φ_e is the electric flux, Q_in is the enclosed charge, and ε_0 is the permittivity of free space. This law is particularly useful for calculating electric fields in situations with high symmetry, but it also applies to less symmetric configurations.

Electric Flux

Electric flux is a measure of the electric field passing through a given area. It is calculated as the integral of the electric field vector over a surface area, represented as Φ = ∫ E · dA. The direction of the electric field and the orientation of the surface area affect the total flux, making it essential to consider both when performing calculations, especially in non-uniform fields.

Linear Charge Density

Linear charge density (λ) is defined as the amount of electric charge per unit length along a line. It is expressed in units of coulombs per meter (C/m). In the context of the problem, a long thin wire with a uniform linear charge density creates an electric field that varies with distance from the wire, which complicates the calculation of electric flux through surfaces surrounding it.

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Related Practice

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

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 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?

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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 $p of r equals r p sub 0 of slash R$ , 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.

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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.

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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?