1. The problem statement, all variables and given/known data A washer made of nonconducting material lies in the x − y plane, with the center at the coordinate origin. The washer has an inner radius a and an outer radius b (so it looks like a disk of radius b with a concentric circular cut-out of radius a). The surface of the washer is uniformly charged with a surface charge density σ. (a) What is the electric field (as a vector) at a distance z along the z-axis (which coincides with the axis of symmetry of the ring)? [Note: this part requires integration.] b) What happens when you let a → 0? (This corresponds to a charged disk of radius b.) (c) What happens to the answer from part (b) when you let z become very large, i.e., when z>> b? How does this compare to the electric field of a single charge q = πb2σ at a distance r = z? [Note: in this part, you may find the following approximation useful: (1 + x) α ≈ 1 + αx, where x is very small (i.e., x <<1) and α is a real number. (This formula comes from a Taylor expansion, after ignoring terms of order x 2 and higher.) This kind of tool is employed in problems where you have two expressions which nearly cancel, and are trying to tease out the leftover. ] 2. Relevant equations E = kq/r2 σ = charge/unit volume Φ = ∫E→ ⋅ dA→ = Qenclosed/εO 3. The attempt at a solution (a.) Conceptually, I think the E-field would become larger as z → 0 and smaller as z gets larger. Stategically, my strategy here is to find the E-field of a positively-charged disk with radius b and then, from that, subtract a smaller negatively-charged disk with radius a. So I started by trying to solve for the E-field of the disk with radius b for a point at a distance of z along the z-axis: σ = Q/Volume so Q = σVolume Volume: area x height → 2πb x db where db is an infinitesimally small length dQ = σ2πbdb Plug dQ into E = kq/r2 for q, where r here equals √(z2 + b2) (found using Pyth. theorem for the right triangle made between the origin, point Z, and radius b): dEz = kzσ2πbdb/(z2 + b2)(3/2) Ez = kzσ2π ∫ b/(z2 + b2)(3/2) db Ez = kzσ2π [-1/√(z2 + b2)] from 0 to b → Ez = kzσ2π [(-1/√z2) + (-1/√(z2 + b2)) At this point I get confused. I feel like I'm missing something here and I don't know what it is–I'm totally lost. Any help would be appreciated! (b.), (c.) As a→0, the E-field should become larger as z approaches 0, because it should be proportional to the distance away from the washer, and vice versa (c.).