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 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 .
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24. Electric Force & Field; Gauss' Law
Electric Flux
Problem 57b
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.


1
Step 1: Understand the problem and identify the key concepts. The problem involves calculating the electric flux (Φₑ) through one face of a cube using Gauss's law. The cube is centered on a long thin wire with linear charge density λ. The electric field varies in strength and direction, so we need to integrate the flux over the surface of the face.
Step 2: Recall the formula for electric flux. Electric flux is given by Φₑ = ∫E · dA, where E is the electric field vector and dA is the infinitesimal area vector perpendicular to the surface. Since the electric field varies, we will need to integrate over the entire face of the cube.
Step 3: Express the electric field due to the wire. For 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)). The direction of the field is radial, pointing outward from the wire.
Step 4: Set up the integral for the flux through one face of the cube. Divide the face into infinitesimal area elements dA. For each element, calculate the electric field E and its component perpendicular to the face. The flux through the face is then Φₑ = ∫(E · dA), where the dot product accounts for the angle between E and dA.
Step 5: Perform the integration. Use the geometry of the cube and the symmetry of the problem to express r (the distance from the wire to the area element) and the angle between E and dA in terms of the coordinates of the face. Integrate over the entire face to find the total flux. This step involves careful evaluation of the integral, but the result will confirm that the flux is consistent with Gauss's law.

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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 net electric flux Φₑ is equal to the enclosed charge Qᵢₙ divided by the permittivity of free space ϵ₀. This law is particularly useful for calculating electric fields in situations with high symmetry, but it also holds true for more complex charge distributions.
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Gauss' Law
Electric Flux
Electric flux is a measure of the electric field passing through a given area. It is defined as the integral of the electric field E over a surface A, expressed as Φ = ∫ E · dA. The direction of the electric field and the orientation of the surface affect the total flux, making it crucial to consider both when calculating flux through surfaces that are not symmetric.
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Electric Flux
Surface Integral
A surface integral is a mathematical tool used to calculate the total quantity passing through a surface. In the context of electric flux, it involves integrating the electric field over a specified surface area. For non-uniform electric fields, this requires breaking the surface into differential elements and summing their contributions, which can be complex but is essential for accurate calculations in scenarios like the one described with the cube and wire.
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Finding Moment Of Inertia By Integrating
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