- #1

Anne Armstrong

- 11

- 1

## Homework Statement

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. ]

## Homework Equations

E = kq/r

^{2}

σ = charge/unit volume

Φ = ∫E

^{→}⋅

*d*A

^{→}= Q

_{enclosed}/ε

_{O}

## The Attempt at a Solution

[/B]

(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

*d*b where

*d*b is an infinitesimally small length

*d*Q = σ2πb

*d*b

Plug

*d*Q into E = kq/r

^{2}for q, where r here equals √(z

^{2}+ b

^{2}) (found using Pyth. theorem for the right triangle made between the origin, point Z, and radius b):

*d*E

_{z}= kzσ2πb

*d*b/(z

^{2}+ b

^{2})

^{(3/2)}

E

_{z}= kzσ2π ∫ b/(z

^{2}+ b

^{2})

^{(3/2)}

*d*b

E

_{z}= kzσ2π [-1/√(z

^{2}+ b

^{2})] from 0 to b → E

_{z}= kzσ2π [(-1/√z

^{2}) + (-1/√(z

^{2}+ b

^{2}))

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.).