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.

Name: Physics for Scientists and Engineers
Author: Knight Calc
ISBN: 9780137344796

Verified step by step guidance

Step 1: Recall the expression for the electric field due to a long thin wire with linear charge density λ. The electric field at a distance r from the wire is given by E = (λ / (2πϵ₀r)), where r is the perpendicular distance from the wire to the point of interest.

Step 2: For the face parallel to the yz-plane, the distance r from the wire to a strip at position y is simply |y|. The electric field at this strip is therefore E = (λ / (2πϵ₀|y|)).

Step 3: The area element dA for the strip is defined as dA = L * dy, where L is the height of the strip and dy is its width. The vector dA points in the x-direction, which is the same direction as the electric field.

Step 4: The electric flux dΦ through this strip is given by dΦ = E * dA. Substituting the expressions for E and dA, we get dΦ = (λ / (2πϵ₀|y|)) * (L * dy).

Step 5: Combine the terms to express the flux through the strip as dΦ = (λL / (2πϵ₀|y|)) * dy. This is the differential flux through the strip in terms of y, λ, L, and ϵ₀.

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 electric flux Φ through a surface is equal to the enclosed charge Q divided by the permittivity of free space ε₀. This law is particularly useful for calculating electric fields in situations with high symmetry, but it can also be applied in less symmetric cases, as shown in the problem.

Electric Flux

Electric flux is a measure of the electric field passing through a given area. It is calculated as the dot product of the electric field E and the area vector A, expressed mathematically as Φ = ∫ E · dA. In cases where the electric field varies, as with the wire in the problem, the flux must be calculated by integrating over the area, taking into account the direction and magnitude of the electric field at each point.

Linear Charge Density

Linear charge density (λ) is defined as the amount of charge per unit length along a line, typically expressed in coulombs per meter (C/m). In the context of the problem, the wire has a uniform linear charge density, which creates an electric field that varies with distance from the wire. This concept is crucial for determining the electric field strength at different points around the wire, which directly affects the calculation of electric flux through the cube's face.

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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.

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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 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

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 $Φ_{e} = Q_{i n} / ϵ_{0}$ .

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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?