Electric field of a washer (hollow disk)

[Anne Armstrong]

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/r²
σ = charge/unit volume
Φ = ∫E^→ ⋅ dA^→ = 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 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):
dE_z = kzσ2πbdb/(z² + b²)^(3/2)
E_z = kzσ2π ∫ b/(z² + b²)^(3/2) db
E_z = kzσ2π [-1/√(z² + b²)] from 0 to b → E_z = 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.).

 

[cnh1995]

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.

You can use another strategy here. Think of the washer being made of infinite no of rings of infinitesimal width dr, from a to b.
You can directly use the formula for E-field along the axis of a ring and then simply integrate it between a to b to get the total electric field due to the washer.
This way, you can also see what happens to the field when a=0.

 

[Anne Armstrong]

Ok so if I use E = k ∫(σ/b²) dA (from kq/r² and Guass' Law using r=b) where dA=dθdb and θ is the angle between the point z and a point along radius a, I can differentiate it from a →b and from θ=0→360°. Is that what you mean?

 

[haruspex]

→ Ez = kzσ2π [(-1/√z²) + (-1/√(z²+ b²))

You have a sign error.
Why are you taking a=0 here? Why not get the general expression as a function of a and b, then worry about the limits?

Ok so if I use E = k ∫(σ/b²) dA (from kq/r² and Guass' Law using r=b) where dA=dθdb and θ is the angle between the point z and a point along radius a, I can differentiate it from a →b and from θ=0→360°. Is that what you mean?

Seems to me that cnh1995 is describing the method you used in the first place. Stick with it.
It was perhaps confusing that you used b as both the integration variable and the upper limit, but it's not unusual.

 

[Anne Armstrong]

You have a sign error.
Why are you taking a=0 here? Why not get the general expression as a function of a and b, then worry about the limits?

I was attempting to get the E field of a disc with radius b--aka integrating from radius of 0 to b, but I don't think that's right. I'm confused by what you mean by "the general expression as a function of a and b." I don't understand how I can get a and b in the same equation...
From the E = k ∫(σ/r²) dA equation (I'm going to switch out b for r here to simplify...) I got E = k ∫(σ/r²) dθdr (integrating from a to b as cnh1995 suggested) ⇒ kσ ∫ (1/r²)dr
At this point I get lost again! I was trying to get an expression that replaces r with a and b, but I'm not sure how to get there.

 

[haruspex]

I don't understand how I can get a and b in the same equation...

It's a matter of choosing the limits.

I got E = k ∫(σ/r²) dθdr

You seem to be using r in two different ways. Inside the integrand it is the distance from the ring element to (0,0,z), but in dθdr it is a radius from the centre of the disc.
I suggest sticking to r as a radius, replacing your variable b. Keep b as the max radius only. What is the range for r?