In this project we use Gauss' Law to calculate the electric field of a uniformly charged wire in part (a) and a flat sheet of charge in part (b). (a) Consider the electric field due to an infinitely...

In this project we use Gauss' Law to calculate the electric field of a uniformly charged wire in part (a) and a flat sheet of charge in part (b). (a) Consider the electric field due to an infinitely long, straight, uniformly charged wire. (There is no current running through the wire—all charges are fixed.) Assuming that the wire is infinitely long means that we can assume that the electric field is perpendicular to any cylinder that has the wire as an axis and that the magnitude of the field is constant on any such cylinder. Denote by E, the magnitude of the electric field due to the wire on a cylinder of radius r. (See Figure 19.32.) Imagine a closed surface S made up of two cylinders, one of radius a and one of larger radius b. both coaxial with the wire, and the two washers that cap the ends. (See Figure 19.33.) The outward orientation of S means that a normal on the outer cylinder points away from the wire and a normal on the inner cylinder points toward the wire. (i) Explain why the flux of ft' the electric field, through the washers is 0. (ii) Explain why Gauss's Law implies that the flux through the inner cylinder is the same as the flux through the outer cylinder. Mint: The charge on the wire is not inside the surface SI. (iii) Use part (ii) to show that Eb/E„ = a/b. (iv) Explain why part (iii) shows that the strength of the field due to an infinitely long uniformly charged wire is proportional to 1/r.