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Homework Help: Electric field of a washer (hollow disk)

  1. Feb 9, 2017 #1
    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 = QenclosedO

    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.).
  2. jcsd
  3. Feb 9, 2017 #2


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    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.
  4. Feb 9, 2017 #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?
  5. Feb 9, 2017 #4


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    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?
    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.
  6. Feb 9, 2017 #5
    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.
  7. Feb 9, 2017 #6


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    It's a matter of choosing the limits.
    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?
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