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_in/Ñ”₀is 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.

a. Consider the face parallel to the yz-plane. Define area dA 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.
b. Now integrate dΦ to find the total flux through this face.
c. Finally, show that the net flux through the cube is Φ_e = Q_in/Ñ”₀.

Transcribed Image Text:

Linear charge -density A dy dA FIGURE CP24.57