Question

Linear charge density λ Problem 3; ll 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 flux Ф,-Qin/eo is independent of the surface. The figure on the right shows a cube of edge length L centered on a long thin wire with linear charge density A. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and dy 1 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 (see figure). Use the known electric of an infinite wire to calculate the electric flux dD 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 dD to find the total flux through this face. Finally, show that the net flux through the cube is Ф.-Qin/G

Please show how to get each answers nicely. Thank you.

Answer 1