Electric field of a washer (hollow disk)

In summary: I got E = k ∫(σ/r²) dθdrSeems like you are already integrating over ##dθ##, so what's left is to integrate over ##dr##. The limits of ##r## are from the inner radius to the outer radius. So, what are the limits of ##r##?I got E = k ∫(σ/r²) dθdrYou 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
  • #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/r2
σ = charge/unit volume
Φ = ∫E dA = QenclosedO

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):
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.).
 
Physics news on Phys.org
  • #2
Anne Armstrong said:
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.
 
  • Like
Likes SammyS
  • #3
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?
 
  • #4
Anne Armstrong said:
→ 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?
Anne Armstrong said:
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.
 
  • #5
haruspex said:
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.
 
  • #6
Anne Armstrong said:
I don't understand how I can get a and b in the same equation...
It's a matter of choosing the limits.
Anne Armstrong said:
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?
 

1. What is an electric field?

An electric field is a physical quantity that describes the strength and direction of the force exerted on a charged particle by an electric charge. It is represented by vector arrows pointing in the direction of the force at different points in space.

2. How is the electric field of a washer (hollow disk) calculated?

The electric field of a washer (hollow disk) can be calculated using the formula E = k * Q * (1/r1 - 1/r2), where k is the Coulomb's constant, Q is the charge on the disk, r1 is the distance from the center of the disk to the point of observation, and r2 is the distance from the edge of the disk to the point of observation.

3. What factors affect the electric field of a washer (hollow disk)?

The electric field of a washer (hollow disk) is affected by the magnitude of the charge on the disk, the distance from the disk to the point of observation, and the size and shape of the disk.

4. How does the electric field of a washer (hollow disk) differ from that of a solid disk?

The electric field of a washer (hollow disk) differs from that of a solid disk because the charge is distributed over a larger area in a hollow disk, resulting in a weaker electric field at a given distance from the disk. Additionally, the electric field inside the hollow part of the disk is zero, while the electric field inside a solid disk is non-zero.

5. What are some real-life applications of the electric field of a washer (hollow disk)?

The electric field of a washer (hollow disk) can be used in various technological applications, such as in particle accelerators, electric motors, and generators. It is also important in understanding the behavior of charged particles in plasma physics and in designing electronic circuits.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
328
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
362
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
992
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
215
  • Introductory Physics Homework Help
Replies
2
Views
511
  • Introductory Physics Homework Help
Replies
7
Views
6K
Back
Top